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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 333 16 "Elliptic curves." }}
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{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "fx := x*(x-1)*(x-2);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%#fxG*(%\"xG\"\"\",&F&F'!\"\"F'F',&F&F'!\"#F'F'
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "fx has 3 real roots." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "f := y^2 - fx;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%\"fG,&*$)%\"yG\"\"#\"\"\"F**(%\"xGF*,&F,F*!\"
\"F*F*,&F,F*!\"#F*F*F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "Becaus
e of the three real roots, the elliptic curve given by the solutions (
x,y) of  f=0 looks like this:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 50 "with(plots): implicitplot(f,x=-0.5..2.7, y=-3..3);" }}{PARA 13 "
" 1 "" {INLPLOT "6$-%'CURVESG6bp7$7$$\"1,++++++7!#<$!1(*******f.\"R\"!
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s)ovf_z/'F-7$7$$\"1K\\!=IguS&F-F_rFdu7$Fju7$$\"1WB]r$=KG'F-$!1')ocYW.c
bF-7$7$$\"1++++++?lF-$!1,+++!)*fV&F-F^v7$7$Fev$\"1********z*fV&F-7$$\"
1$*[!=IguS&F-F\\s7$F]wF`u7$Fdv7$$\"1oT#\\+M,@(F-$!11GBf7+%*[F-7$7$$\"1
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7$7$$\"1,+++++!3*F-$!1,+++SW+GF-Fey7$7$$\"1************z!*F-$\"1******
**RW+GF-7$$\"1drFt$\\Tm)F-F^o7$FfzFdx7$7$$\"1++++++!3*F-$!1++++SW+GF-7
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$7$$\"1,+++++g?!#:$!1@++++v9OF-7$$\"1&3#\\+\\Gf?Fi\\lFW7$F\\]l7$$\"1(R
!oj!*oe?Fi\\l$!14u+W*>ad$F-7$7$$\"15vS8t5,?Fi\\lF2F`]l7$Ff]l7$Fg]l$!1
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l$\"1B.'>aBqg$F-7$7$Fg\\l$\"1E++++v9OF-F]_l7$7$$\"1,+++++)=#Fi\\l$!1B+
++!)e$*oF-7$$\"1a9=\\!\\>9#Fi\\lF_r7$F\\`lFf\\l7$Fc_l7$$\"1&o\"H9<%)4@
Fi\\l$\"1J4.dyYl]F-7$7$$\"1b9=\\!\\>9#Fi\\lF\\sFa`l7$Fg`l7$$\"1;W#ekzX
;#Fi\\l$\"1OAzS;8RkF-7$7$Fh_l$\"1C+++!)e$*oF-F[al7$7$$\"1,+++++;BFi\\l
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7$$\"1b@E5$f=A#Fi\\l$\"1?YeKz8lxF-7$7$$\"1*GXoSBlC#Fi\\l$\"12++++++%)F
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IET(*F-F^cl7$7$$\"1,+++++WCFi\\l$!1.++g,*oC\"Fi\\l7$$\"1hQPa>4fBFi\\l$
!1************z5Fi\\l7$F]dlFeal7$Fdcl7$$\"1.nTN.WSBFi\\l$\"1%o$4rV<M5F
i\\l7$7$$\"1iQPa>4fBFi\\l$\"1,+++++!3\"Fi\\lFddl7$Fjdl7$$\"1#>u;8X0S#F
i\\l$\"1$*36yyZh6Fi\\l7$7$Ficl$\"1.++g,*oC\"Fi\\lF`el7$7$$\"1,+++++sDF
i\\l$!1pmmm7-=:Fi\\l7$$\"1f=4]e6uCFi\\l$!1************>8Fi\\l7$F_flFhc
l7$Ffel7$$\"1$\\Y^$>*4Y#Fi\\l$\"1H.g@,9)G\"Fi\\l7$7$F`fl$\"1,+++++?8Fi
\\lFffl7$F\\gl7$$\"1Zxy\"[*>@DFi\\l$\"1xzkr4D:9Fi\\l7$7$F[fl$\"1qmmm7-
=:Fi\\lF`gl7$7$$\"1-++++++FFi\\l$!1*G9dGk>z\"Fi\\l7$$\"1vRT$y)=*e#Fi\\
l$!1************f:Fi\\l7$F_hlFjel7$Ffgl7$$\"1#3zTcB<e#Fi\\l$\"1CU;<$o<
a\"Fi\\l7$7$F`hl$\"1,+++++g:Fi\\lFfhl7$F\\il7$$\"1TL$*)4Y:k#Fi\\l$\"1*
)**\\k5gp;Fi\\l7$7$F[hl$\"1*G9dGk>z\"Fi\\lF`il-%'COLOURG6&%$RGBG\"\"\"
\"\"!F^jl-%+AXESLABELSG6$%\"xG%\"yG" 2 365 365 365 2 0 1 0 2 9 0 4 2 
1 0.000000 45.000000 45.000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 
0 0 0 0 0 0 0 0 0 0 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 "Here
 is another example, but now the polynomial fx has only 1 real root an
d 2 complex roots:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "fx :=
 4*x^3-8*x^2+4*x+1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#fxG,**$)%\"x
G\"\"$\"\"\"\"\"%*$)F(\"\"#F*!\")F(F+F*F*" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 14 "f := y^2 - fx;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%\"fG,,*$)%\"yG\"\"#\"\"\"F**$)%\"xG\"\"$F*!\"%*$)F-F)F*\"\")F-F/!\"\"
F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Now the curve looks like:" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "implicitplot(f,x=-0.5..2.
7, y=-3..3);" }}{PARA 13 "" 1 "" {INLPLOT "6$-%'CURVESG6gq7$7$$!1+++++
+g6!#;$!1VWWW%38V'F*7$$!1w\\d]IDX7F*$!1#*************fF*7$F-7$$!17l<7%
HKF\"F*$!1K!p@&[p(y&F*7$7$$!1yoLj8^h:F*$!1#************f$F*F37$7$$!1xo
Lj8^h:F*F<7$$!1.L(G\"\\5=;F*$!1]DO)G`5u#F*7$7$$!1Gyr>0k><F*$!1#*******
*****>\"F*FB7$FH7$$!1Fyr>0k><F*$!1e1Hb-u1:!#<7$7$FO$\"13++++++7F*FN7$F
U7$$!1S#RnP0\"e;F*$\"1$3'Q\"eZR8#F*7$7$$!1woLj8^h:F*$\"13++++++OF*FY7$
Fin7$$!1Z%*R;?'>[\"F*$\"1s*[2yyO?%F*7$7$$!1u\\d]IDX7F*$\"13++++++gF*F_
o7$Feo7$$!1?!oBd+AA\"F*$\"1X+>tgi;hF*7$7$F($\"1YWWW%38V'F*F[p7$7$$\"1,
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ap7$$!1$yVV&fTgyFS$\"1n%R9*z#))p(F*7$7$$!1px#3\\O<z&FS$\"12++++++%)F*F
bq7$Fhq7$$!1$e57mPQ#GFS$\"1#*pZh&pW:*F*7$7$Ffp$\"1LLLLpt<5FjpF^r7$7$$
\"1+++++++9F*$!1+++++C$=\"Fjp7$$\"1(>7&H`xl`FS$!1************z5Fjp7$7$
$\"1)>7&H`xl`FSF`sFep7$Fdr7$$\"1'Q>qypx/$FS$\"1(Q$\\;VNX5Fjp7$7$$\"19B
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LLLL)>2\"Fjp7$$\"1+!=v8t/o(F*F`t7$Fd\\lFdz7$F[\\l7$$\"1!paRzk6y)F*$!1.
l'))RoR-\"Fjp7$7$$\"1************z!*F*$!1++++WM45FjpFi\\l7$Fa\\l7$$\"1
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7$$\"1!f<HbQg$**F*$!1Z?nGs]+5Fjp7$7$$\"1++++++O5Fjp$!1KLLL`Ih**F*Fa^l7
$F[^l7$$\"1.]lXhyy%*F*$\"1)o*o(fF_+\"Fjp7$7$Fh^l$\"1NLLL`Ih**F*F]_l7$F
g^l7$$\"1Bk)GtU&[6Fjp$!1T?;Cw,^5Fjp7$7$$\"1++++++k6Fjp$!1LLLLhbe5FjpFg
_l7$Fc_l7$$\"1\"o)G)H(\\r5Fjp$\"1B(3dcUM,\"Fjp7$7$F^`l$\"1LLLLhbe5FjpF
c`l7$7$$\"1++++++#H\"Fjp$!1******z%oU>\"Fjp7$$\"1*[bX?32=\"FjpF`s7$Fba
lF]`l7$Fi`l7$$\"1Az!)*R*yq6Fjp$\"1Z,Ov)ps1\"Fjp7$7$$\"1!\\bX?32=\"FjpF
`tFgal7$F]bl7$$\"1KE\"Rz?ED\"Fjp$\"1;QT6g$Q:\"Fjp7$7$F^al$\"1+++![oU>
\"FjpFabl7$7$$\"1++++++?9Fjp$!1BAAAA759Fjp7$$\"1Lq`P?\"3O\"Fjp$!1*****
*******>8Fjp7$F`clF]al7$Fgbl7$$\"1'p4LahfK\"Fjp$\"1AVH1@Kc7Fjp7$7$$\"1
Mq`P?\"3O\"Fjp$\"1,+++++?8FjpFgcl7$F]dl7$$\"1&R:t3vMR\"Fjp$\"1NOG6Utp8
Fjp7$7$F\\cl$\"1CAAAA759FjpFcdl7$7$$\"1++++++[:Fjp$!11>w/2v'o\"Fjp7$$
\"1_I)fIIW[\"Fjp$!1************f:Fjp7$FbelF[cl7$Fidl7$$\"1M-L#*\\nb9Fj
p$\"183Q*Q4J\\\"Fjp7$7$$\"1`I)fIIW[\"Fjp$\"1,+++++g:FjpFiel7$F_fl7$$\"
1i6!pWQ_^\"Fjp$\"1@.17zU@;Fjp7$7$F^el$\"11>w/2v'o\"FjpFefl7$7$$\"1,+++
++w;Fjp$!1-+++1Z9?Fjp7$$\"1m2'[&3X)e\"Fjp$!1*************z\"Fjp7$Fdgl7
$F^el$!10>w/2v'o\"Fjp7$F[gl7$$\"1a$zh7EAd\"Fjp$\"19PT8gda<Fjp7$7$$\"1n
2'[&3X)e\"Fjp$\"1,++++++=FjpF^hl7$7$FeglFghl7$$\"1A^<t\"3ki\"Fjp$\"1\\
mHvY)H*=Fjp7$7$F`gl$\"1,+++1Z9?FjpF[il7$7$$\"1,+++++/=Fjp$!1NLL`?6xBFj
p7$$\"1aR,$eJnw\"Fjp$!1************zAFjp7$Fjil7$$\"1Wgum=C)p\"Fjp$!1H)
[,].<3#Fjp7$7$$\"1,PGwy$Qo\"Fjp$!1************R?FjpF`jl7$FfjlF_gl7$Fai
l7$$\"1q#HTPu4o\"Fjp$\"1@w][InI?Fjp7$7$$\"1-PGwy$Qo\"Fjp$\"1,+++++S?Fj
pF][m7$Fc[m7$$\"1&R(p,(o5t\"Fjp$\"1'Q<V=Yn<#Fjp7$7$F[jl$\"1,+++++!G#Fj
pFi[m7$F_\\m7$$\"1B&yX\"oK#y\"Fjp$\"1s_mAsj?BFjp7$7$Ffil$\"1NLL`?6xBFj
pFc\\m7$7$$\"1,+++++K>Fjp$!1\"yxx(\\^wFFjp7$$\"1Zd#\\:bg#>Fjp$!1++++++
gFFjp7$Fb]m7$$\"1AsR\"H4k\">Fjp$!1*yp8Un2t#Fjp7$7$$\"1galSC'o%=Fjp$!1+
+++++?DFjpFh]m7$F^^mFeil7$Fi\\m7$$\"1&e70(GVJ=Fjp$\"11*)ynLcoCFjp7$7$F
_^m$\"1,+++++?DFjpFe^m7$7$$\"1halSC'o%=FjpF\\_m7$$\"1wdI>NSz=Fjp$\"1t;
I,%='=EFjp7$7$Fc]m$\"1-+++++gFFjpFb_m7$Fh_m7$$\"1R]=22XG>Fjp$\"1c!GS#
\\lmFFjp7$7$F^]m$\"1#yxx(\\^wFFjpF\\`m7$7$$\"1h#*HyUs'*>Fjp$!\"$\"\"!F
]]m7$Fb`m7$$\"1ekG\\j%R(>Fjp$\"1&*y3&f]8#HFjp7$7$Fg`m$\"1-++++++IFjpF]
am-%'COLOURG6&%$RGBG\"\"\"F[amF[am-%+AXESLABELSG6$%\"xG%\"yG" 2 365 
365 365 2 0 1 0 2 9 0 4 2 1 0.000000 45.000000 45.000000 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 332 "We see that an elliptic curve of the form:  f \+
= y^2 - fx where fx is a polynomial in x of degree 3, can have either \+
one or two components over the real numbers. The plots suggests that o
ne of the components is topologically equivalent to a line and the oth
er (if there is a second component) is topologically equivalent to a c
ircle." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 
"However: don't forget the point at infinity!" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 343 "The one point at infinit
y makes that the component that looks like a line, is in fact also top
ologically equivalent to a circle. So: an elliptic curve f=y^2 - polyn
omial fx in x of degree 3, over the real numbers, is topologically equ
ivalent to 1 circle or 2 circles, depending on whether fx has 3 real r
oots, or 1 real root + 2 complex roots." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 58 "We'll continue with the second curve
. Here are two points:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "P
1 := [0,1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#P1G7$\"\"!\"\"\"" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "P2 := [1, -1];" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#P2G7$\"\"\"!\"\"" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 73 "How can you check if P1 = (x=0,y=1) is a point on the cur
ve? That's easy:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "subs(x=
0, y=1, f);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 13 "OK. Likewise:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 19 "subs(x=1, y=-1, f);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "So P2 is also a point \+
on the curve." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 76 "Remember that over the complex numbers, the elliptic curv
e is equivalent to:" }}{PARA 0 "" 0 "" {TEXT -1 3 "   " }{TEXT 256 1 "
C" }{TEXT -1 2 "/(" }{TEXT 257 2 "Z " }{TEXT -1 5 "w1 + " }{TEXT 258 
2 "Z " }{TEXT -1 3 "w2)" }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{TEXT 
259 1 "C" }{TEXT -1 37 " is the field of complex numbers and " }{TEXT 
260 1 "Z" }{TEXT -1 116 " is the set of all integers, and where w1 and
 w2 are two complex numbers. We saw last week that there was a 1-1 map
:" }}{PARA 0 "" 0 "" {TEXT -1 18 "   curve  < --->  " }{TEXT 261 1 "C
" }{TEXT -1 2 "/(" }{TEXT 262 2 "Z " }{TEXT -1 5 "w1 + " }{TEXT 263 2 
"Z " }{TEXT -1 3 "w2)" }}{PARA 0 "" 0 "" {TEXT -1 189 "and that this m
ap depended on the choice of a base-point P0 on the curve (the base-po
int is the starting point for the integral). To compute the image of a
 point P of the curve in the set  " }{TEXT 334 1 "C" }{TEXT -1 3 " /(
" }{TEXT 264 2 "Z " }{TEXT -1 5 "w1 + " }{TEXT 265 2 "Z " }{TEXT -1 
36 "w2)  we need to compute the integral" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 13 "Int( 1/y, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$I
ntG6$*&\"\"\"F'%\"yG!\"\"%\"xG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 
"over a path that starts at the base-point P0 and ends at P." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 190 "So we need to \+
fix a base-point on the curve, one usually takes the point P0 as the o
ne point at infinity. In homogeneous coordinates, in other words: coor
dinates in the projective plane P^2(" }{TEXT 266 1 "C" }{TEXT -1 46 ")
, this point P0 can be denoted as P0=(0:1:0)." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "With this choice of base-
point, we may now wonder:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 112 "Suppose we take two points on the curve, say P1 a
nd P2, we compute the images Z1 and Z2 of P1 and P2 in the set " }
{TEXT 267 1 "C" }{TEXT -1 3 " /(" }{TEXT 268 2 "Z " }{TEXT -1 5 "w1 + \+
" }{TEXT 269 2 "Z " }{TEXT -1 160 "w2) by computing the integrals of (
1/y)*dx from the basepoint to those points. Then add up those images (
those integrals) so that we get a third point Z1+Z2 in " }{TEXT 270 1 
"C" }{TEXT -1 3 " /(" }{TEXT 271 2 "Z " }{TEXT -1 5 "w1 + " }{TEXT 
272 2 "Z " }{TEXT -1 176 "w2),. Then we may wonder, for which point Q \+
on the curve is it so that if we compute the integral of (1/y)*dx from
 the base-point to Q, that this integral then is Z1+Z2 modulo " }
{TEXT 335 2 "Z " }{TEXT -1 5 "w1 + " }{TEXT 336 2 "Z " }{TEXT -1 3 "w2
." }}{PARA 0 "" 0 "" {TEXT -1 34 "Lets denote that point Q by P1+P2." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "Is ther
e any easy way to find P1+P2?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 176 "It turns out that the answer is yes, the
re is a nice geometric construction that gives you P1+P2, where the on
ly thing we have to do is intersect the curve with straight lines." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 273 77 "Claim: I
f 3 points P1, P2, P3 are on the same straight line then: P1+P2+P3=0.
" }}{PARA 0 "" 0 "" {TEXT -1 259 "More generally: if we intersect the \+
curve with any other curve, compute the intersection points (with mult
iplicity) then the sum of those points is zero. We will only intersect
 the curve with straight lines though, because that is the easiest to \+
compute with." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 249 "What is 0 in this setting? Remember that the integral is
 over a path from the basepoint (0:1:0) to the point P. If P=(0:1:0), \+
i.e. P is the point at infinity, then we are integrating from (0:1:0) \+
to itself and so the integral is the zero element in " }{TEXT 274 1 "C
" }{TEXT -1 3 " /(" }{TEXT 275 2 "Z " }{TEXT -1 5 "w1 + " }{TEXT 276 
2 "Z " }{TEXT -1 4 "w2)." }}{PARA 0 "" 0 "" {TEXT -1 71 "So, the 0 ele
ment of this addition must be the base-point P0 = (0:1:0)." }}{PARA 0 
"" 0 "" {TEXT -1 277 "The zero element in the group of points is the p
oint at infinity because that is the point we took as base-point for t
he integrals. We could have taken any point as base-point (and hence a
s zero in the group). But one always takes the point at infinity unles
s stated otherwise." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 99 "We won't prove the claim, but assuming that it is true,
 it is now easy to compute the sum of P1+P2:" }}{PARA 0 "" 0 "" {TEXT 
-1 64 "Step 1) Compute the straight line L that goes through P1 and P2
." }}{PARA 0 "" 0 "" {TEXT -1 126 "Step 2) Compute the intersection po
ints of L and the curve. There must be 3 intersection points, two of w
hich we already know." }}{PARA 0 "" 0 "" {TEXT -1 138 "Step 3) Compute
 the third intersection point, say that is P3. Now P1+P2+P3=0, so P1+P
2 = -P3. We now know P3 and we have to calculate -P3." }}{PARA 0 "" 0 
"" {TEXT -1 86 "Step 4) Compute the line M that goes through: the base
-point P0=(0:1:0), the point P3." }}{PARA 0 "" 0 "" {TEXT -1 261 "Step
 5) Compute the intersection points of M and the curve, we already kno
w P0 and P3, and there must be one more intersection point, lets call \+
that P4. Then we have:  P0+P3+P4=0, and the base-point P0 corresponds \+
to 0, so P3+P4=0 so P4=-P3, which equals P1+P2." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 425 "Remark on step 4 and ste
p 5. A line that goes through (0:1:0) must be vertical. Because of tha
t, it is easy to see that if P3=(a,b) then P4 must have the same x-coo
rdinate as P3, therefore P4 must be (a,-b). We see now why taking infi
nity as base-point is a good choice, because steps 4 and 5 are computa
tionally trivial. With infinity as zero, we really only have to do ste
ps 1,2,3, and then multiply the y-coordinate by -1." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "Step 1: Take an equatio
n of a line:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "L := a*x+b*
y+c;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,(*&%\"aG\"\"\"%\"xGF(F(
*&%\"bGF(%\"yGF(F(%\"cGF(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Now \+
compute the equations that say that P1 and P2 are on the line L:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "subs(x=P1[1],y=P1[2],L) = 0,
  subs(x=P2[1],y=P2[2],L) = 0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$/,&%
\"bG\"\"\"%\"cGF&\"\"!/,(%\"aGF&F%!\"\"F'F&F(" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 54 "Solve the equations, so that L goes through P1 and P2.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "solve(\{%\},\{a,b,c\});
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<%/%\"cGF%/%\"bG,$F%!\"\"/%\"aG,$F
%!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "L:=subs(%, L);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,(*&%\"cG\"\"\"%\"xGF(!\"#*&F'F(
%\"yGF(!\"\"F'F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "Do one more s
ubstitution to get rid of this variable" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "L:=subs(a=1,b=1,c=1,L);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"LG,(%\"xG!\"#%\"yG!\"\"\"\"\"F*" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 50 "Solve f=0 and L=0 to find the intersection points:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "solve(\{L=0, f=0\},\{x,y\});
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%<$/%\"xG\"\"!/%\"yG\"\"\"<$/F%F)/F
(!\"\"<$/F%\"\"#/F(!\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "
intersectionPoints:=\{seq(subs(i,[x,y]),i=[%])\};" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%3intersectionPointsG<%7$\"\"#!\"$7$\"\"\"!\"\"7$\"\"!
F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "We already knew 2 of these \+
points, take the third one:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
39 "P3 := intersectionPoints minus \{P1,P2\};" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#P3G<#7$\"\"#!\"$" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 13 "P3 := op(P3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#P
3G7$\"\"#!\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "Steps 4 and 5:" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "P4 := [P3[1],-P3[2]];" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#P4G7$\"\"#\"\"$" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 17 "So P1+P2 = [2,3]." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 35 "Here is a plot of the construction." 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "implicitplot(\{f,L,x-2\},
x=-0.5..2.7, y=-4..4);" }}{PARA 13 "" 1 "" {INLPLOT "6&-%'CURVESG6U7$7
$$\"\"#\"\"!$!\"%F*7$F($!1)***********\\Q!#:7$7$F($!1,+++++!o$F0F-7$7$
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1)***********4KF07$7$F($!1++++++SIF0FA7$FE7$F($!1*************)GF07$7$
F($!1-+++++?FF0FI7$7$F($!1,+++++?FF07$F($!1************pDF07$7$F($!1,+
+++++CF0FT7$FX7$F($!1************\\AF07$7$F($!1,+++++!3#F0Ffn7$Fjn7$F(
$!1************H>F07$7$F($!1,+++++g<F0F^o7$Fbo7$F($!1************4;F07
$7$F($!1,+++++S9F0Ffo7$Fjo7$F($!1*************G\"F07$7$F($!1,+++++?6F0
F^p7$Fbp7$F($!1))***********p*!#;7$7$F($!12++++++!)FipFfp7$F[q7$F($!1'
)***********\\'Fip7$7$F($!13++++++[FipF_q7$7$F($!12++++++[Fip7$F($!1&)
***********H$Fip7$7$F($!12++++++;FipFjq7$F^r7$F($!0[)************!#<7$
7$F($\"1$************f\"FipFbr7$Fgr7$F($\"1;++++++JFip7$7$F($\"1$*****
*******z%FipF[s7$F_s7$F($\"1<++++++jFip7$7$F($\"1%*************zFipFcs
7$Fgs7$F($\"1:++++++&*Fip7$7$F($\"1************>6F0F[t7$F_t7$F($\"1-++
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++++IAF07$7$F($\"1*************R#F0F[v7$F_v7$F($\"1-+++++]DF07$7$F($\"
1************>FF0Fcv7$Fgv7$F($\"1,+++++qGF07$7$F($\"1************RIF0F
[w7$F_w7$F($\"1,+++++!>$F07$7$F($\"1************fLF0Fcw7$Fgw7$F($\"1-+
++++5NF07$7$F($\"1************zOF0F[x7$F_x7$F($\"1,+++++IQF07$7$F($\"1
**************RF0Fcx-%'COLOURG6&%$RGBG\"\"\"F*F*-F$6]r7$7$$!1++++++g6F
ip$!1********>s(H'Fip7$$!1WN]^$[JU\"Fip$!11++++++[Fip7$7$FhyFhq7$$!1\"
G&4cq?(\\\"Fip$!10=wfB)p&RFip7$7$$!1Yjt^oE/<FipF_rF^z7$Fdz7$Fez$!1<9f1
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YN]^$[JU\"FipF`sF^[l7$Fd[l7$$!1g%QhVPGM\"Fip$\"1XhM!f$4d_Fip7$7$Fcy$\"
1++++?s(H'FipFh[l7$7$$\"1,++++++7Fer$!1++++O!>,\"F07$$!1&fu#R'pd8(FerF
\\q7$Fg\\l7$Fcy$!1++++?s(H'Fip7$7$Fcy$\"1,+++?s(H'Fip7$$!15^MH'*f.*)Fe
r$\"1siL2***eK(Fip7$7$$!1QYFR'pd8(FerFhsFb]l7$Fh]l7$$!1EW-*Q&o-IFer$\"
1)4csMr10*Fip7$7$Fc\\l$\"1++++O!>,\"F0F\\^l7$7$$\"1+++++++9Fip$!1++++D
T#=\"F07$$\"1_?*oB-BV)FerFcp7$F[_lFb\\l7$Fb^l7$$\"1gwZ/1/1RFer$\"1e!))
[)*[B0\"F07$7$$\"1\\>*oB-BV)FerF`tF`_l7$7$$\"1_>*oB-BV)FerF`t7$$\"1&Qc
H?OwA\"Fip$\"1.4E\\44j6F07$7$Fg^l$\"1++++DT#=\"F0F]`l7$Ff^l7$$\"12\"Hj
o^.r\"Fip$!1yAe@ze(>\"F07$7$$\"1++++++!o#Fip$!1++++8+X7F0Fg`l7$Fc`l7$$
\"1Rd\"f3#zhAFip$\"1l5_y>bC7F07$7$F^al$\"1++++8+X7F0Fcal7$F]al7$$\"1wO
a1#H@=$Fip$!1?fj,B`X7F07$7$$\"1++++++gRFip$!1++++\\NY7F0F]bl7$Fial7$$
\"1$=y&>$4nX$Fip$\"1aa5qE#eC\"F07$7$Fdbl$\"1++++\\NY7F0Fibl7$Fcbl7$$\"
1\"=(G)[')*3WFip$!1'zr?iYAB\"F07$7$$\"1++++++S_Fip$!1++++T817F0Fccl7$F
_cl7$$\"1!3'fkL$f%[Fip$\"1z4&)em^=7F07$7$Fjcl$\"1++++T817F0F_dl7$Ficl7
$$\"1%Q30P:&GbFip$!1(4FE%)G@>\"F07$7$$\"1************>lFip$!1++++(**R9
\"F0Fidl7$7$Fjcl$\"1*******4Mh?\"F07$$\"1GdLi'p3S'Fip$\"1ngT%e#y\\6F07
$7$$\"1++++++?lFip$\"1++++(**R9\"F0Fhel7$7$F_flFbel7$$\"1.wf:%=**f'Fip
$!1-%**Qgz*R6F07$7$$\"1Usi*\\2r*pFipFcpFefl7$F[gl7$$\"1WAD'oj,j(Fip$!1
iIc@4ax5F07$7$$\"1+++++++yFip$!1+++++:m5F0F_gl7$7$Ffgl$\"1+++++:m5F07$
$\"1ksi*\\2r*pFipF`t7$F^hlF^fl7$Fegl7$$\"1Qk#[OA.p)Fip$!15m?\"f!eA5F07
$7$$\"1,+++++!3*Fip$!1nmmm5^.5F0Fchl7$F[hl7$$\"1.V7eg#y1)Fip$\"1B*oa[V
I0\"F07$7$$\"1++++++!3*Fip$\"1mmmm5^.5F0F_il7$Fihl7$$\"1<WuzEwh)*Fip$!
1_5O*p1W&**Fip7$7$$\"1++++++O5F0$!1.+++?(H!**FipF[jl7$Feil7$$\"1_PpR`-
m&*Fip$\"17cw]m$\\)**Fip7$7$Fbjl$\"1,+++?(H!**FipFgjl7$Fajl7$$\"1W3(G,
n08\"F0$!16r<KvTO5F07$7$$\"1++++++k6F0$!1++++Gs_5F0Fa[m7$7$Fbjl$\"1***
*****>(H!**Fip7$$\"1R+#pD7%z5F0$\"1.**pd$p9,\"F07$7$Fh[m$\"1++++Gs_5F0
F`\\m7$7$$\"1++++++#H\"F0$!1++++<v#>\"F07$$\"1c_b2(>k@\"F0Fcp7$F_]mFg[
m7$Ff\\m7$$\"1hO\")\\?y\"=\"F0$\"1YeYv[av5F07$7$$\"1b_b2(>k@\"F0F`tFd]
m7$Fj]m7$$\"1w$4Q8*)4F\"F0$\"1glZlr_s6F07$7$F[]m$\"1++++<v#>\"F0F^^m7$
Fj\\m7$$\"1n&))y')QWQ\"F0$!1>9spr4^8F07$7$$\"1++++++?9F0$!1,+++D,79F0F
h^m7$Fd^m7$$\"17L*3M(o]8F0$\"1@nwZ;G$H\"F07$7$F__m$\"1,+++D,79F0Fd_m7$
7$$\"1++++++[:F0$!1,++S#)e&o\"F07$$\"1H4o*f%pI9F0F[p7$Fc`mF^_m7$Fj_m7$
$\"1Y\"4,apaU\"F0$\"1Mrs\\hKE9F07$7$Fd`mFhtFh`m7$F^am7$$\"1,O\"fjOX\\
\"F0$\"1)*f@5%eOd\"F07$7$F_`m$\"1-++S#)e&o\"F0F`am7$7$$\"1,+++++w;F0$!
1NLLLRb6?F07$$\"1\"[3)\\IJt:F0Fco7$7$$\"1#[3)\\IJt:F0FcoF^`m7$Ffam7$$
\"1**3K([z;c\"F0$\"1axp\"G,es\"F07$7$F`bmF`uFgbm7$F]cm7$$\"1:Ad9\"y^i
\"F0$\"1k%pNraq)=F07$7$F[bm$\"1NLLLRb6?F0F_cm7$7$$\"1,+++++/=F0$!1(G9d
[\"pyBF07$$\"1Iu8#)Q,(p\"F0F[o7$F^dmFjam7$Fecm7$$\"1')*o8\")))yo\"F0$
\"1PvdrzF]?F07$7$F_dmFhuFcdm7$Fidm7$$\"1JH**[lbZ<F0$\"1ww^F'36A#F07$7$
Fjcm$\"1(G9d[\"pyBF0F[em7$7$$\"1,+++++K>F0$!1PLLL0duFF07$$\"1,)4_TcB\"
>F0FR7$Fjem7$$\"1Jji.#=Q$=F0$!1Ee14bauCF07$7$$\"1oJ9Ne'*4=F0FYF^fm7$Fd
fmFicm7$Faem7$$\"11SsZ%4v!=F0$\"1')**o!QE7R#F07$7$$\"1nJ9Ne'*4=F0F`vFi
fm7$F_gm7$$\"1pPvp/?k=F0$\"1!e:c#))\\pDF07$7$$\"1+)4_TcB\">F0FhvFcgm7$
Figm7$$\"1[Rj?3m@>F0$\"1I^T[z%eu#F07$7$Ffem$\"1OLLL0duFF0F]hm7$7$$\"1,
+++++g?F0$!1-+++])))>$F07$$\"1P\"Qa&>()3?F0FF7$F\\imFeem7$Fchm7$$\"1![
_(G%)ew>F0$\"1,)=\"G*G&GHF07$7$F]imF`wFaim7$Fgim7$$\"1F2:\"3m;.#F0$\"1
'=BrzM36$F07$7$Fhhm$\"1.+++])))>$F0Fiim7$7$$\"1,+++++)=#F0$!1&444HmHl$
F07$$\"1SL*3wjE5#F0$!1************fLF07$FhjmFghm7$F_jm7$$\"1$yX4,qc3#F
0$\"1Xbjs\\#eH$F07$7$FijmFhwF_[n7$Fe[n7$$\"1LTnl:iQ@F0$\"1qY\"e3YM[$F0
7$7$Fdjm$\"1%444HmHl$F0Fg[n7$7$$\"1/O.7j!)zAF0F+7$$\"14pL![Px?#F0$!1qA
%3qV$HPF07$7$$\"15K==&)f%>#F0F7Fd\\n7$7$F[]nF37$Fdjm$!1%444HmHl$F07$F]
\\n7$$\"1r3t=z4#>#F0$\"1BG<._vpOF07$7$F[]nF`xFc]n7$Fi]n7$$\"11/E@i6VAF
0$\"1()*[oW4A'QF07$7$$\"1.O.7j!)zAF0FhxF[^nFjx-F$6R7$7$$!1++++++?PFip$
\"1++++++W<F07$$!1)************z$FipF`u7$F\\_n7$$!1+++++++]Fip$\"1++++
+++?F07$Fg^n7$$!14++++++MFip$\"1-+++++!o\"F07$7$$!1++++++SCFip$\"1++++
++)[\"F0Ff_n7$7$Fcy$\"1++++++K7F07$$!1)************>#FipFht7$Fe`nF\\`n
7$7$Fc\\l$\"1)***********f(*Fip7$$!1t************fFerF`t7$7$$!1u******
******fFerF`tFb`n7$7$Fg^l$\"1+++++++sFip7$$\"1.++++++5FipFhs7$FhanFj`n
7$7$F^al$\"1++++++SYFip7$$\"1.++++++EFipF`s7$F`bnFean7$F]bn7$$\"1()***
*********HFip$\"1E++++++SFip7$7$Fdbl$FiuFipFebn7$7$Fjcl$!1************
*z%Fer7$$\"1/++++++UFipFhr7$FacnF[cn7$7$F_fl$!1************RIFip7$$\"1
/++++++eFipF_r7$Ficn7$Fjcl$!1-++++++[Fer7$7$$\"1,++++++yFip$!1,++++++c
Fip7$$\"1/++++++uFipFdq7$Ffdn7$F_fl$FGFip7$7$Ffil$!1,+++++g\")Fip7$$\"
1/++++++!*FipF\\q7$F`en7$FfglFddn7$F]en7$$\"1')***********R*Fip$!1v***
********z)Fip7$7$Fbjl$!1++++++s5F0Ffen7$7$Fh[m$!1,+++++G8F07$$\"1+++++
+g5F0Fcp7$FcfnF\\fn7$7$F[]m$!1,+++++%e\"F07$$\"1++++++?7F0F[p7$F[gn7$F
h[m$!1++++++G8F07$7$F__m$!1,+++++S=F07$$\"1++++++!Q\"F0Fco7$FfgnFhfn7$
7$F_`m$!1,+++++'4#F07$$\"1++++++S:F0F[o7$F^hnFcgn7$F[hn7$$\"1,+++++!e
\"F0$!1,+++++g@F07$7$F[bm$!1,+++++_BF0Fchn7$7$Fjcm$!1-+++++3EF07$$\"1,
++++++<F0FY7$F`inFihn7$7$Ffem$!1-+++++kGF07$$\"1,+++++g=F0FR7$FhinF]in
7$7$Fhhm$!1,+++++?JF07$$\"1++++++??F0FF7$F`jnFein7$7$Fdjm$!1-+++++wLF0
7$$\"1++++++!=#F0F>7$FhjnF]jn7$Fejn7$$\"10+++++?AF0$!14+++++SMF07$7$$
\"1,+++++;BF0$!1-+++++KOF0F][o7$7$$\"1,+++++WCF0$!1.+++++))QF07$$\"1++
++++SBF0F37$F^\\oFc[o7$7$$\"1*************\\#F0F+Fi[oFjx-%+AXESLABELSG
6$%\"xG%\"yG" 2 365 365 365 2 0 1 0 2 9 0 4 2 1 0.000000 45.000000 
45.000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 19 0 0 0 0 0 
0 0 "" "" "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "Here is a little p
rocedure to add points." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
1570 "addPoints:=proc(P1,P2,f,x,y)\n  local su1,su2,L,a,b,c,i,P3;\n  i
f P1=[0,1,0] then RETURN(P2) fi;\n  if P2=[0,1,0] then RETURN(P1) fi;
\n  if P1[1]=P2[1] and P1[2]=-P2[2] then\n      # P1 and P2 are on a v
ertical line, sum=0\n      RETURN([0,1,0])\n  fi;\n  su1:=x=P1[1], y=P
1[2]:\n  su2:=x=P2[1], y=P2[2]:\n  if P1=P2 then\n      # If P1=P2 the
n we need to take a tangent line.\n      # The following line has the \+
correct angle:\n      L:=subs(su1,diff(f,x))*x + subs(su1,diff(f,y))*y
 + c;\n      # And now it also goes through the point P1:\n      L:=su
bs(c=solve(subs(su1,L)=0,c),L);\n      # Now compute the intersection \+
points, and\n      # throw away P1:\n      P3:=\{seq(subs(i,[x,y]), i=
\{solve(\{L,f\},\{x,y\})\})\};\n      if nops(P3)=1 then\n         # t
here is only 1 intersection point with the line\n         # and the cu
rve, so that must be an inflection point.\n         P3:=P1\n      else
\n         P3:=op(P3 minus \{P1\})\n      fi\n  else\n      L:=a*x+b*y
+c;\n      L:=subs(solve(\{subs(su1,L),subs(su2,L)\},\{a,b,c\}),L);\n \+
     L:=subs(a=1,b=1,c=1,L);\n      P3:=\{seq(subs(i,[x,y]), i=\{solve
(\{L,f\},\{x,y\})\})\};\n      if nops(P3)=2 then\n         # there ar
e only 2 points on this line, so one\n         # of the two must be a \+
tangent point.\n         L:=normal(subs(solve(\{L\},\{y\}),f)/((x-P1[1
])*(x-P2[1])));\n         if subs(x=P1[1],L)=0 then P3:=P1\n         e
lif subs(x=P2[1],L)=0 then P3:=P2 else ERROR()\n         fi\n      els
e\n         # there are 3 points on the line:\n         P3:=op(P3 minu
s \{P1,P2\});\n      fi\n  fi;\n      # Now put in the -sign at the y-
coordinate:\n      [P3[1],-P3[2]];\nend:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "addPoints(P1,P2,f,x,y);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#7$\"\"#\"\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "So we see t
hat the procedure returns the same result as we had before." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "Now lets add a \+
point to itself:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "P2;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#7$\"\"\"!\"\"" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 23 "addPoints(P2,P2,f,x,y);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#7$\"\"!\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "A
nd the plot for that is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 
"implicitplot(\{f,y+1,x\},x=-0.5..2.7, y=-3..3);" }}{PARA 13 "" 1 "" 
{INLPLOT "6&-%'CURVESG6U7$7$$\"1NS))ztht')!#L$!\"$\"\"!7$$!12o(fZBZt\"
F*$!1+++++]xH!#:7$7$F/$!1************fFF3F.7$7$F($!1++++++gFF37$F/$!1*
*********\\PFF37$7$F/$!1++++++?DF3F<7$7$F(FA7$F/$!1+++++](\\#F37$7$F/$
!1************zAF3FE7$7$F($!1)***********zAF37$F/$!1**********\\dAF37$
7$F/$!1************R?F3FP7$7$F(FU7$F/$!1**********\\<?F37$7$F/$!1*****
********z\"F3FY7$7$F(Fhn7$F/$!1**********\\x<F37$7$F/$!1************f:
F3F\\o7$7$F(Fao7$F/$!1**********\\P:F37$7$F/$!1************>8F3Feo7$7$
F(Fjo7$F/$!1**********\\(H\"F37$7$F/$!1************z5F3F^p7$7$F(Fcp7$F
/$!1**********\\d5F37$7$F/$!1#************R)!#;Fgp7$7$F(F\\q7$F/$!1#**
********\\<)F^q7$7$F/$!1#*************fF^qFaq7$7$F(Ffq7$F/$!1\"*******
***\\x&F^q7$7$F/$!1#************f$F^qFjq7$7$F(F_r7$F/$!1#**********\\P
$F^q7$7$F/$!1#************>\"F^qFcr7$7$F(Fhr7$F/$!1=**********\\(*!#<7
$7$F/$\"13++++++7F^qF\\s7$7$F(Fbs7$F/$\"13+++++D9F^q7$7$F/$\"13++++++O
F^qFfs7$7$F(F[t7$F/$\"13+++++DQF^q7$7$F/$\"13++++++gF^qF_t7$7$F(Fdt7$F
/$\"12+++++DiF^q7$7$F/$\"11++++++%)F^qFht7$7$F($\"12++++++%)F^q7$F/$\"
12+++++D')F^q7$7$F/$\"1,+++++!3\"F3Fcu7$7$F(Fhu7$F/$\"1,++++]-6F37$7$F
/$\"1,+++++?8F3F\\v7$7$F(Fav7$F/$\"1,++++]U8F37$7$F/$\"1,+++++g:F3Fev7
$7$F(Fjv7$F/$\"1,++++]#e\"F37$7$F/$\"1,++++++=F3F^w7$7$F(Fcw7$F/$\"1,+
+++]A=F37$7$F/$\"1,+++++S?F3Fgw7$7$F(F\\x7$F/$\"1,++++]i?F37$7$F/$\"1,
+++++!G#F3F`x7$7$F(Fex7$F/$\"1,++++]-BF37$7$F/$\"1++++++?DF3Fix7$7$F($
\"1,+++++?DF37$F/$\"1,++++]UDF37$7$F/$\"1-+++++gFF3Fdy7$7$F(Fiy7$F/$\"
1-++++]#y#F37$7$F/$\"1-++++++IF3F]z-%'COLOURG6&%$RGBG\"\"\"F-F--F$6gq7
$7$$!1++++++g6F^q$!1VWWW%38V'F^q7$$!1w\\d]IDX7F^qFfq7$Fa[l7$$!17l<7%HK
F\"F^q$!1K!p@&[p(y&F^q7$7$$!1yoLj8^h:F^qF_rFe[l7$7$$!1xoLj8^h:F^qF_r7$
$!1.L(G\"\\5=;F^q$!1]DO)G`5u#F^q7$7$$!1Gyr>0k><F^qFhrFb\\l7$Fh\\l7$$!1
Fyr>0k><F^q$!1e1Hb-u1:F_s7$7$F]]lFbsF\\]l7$Fb]l7$$!1S#RnP0\"e;F^q$\"1$
3'Q\"eZR8#F^q7$7$$!1woLj8^h:F^qF[tFd]l7$Fj]l7$$!1Z%*R;?'>[\"F^q$\"1s*[
2yyO?%F^q7$7$$!1u\\d]IDX7F^qFdtF^^l7$Fd^l7$$!1?!oBd+AA\"F^q$\"1X+>tgi;
hF^q7$7$F][l$\"1YWWW%38V'F^qFh^l7$7$$\"1,++++++7F_s$!1LLLLpt<5F37$$!1@
y#3\\O<z&F_sF\\q7$Fg_lF\\[l7$F^_l7$$!1$yVV&fTgyF_s$\"1n%R9*z#))p(F^q7$
7$$!1px#3\\O<z&F_sFauF\\`l7$Fb`l7$$!1$e57mPQ#GF_s$\"1#*pZh&pW:*F^q7$7$
Fc_l$\"1LLLLpt<5F3Ff`l7$7$$\"1+++++++9F^q$!1+++++C$=\"F37$$\"1(>7&H`xl
`F_sFcp7$7$$\"1)>7&H`xl`F_sFcpFb_l7$F\\al7$$\"1'Q>qypx/$F_s$\"1(Q$\\;V
NX5F37$7$$\"19B^H`xl`F_sFhuF]bl7$7$$\"1:B^H`xl`F_sFhu7$$\"1FcBX2zj5F^q
$\"1@$o_BRI9\"F37$7$Faal$\"1+++++C$=\"F3Fjbl7$F`al7$$\"1%zZc`/G;#F^q$!
1gRV]e-B7F37$7$$\"1++++++!o#F^q$!1mmm'Q,+D\"F3Fdcl7$F`cl7$$\"1B!)*>4[1
(>F^q$\"1s.v#[.I@\"F37$7$F[dl$\"1nmm'Q,+D\"F3F`dl7$Fjcl7$$\"1ewcE;;#f$
F^q$!1N9)\\II5D\"F37$7$$\"1++++++gRF^q$!1nmmE_W^7F3Fjdl7$Ffdl7$$\"1z[k
RM4^IF^q$\"1'em+)*>/D\"F37$7$Fael$\"1nmmE_W^7F3Ffel7$F`el7$$\"1HgX;%4d
t%F^q$!1/be:bWD7F37$7$$\"1++++++S_F^q$!1+++SIa37F3F`fl7$F\\fl7$$\"1)=F
h(4@0WF^q$\"1`h(>%H_O7F37$7$Fgfl$\"1+++SIa37F3F\\gl7$Fffl7$$\"1&[_jh5s
x&F^q$!1!4m0*ps!=\"F37$7$$\"1************>lF^q$!1mmmYjEU6F3Ffgl7$Fbgl7
$$\"1q\"fZ;>71'F^q$\"1dwg19-m6F37$7$$\"1++++++?lF^q$\"1nmmYjEU6F3Fbhl7
$7$FihlF_hl7$$\"1,\\U]%*>ynF^q$!1oY&pR7%G6F37$7$$\"1F!=v8t/o(F^qFcpF_i
l7$Feil7$$\"1eT,*)*3&oxF^q$!1a,WNa4u5F37$7$$\"1+++++++yF^q$!1LLLLL)>2
\"F3Fiil7$7$F`jl$\"1LLLLL)>2\"F37$$\"1+!=v8t/o(F^qFhu7$FhjlFhhl7$F_jl7
$$\"1!paRzk6y)F^q$!1.l'))RoR-\"F37$7$$\"1************z!*F^q$!1++++WM45
F3F][m7$Fejl7$$\"1)=eNhby&yF^q$\"1TyXs?:p5F37$7$$\"1++++++!3*F^q$\"1++
++WM45F3Fi[m7$Fc[m7$$\"1!f<HbQg$**F^q$!1Z?nGs]+5F37$7$$\"1++++++O5F3$!
1KLLL`Ih**F^qFe\\m7$F_\\m7$$\"1.]lXhyy%*F^q$\"1)o*o(fF_+\"F37$7$F\\]m$
\"1NLLL`Ih**F^qFa]m7$F[]m7$$\"1Bk)GtU&[6F3$!1T?;Cw,^5F37$7$$\"1++++++k
6F3$!1LLLLhbe5F3F[^m7$Fg]m7$$\"1\"o)G)H(\\r5F3$\"1B(3dcUM,\"F37$7$Fb^m
$\"1LLLLhbe5F3Fg^m7$7$$\"1++++++#H\"F3$!1******z%oU>\"F37$$\"1*[bX?32=
\"F3Fcp7$Ff_mFa^m7$F]_m7$$\"1Az!)*R*yq6F3$\"1Z,Ov)ps1\"F37$7$$\"1!\\bX
?32=\"F3FhuF[`m7$Fa`m7$$\"1KE\"Rz?ED\"F3$\"1;QT6g$Q:\"F37$7$Fb_m$\"1++
+![oU>\"F3Fe`m7$7$$\"1++++++?9F3$!1BAAAA759F37$$\"1Lq`P?\"3O\"F3Fjo7$F
damFa_m7$F[am7$$\"1'p4LahfK\"F3$\"1AVH1@Kc7F37$7$$\"1Mq`P?\"3O\"F3FavF
iam7$F_bm7$$\"1&R:t3vMR\"F3$\"1NOG6Utp8F37$7$F`am$\"1CAAAA759F3Fcbm7$7
$$\"1++++++[:F3$!11>w/2v'o\"F37$$\"1_I)fIIW[\"F3Fao7$FbcmF_am7$Fibm7$$
\"1M-L#*\\nb9F3$\"183Q*Q4J\\\"F37$7$$\"1`I)fIIW[\"F3FjvFgcm7$F]dm7$$\"
1i6!pWQ_^\"F3$\"1@.17zU@;F37$7$F^cm$\"11>w/2v'o\"F3Fadm7$7$$\"1,+++++w
;F3$!1-+++1Z9?F37$$\"1m2'[&3X)e\"F3Fhn7$F`em7$F^cm$!10>w/2v'o\"F37$Fgd
m7$$\"1a$zh7EAd\"F3$\"19PT8gda<F37$7$$\"1n2'[&3X)e\"F3FcwFhem7$7$FaemF
cw7$$\"1A^<t\"3ki\"F3$\"1\\mHvY)H*=F37$7$F\\em$\"1,+++1Z9?F3Fcfm7$7$$
\"1,+++++/=F3$!1NLL`?6xBF37$$\"1aR,$eJnw\"F3FJ7$Fbgm7$$\"1Wgum=C)p\"F3
$!1H)[,].<3#F37$7$$\"1,PGwy$Qo\"F3FUFfgm7$F\\hmF[em7$Fifm7$$\"1q#HTPu4
o\"F3$\"1@w][InI?F37$7$$\"1-PGwy$Qo\"F3F\\xFahm7$Fghm7$$\"1&R(p,(o5t\"
F3$\"1'Q<V=Yn<#F37$7$FcgmFexF[im7$Faim7$$\"1B&yX\"oK#y\"F3$\"1s_mAsj?B
F37$7$F^gm$\"1NLL`?6xBF3Fcim7$7$$\"1,+++++K>F3$!1\"yxx(\\^wFF37$$\"1Zd
#\\:bg#>F3F:7$Fbjm7$$\"1AsR\"H4k\">F3$!1*yp8Un2t#F37$7$$\"1galSC'o%=F3
FAFfjm7$F\\[nF]gm7$Fiim7$$\"1&e70(GVJ=F3$\"11*)ynLcoCF37$7$F][nFbyFa[n
7$7$$\"1halSC'o%=F3Fby7$$\"1wdI>NSz=F3$\"1t;I,%='=EF37$7$FcjmFiyF\\\\n
7$Fb\\n7$$\"1R]=22XG>F3$\"1c!GS#\\lmFF37$7$F^jm$\"1#yxx(\\^wFF3Fd\\n7$
7$$\"1h#*HyUs'*>F3F+F]jm7$Fj\\n7$$\"1ekG\\j%R(>F3$\"1&*y3&f]8#HF37$7$F
_]nFbzFb]nFdz-F$6U7$7$$!1+++++++]F^q$!1+++++++5F37$$!1immmmmYTF^qF_^n7
$7$$!1++++++?PF^qF_^nFa^n7$Fe^n7$$!1immmmmmGF^qF_^n7$7$$!1++++++SCF^qF
_^nFi^n7$F]_n7$$!1immmmm'e\"F^qF_^n7$7$F][lF_^nFa_n7$Fe_n7$$!1AmmmmmmI
F_sF_^n7$7$Fc_lF_^nFg_n7$F[`n7$$\"1zLLLLLL(*F_sF_^n7$7$FaalF_^nF]`n7$F
a`n7$$\"1QLLLLL`AF^qF_^n7$7$F[dlF_^nFc`n7$Fg`n7$$\"1QLLLLLLNF^qF_^n7$7
$FaelF_^nFi`n7$F]an7$$\"1PLLLLL8[F^qF_^n7$7$FgflF_^nF_an7$Fcan7$$\"1QL
LLLL$4'F^qF_^n7$7$FihlF_^nFean7$Fian7$$\"1QLLLLLttF^qF_^n7$7$F`jlF_^nF
[bn7$F_bn7$$\"1QLLLLL`')F^qF_^n7$7$$\"1,+++++!3*F^qF_^nFabn7$Febn7$$\"
1QLLLLLL**F^qF_^n7$7$F\\]mF_^nFibn7$F]cn7$$\"1MLLLLL@6F3F_^n7$7$Fb^mF_
^nF_cn7$Fccn7$$\"1MLLLLL\\7F3F_^n7$7$Fb_mF_^nFecn7$Ficn7$$\"1MLLLLLx8F
3F_^n7$7$F`amF_^nF[dn7$F_dn7$$\"1MLLLLL0:F3F_^n7$7$F^cmF_^nFadn7$Fedn7
$$\"1NLLLLLL;F3F_^n7$7$F\\emF_^nFgdn7$F[en7$$\"1NLLLLLh<F3F_^n7$7$F^gm
F_^nF]en7$Faen7$$\"1NLLLLL*)=F3F_^n7$7$F^jmF_^nFcen7$Fgen7$$\"1NLLLLL<
?F3F_^n7$7$$\"1,+++++g?F3F_^nFien7$F]fn7$$\"1NLLLLLX@F3F_^n7$7$$\"1,++
+++)=#F3F_^nFafn7$Fefn7$$\"1NLLLLLtAF3F_^n7$7$$\"1,+++++;BF3F_^nFifn7$
F]gn7$$\"1MLLLLL,CF3F_^n7$7$$\"1,+++++WCF3F_^nFagn7$Fegn7$$\"1NLLLLLHD
F3F_^n7$7$$\"1,+++++sDF3F_^nFign7$F]hn7$$\"1OLLLLLdEF3F_^n7$7$$\"1-+++
+++FF3F_^nFahnFdz-%+AXESLABELSG6$%\"xG%\"yG" 2 365 365 365 2 0 1 0 2 
9 0 4 2 1 0.000000 45.000000 45.000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 "" "" "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 17 "One more example:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 3 "P1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$\"\"!\"\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "addPoints(P1,P1,f,x,y);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#7$\"\"$!\"(" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 49 "implicitplot(\{f,y-2*x-1,x-3\},x=-1..4, y=-10..10)
;" }}{PARA 13 "" 1 "" {INLPLOT "6&-%'CURVESG6go7$7$$!\"\"\"\"!$!1+++++
++5!#:7$$!1***************)!#;$!1**************zF17$7$$!1+++++++!)F1$!
1-++++++gF1F.7$7$F6$!1,++++++gF17$$!1nmmmmmmwF1$!1LLLLLLL`F17$7$$!1)**
***********pF1$!1(*************RF1F>7$FD7$$!1MLLLLLLjF1$!1mmmmmmmEF17$
7$F<$!1-++++++?F1FJ7$FP7$$!1+++++++]F1$\"1\"*GchvbvF!#K7$7$$!1+++++++S
F1$\"1)*************>F1FT7$7$$!1,++++++SF1Fhn7$$!1nmmmmmmOF1$\"1ommmmm
mEF17$7$$!1)*************HF1$\"1/++++++SF1F^o7$Fdo7$$!1MLLLLLLBF1$\"1M
LLLLLL`F17$7$$!1,++++++?F1$\"1**************fF1Fjo7$F`p7$$F,F1$\"1)***
**********zF17$7$$!1$yDJ7:6b&FY$\"1+++++++5F-Ffp7$7$F\\q$\"0**********
*****F-7$$\"1PLLLLLLL!#<$\"1mmmmmmm5F-7$7$$\"1-++++++5F1$\"1+++++++7F-
Fdq7$F[r7$$\"1nmmmmmm;F1$\"1LLLLLLL8F-7$7$$\"1+++++++?F1$\"1**********
***R\"F-Far7$Fgr7$$\"1,++++++IF1$\"1+++++++;F-7$7$$\"1+++++++SF1$\"1**
***********z\"F-F]s7$Fcs7$$\"1NLLLLLLVF1$\"1mmmmmmm=F-7$7$$\"1-++++++]
F1$\"\"#F*Fis7$F_t7$$\"1nmmmmmmcF1$\"1LLLLLLL@F-7$7$$\"1+++++++gF1$\"1
+++++++AF-Fet7$F[u7$$\"1,++++++qF1$\"1+++++++CF-7$7$$\"1+++++++!)F1$\"
1,++++++EF-Fau7$Fgu7$$\"1NLLLLLL$)F1$\"1nmmmmmmEF-7$7$$\"1/++++++!*F1$
\"1,++++++GF-F]v7$Fcv7$$\"1ommmmmm'*F1$\"1MLLLLLLHF-7$7$$\"\"\"F*$\"\"
$F*Fiv7$7$F`w$\"1+++++++IF-7$$\"1+++++++6F-$\"1,++++++KF-7$7$F^r$\"1++
+++++MF-Fhw7$F^x7$$\"1LLLLLLL7F-$\"1nmmmmmmMF-7$7$$\"1+++++++8F-$\"1,+
+++++OF-Fbx7$Fhx7$$\"1nmmmmmm8F-$\"1MLLLLLLPF-7$7$$\"1+++++++9F-$\"1++
+++++QF-F^y7$Fdy7$$\"1+++++++:F-$\"\"%F*7$7$F`s$\"1*************>%F-Fj
y7$F`z7$$\"1MLLLLLL;F-$\"1nmmmmmmUF-7$7$$\"1+++++++<F-$\"1+++++++WF-Fd
z7$Fjz7$$\"1nmmmmmm<F-$\"1LLLLLLLXF-7$7$$\"1+++++++=F-$\"1+++++++YF-F`
[l7$Ff[l7$$\"1+++++++>F-$\"1+++++++[F-7$7$Fbt$\"1**************\\F-F\\
\\l7$7$Fbt$\"\"&F*7$$\"1MLLLLLL?F-$\"1mmmmmmm]F-7$7$$\"1,++++++@F-$\"1
+++++++_F-Fi\\l7$F_]l7$$\"1nmmmmmm@F-$\"1LLLLLLL`F-7$7$F^u$\"1********
*****R&F-Fe]l7$7$F^u$\"1+++++++aF-7$$\"1+++++++BF-$\"1+++++++cF-7$7$Fd
u$\"1+++++++eF-Fb^l7$7$Fdu$\"1*************z&F-7$$\"1LLLLLLLCF-$\"1nmm
mmmmeF-7$7$$\"1+++++++DF-$\"\"'F*F__l7$Fe_l7$$\"1mmmmmmmDF-$\"1MLLLLLL
hF-7$7$$\"1+++++++EF-$\"1*************>'F-F[`l7$7$Fb`l$\"1+++++++iF-7$
$\"1+++++++FF-$\"1+++++++kF-7$7$$\"1+++++++GF-$\"1,++++++mF-Fj`l7$F`al
7$$\"1LLLLLLLGF-$\"1nmmmmmmmF-7$7$$\"1**************GF-$\"1+++++++oF-F
fal7$F\\bl7$$\"1mmmmmmmHF-$\"1MLLLLLLpF-7$7$Fbw$FcuF-Fbbl7$Fhbl7$$\"1,
++++++JF-$\"1,++++++sF-7$7$F[x$\"1-++++++uF-F[cl7$Facl7$$\"1MLLLLLLKF-
$\"1ommmmmmuF-7$7$$\"1+++++++LF-$\"1+++++++wF-Fecl7$F[dl7$$\"1nmmmmmmL
F-$\"1MLLLLLLxF-7$7$$\"1,++++++MF-$\"1-++++++yF-Fadl7$Fgdl7$$\"1,+++++
+NF-$\"1-++++++!)F-7$7$F[y$\"1,++++++#)F-F]el7$Fcel7$$\"1MLLLLLLOF-$\"
1nmmmmmm#)F-7$7$$\"1+++++++PF-$\"1+++++++%)F-Fgel7$7$F^fl$\"1*********
****R)F-7$$\"1nmmmmmmPF-$\"1NLLLLLL&)F-7$7$$\"1,++++++QF-$\"1,++++++')
F-Fffl7$7$$\"1-++++++QF-F_gl7$$\"1,++++++RF-$\"1,++++++))F-7$7$$\"1,++
++++SF-$\"1-++++++!*F-Fegl-%'COLOURG6&%$RGBGFawF*F*-F$6U7$7$Fbw$!1****
*********>*F-7$Fbw$!#5F*7$7$Fbw$!1*************R)F-Fghl7$7$Fbw$!1*****
********f(F-F^il7$7$Fbw$!1*************z'F-Fbil7$7$Fbw$!1*************
*fF-Ffil7$7$Fbw$!1*************>&F-Fjil7$7$Fbw$!1*************R%F-F^jl
7$7$Fbw$!1+++++++OF-Fbjl7$7$Fbw$!1+++++++GF-Ffjl7$7$Fbw$!\"#F*Fjjl7$7$
Fbw$!1+++++++7F-F^[m7$7$FbwFGFb[m7$7$FbwFgoFf[m7$7$FbwF^rFh[m7$7$FbwFb
tFj[m7$7$FbwFfvF\\\\m7$7$FbwF[yF^\\m7$7$FbwF][lF`\\m7$7$FbwFb]lFb\\m7$
7$FbwFh_lFd\\m7$7$FbwF_blFf\\m7$7$FbwF^dlFh\\m7$7$FbwF`flFj\\m7$7$Fbw$
\"1,++++++#*F-F\\]m7$7$Fbw$\"#5F*F^]m7$FjhlFghl7$FghlF^il7$F^ilFbil7$F
bilFfil7$FfilFjil7$FjilF^jl7$F^jlFbjl7$FbjlFfjl7$FfjlFjjl7$FjjlF^[m7$F
^[mFb[m7$Fb[mFf[m7$Ff[mFh[m7$Fh[mFj[m7$Fj[mF\\\\m7$F\\\\mF^\\m7$F^\\mF
`\\m7$F`\\mFb\\m7$Fb\\mFd\\m7$Fd\\mFf\\m7$Ff\\mFh\\m7$Fh\\mFj\\m7$Fj\\
mF\\]m7$F\\]mF^]m7$F^]mFb]mF`hl-F$6gq7$7$F\\q$!1++++++]#*F17$$!1MLLLLL
e9F1FG7$Fd_m7$$!1LLLLLLe9F1$\"1LLLLLLL=F17$7$Fi_m$\"10++++++SF1Fh_m7$7
$Fi_mFgo7$$!14@%ot%*y!pFgq$\"1Uot%*y:jnF17$7$F\\q$\"1%***********\\#*F
1Fc`m7$7$Fhr$!1+++++]A7F-7$$\"1)********\\(=<F1Fc[m7$F`amFa_m7$7$F\\q$
\"1'***********\\#*F17$$\"1$H9dG92\"\\Fgq$\"1G9dG9d.5F-7$7$$\"1/++++v=
<F1F^rFham7$F^bm7$$\"1,+++]7`>F1$\"1+++++v=7F-7$7$Fhr$\"1+++++]A7F-Fbb
m7$F]am7$$\"13Bp2Bpd?F1$!1Bp2Bp2B7F-7$7$Fds$!1+++++]U7F-F\\cm7$Fhbm7$$
\"1NYTj9M'*QF1$\"1YTj9MYT7F-7$7$Fds$\"1+++++]U7F-Ffcm7$Fbcm7$$\"1EBI4s
$))4%F1$!1I4s$)[`R7F-7$7$$\"1smmmmm;aF1Fc[mF`dm7$Ffdm7$$\"1i#yM/8R#fF1
$!1/8R<_cp6F-7$7$F\\u$!1++++++l6F-Fjdm7$7$F\\u$\"1***********\\;\"F-7$
$\"1`mmmmm;aF1F^r7$Fgem7$Fds$\"1**********\\U7F-7$7$FcpFaem7$$\"1,++++
v$f(F1$!1+++++]P5F-7$7$Fhu$!1++++++05F-F`fm7$Fdem7$$\"1-+++]P4hF1$\"1*
********\\i:\"F-7$7$Fhu$\"1***********\\+\"F-Fjfm7$Fffm7$$\"1++++++v$*
F1$!1,++++++&*F17$7$F`wFb_mFdgm7$F`gm7$$\"1pmmmmmT&)F1$\"1FLLLLLL)*F17
$7$F`wFfamF\\hm7$Fjgm7$$\"1#)eqk<Ta6F-$!1HN#)eqk<5F-7$7$F^r$!1++++++X5
F-Fdhm7$Fbhm7$$\"1_cp3Eyf5F-$\"11R<_cp3'*F17$7$F^r$\"1***********\\/\"
F-F^im7$7$Fey$!1+++++]U8F-7$$\"1YXXXXXq7F-Fc[m7$F[jmFjhm7$Fdim7$$\"1++
++++D7F-$\"1*************4\"F-7$7$F\\jmF^rF`jm7$Ffjm7$$\"17%HN#)e?P\"F
-$\"1_B)eqk<J\"F-7$7$Fey$\"1**********\\U8F-Fhjm7$Fhim7$$\"1nmmmm;z9F-
$!1mmmmmm;:F-7$7$F`s$!1+++++]#y\"F-Fb[n7$F^[n7$$\"1un4(Q[g]\"F-$\"1.Hh
^k!ed\"F-7$7$F`s$\"1**********\\#y\"F-F\\\\n7$7$Fg[l$!1***********\\L#
F-7$$\"1nmmmmTg;F-F_[m7$Fi\\nFh[n7$Fb\\n7$$\"1KE0@%='G;F-$\"1t%*y:j_&)
=F-7$7$Fj\\nFbtF^]n7$Fd]n7$$\"1+++]ilZ<F-$\"1++++]P4AF-7$7$Fg[l$\"1***
********\\L#F-Ff]n7$7$Fbt$!1*********\\7)HF-7$$\"11\\etPgJ>F-F[[m7$Fc^
nFf\\n7$F\\^n7$$\"19orjcsh=F-$\"1VF8Xt4`DF-7$7$Fd^nFfvFh^n7$F^_n7$$\"1
cA!4OWF(>F-$\"1y4RcD-4HF-7$7$Fbt$\"1+++++D\")HF-F`_n7$7$F^u$!1********
****)o$F-7$$\"11_%za?&p@F-Fgjl7$7$F^`n$!1,++++++OF-F`^n7$Ff_n7$$\"1x6%
HN#)33#F-$\"1%HN#)eqkF$F-7$7$F^`nF[yFe`n7$F[an7$$\"1+PKG(Qr=#F-$\"1-_q
'3X9l$F-7$7$F^u$\"1++++++*o$F-F]an7$7$Fdu$!1)*********\\ZWF-7$$\"1+++]
i:&Q#F-Fcjl7$FjanFj_n7$Fcan7$$\"1/-^v()o!H#F-$\"1%=fz*[CPSF-7$7$F[bnF]
[lF_bn7$7$$\"1,++]i:&Q#F-F][l7$$\"1yxxxFS$R#F-$\"1*)))))))))QEWF-7$7$F
du$\"1+++++]ZWF-Fjbn7$7$Fju$!1dG9dG9__F-7$$\"1dlg$)4/&e#F-F_jl7$FgcnFg
an7$7$Fdu$\"1,++++]ZWF-7$$\"1Up'R7&G$\\#F-$\"1KA8/&fo#[F-7$7$$\"1elg$)
4/&e#F-Fb]lF_dn7$Fedn7$$\"1R]9^V.$f#F-$\"1[)>afiyA&F-7$7$Fb`l$\"1dG9dG
9__F-Fidn7$7$Faal$!1,+++]i+hF-7$$\"1F4'3PWLx#F-F[jl7$Ffen7$Fb`lFecn7$7
$FjuF`en7$$\"1Z#3OX[**o#F-$\"16qc&=1-k&F-7$7$FgenFh_lF]fn7$Fcfn7$$\"10
N\"y(y0(y#F-$\"1#)fu)[o<0'F-7$7$Faal$\"1,+++]i+hF-Fefn7$7$Ffw$!1nmmmmm
\"*pF-7$$\"1;!fC&)oG&HF-Fgil7$FbgnFcen7$F[gn7$$\"1[Qwm(f:)GF-$\"11Y%H$
4wtkF-7$7$FcgnF_blFggn7$F]hn7$$\"1([@%p'Ri(HF-$\"1]SJA8/&*oF-7$7$Fbw$
\"1ommmmm\"*pF-F_hn7$7$F[x$!1/++++]CzF-7$$\"17a\\%Rtb7$F-$!1)*********
***f(F-7$7$$\"18a\\%Rtb7$F-Fcil7$FbwF`gn7$7$Fbw$\"1pmmmmm\"*pF-7$$\"1w
^.29yoIF-$\"1(Hf=Pu[K(F-7$7$FcinF^dlFjin7$F`jn7$$\"1;j_5U=hJF-$\"1QZ*y
:j_v(F-7$7$F[x$\"1/++++]CzF-Fbjn7$7$Fhdl$!1pjjjjj)*))F-7$$\"1++v$4rGH$
F-F_il7$F_[oFihn7$Fhjn7$$\"1PZ*y:Q@D$F-$\"1a5UotW\">)F-7$7$$\"1,+v$4rG
H$F-F`flFd[o7$7$F`[oFdfl7$$\"18:y`WQULF-$\"1_R([=i/j)F-7$7$Fhdl$\"1pjj
jjj)*))F-F_\\o7$7$F[y$!11++++v8**F-7$$\"1!33333eX$F-Fhhl7$F\\]oF\\[o7$
Fe\\o7$$\"1\"G:8'*f?V$F-$\"1\"))QZ:g<2*F-7$7$$\"1\"33333eX$F-F_]mFa]o7
$Fg]o7$$\"1X![mR^-_$F-$\"1DyS8W**=&*F-7$7$F[y$\"1/++++v8**F-F[^o7$7$$
\"1y\"zI&f<:OF-$F,!#9Fi\\o7$Fa^o7$$\"1B')zcq!*3OF-$\"13b!GxrV'**F-7$7$
$\"1z\"zI&f<:OF-Fc]mF[_oF`hl-%+AXESLABELSG6$%\"xG%\"yG" 2 365 365 365 
2 0 1 0 2 9 0 4 2 1 0.000000 45.000000 45.000000 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 1 1 0 0 0 2129 0 0 0 0 0 0 0 0 "" "" "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 23 "OK, lets try something:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 6 "Q:=P2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"Q
G7$\"\"\"!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "Q:=addPoi
nts(Q,Q,f,x,y); length_expression = length(Q);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"QG7$\"\"!\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
/%2length_expressionG\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
55 "Q:=addPoints(Q,Q,f,x,y); length_expression = length(Q);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"QG7$\"\"$!\"(" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/%2length_expressionG\"\"(" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 55 "Q:=addPoints(Q,Q,f,x,y); length_expression = length
(Q);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"QG7$#\"#g\"#\\#!$$Q\"$V$" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%2length_expressionG\"#@" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "Q:=addPoints(Q,Q,f,x,y); len
gth_expression = length(Q);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"QG7
$#!(x1:\"\"(hx=(#!+t-0kr\",TsQq#>" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/
%2length_expressionG\"#Y" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 
"Q:=addPoints(Q,Q,f,x,y); length_expression = length(Q);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%\"QG7$#\"=?AIU1(3_BE=Aqs)\"<p&Rf0xfC@x)=!*o$#
\"L<Wo,A#QDaNP]Zb]d9OFih:\"I(\\:#=B_yF@G$[zendqJa3(" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#/%2length_expressionG\"$\\\"" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 55 "Q:=addPoints(Q,Q,f,x,y); length_expression = len
gth(Q);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%\"QG7$#\"[r.E)*pUpx_fP4\\
q[R&3NSvwB'>0Z/B5WSTVJ&)fSQb$y]TbzaE+Z1v*Q'>N*yd\"jqTy<f3l_FIS!>aY'[?^
y/:IY/3\"[\"pSOa[*4I`IgY!oj\")e^$Q-S0BF6BLG'**)#\"cu$fI-3l.Z'Q*H3xMJq
\"==]x:eR<@3(Qel&fh(eT[GFhY`Q+()o5?rK2`_;qAr!ou>cVInYsox(fc\\%[R)*y%[!
yyijLBu\"buhB$\\>pp-rD[yyr#=#o4b=5be9\"['>\">*RFFVw0yL!)*[[ua\"pPte52R
v$Q)oGp14'p\"=*H#H]>)z`nI;#RrPE+P3owK)p#" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/%2length_expressionG\"$p&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
101 "If you add Q again to Q then you need close to 4*569, which is mo
re than 2000, digits to represent Q!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 204 "Notice that when the point has already
 gotten big, then whenever you multiply Q by 2 you get about 4 times a
s much digits, and this factor 4 gets more and more accurate when Q be
comes a larger expression." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT 277 254 "Definition: If P is a point on the curve, let n
 be the numerator of the x-coordinate and let d be the denominator of \+
the x-coordinate. Then the logarithmic height (we'll just call it: hei
ght) of the point P is defined as the maximum of log(n) and log(d)." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "Denote t
he height as: " }{TEXT 279 4 "h(P)" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" 
{TEXT -1 96 "So up to some bounded factor, h(P) is more or less the nu
mber of digits you need to represent P." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT 278 53 "Theorem:  h(P+P) is at least 4*h(P)
 minus a constant." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 120 "So whenever h(P) is larger than some constant, then h(P+
P) is about 4 times h(P). This matches our experiment very well." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "with(algcurves):" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 63 "Load the algcurves package so we can run \+
the following command:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "j
_invariant(f,x,y);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##!%'4%\"#V" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#$!+&R\"eD&*!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
4 "The " }{TEXT 280 11 "j invariant" }{TEXT -1 73 " is a number that c
haracterizes elliptic curves over the complex numbers." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 91 "If C1 and C2 are alg
ebraic curves, then they are called birational when there is a 1-1 map
:" }}{PARA 0 "" 0 "" {TEXT -1 13 "  C1  ---> C2" }}{PARA 0 "" 0 "" 
{TEXT -1 235 "defined by rational functions. Two curves are birational
 if and only if they are isomorphic as Riemann surfaces (as compact su
rfaces with a complex structure on it). Two elliptic curves are birati
onal if and only if they have the same " }{TEXT 281 12 "j invariant." 
}}{PARA 0 "" 0 "" {TEXT -1 25 "For every complex number " }{TEXT 282 
3 "j, " }{TEXT -1 141 "there is (up to birational equivalence) precise
ly one elliptic curve with that j invariant. So the set of all possibl
e j invariants (the set " }{TEXT 337 2 "C " }{TEXT -1 162 "of complex \+
numbers) classifies the set of all elliptic curves over the complex nu
mbers, up to equivalence. One says: the moduli space of the elliptic c
urves over " }{TEXT 283 1 "C" }{TEXT -1 66 " is isomorphic to  the set
 of all possible j invariants, which is " }{TEXT 284 1 "C" }{TEXT -1 
1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "T
he j invariant does not classify the elliptic curves over the real num
bers. For example:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "f1 :=
 y^2 - x*(x^2+1);   f2:= y^2 - x*(x^2-1);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#f1G,&*$)%\"yG\"\"#\"\"\"F**&%\"xGF*,&*$)F,F)F*F*F*F*
F*!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f2G,&*$)%\"yG\"\"#\"\"\"
F**&%\"xGF*,&*$)F,F)F*F*!\"\"F*F*F0" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 41 "j_invariant(f1,x,y); j_invariant(f2,x,y);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#\"%G<" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"%
G<" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "As you can see, f1 and f2 h
ave the same j invariant. This means you can construct a map:" }}
{PARA 0 "" 0 "" {TEXT -1 12 "  C1 ---> C2" }}{PARA 0 "" 0 "" {TEXT -1 
572 "that is defined with rational functions. This map is also an isom
orphism between the groups of additive points (over the complex number
s) of C1 and C2. However, in this example this map can not be defined \+
over the real numbers, you have to use complex numbers to write down t
he map. Why? Well, over the real numbers the curve C1 (given by f1=0) \+
has only 1 component and C2 has 2 real components. So over the reals t
here can not be an isomorphism. Over the complex numbers this problem \+
does not appear because both have just 1 connected component, and thes
e are isomorphic." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 143 "So over the real numbers, the j-invariant does not class
ify elliptic curves up to equivalence. But there are still some nice p
roperties though:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 122 "If C is an elliptic curve given by f=y^2-fx with fx a po
lynomial of degree 3 in x with real numbers as coefficients, then:" }}
{PARA 0 "" 0 "" {TEXT -1 19 "j is a real number." }}{PARA 256 "" 0 "" 
{TEXT -1 42 "If C has 1 real component then  j <= 1728." }}{PARA 257 "
" 0 "" {TEXT -1 42 "If C has 2 real components then j >= 1728." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "We saw in
 the example before where f had j=-95.2558... that the curve had just \+
1 real component." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 144 "What about the group of points on an elliptic curve over
 the reals? Topologically such curve is either 1 or 2 circles. You can
 think of the set " }{TEXT 285 4 "R/Z " }{TEXT -1 19 "as a circle, whe
re " }{TEXT 286 2 "R " }{TEXT -1 25 "are the real numbers and " }
{TEXT 287 2 "Z " }{TEXT -1 105 "the integers. This would lead one to b
elieve that the group of points on a real elliptic curve is either:" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "If the n
umber of real components is 1 then the group is isomorphic to:" }
{TEXT 288 6 "  R/Z." }}{PARA 0 "" 0 "" {TEXT 291 0 "" }{TEXT -1 71 "If
 the number of real components is 2 then the group is isomorphic to: \+
" }{TEXT 289 9 " R/Z * C2" }{TEXT -1 17 ", the product of " }{TEXT 
290 4 "R/Z " }{TEXT -1 39 "and a cyclic group with two elemements." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 113 "It is ea
sy to see why this is true if we consider what the possible image coul
d be of the real points in the set " }{TEXT 292 1 "C" }{TEXT -1 3 " /(
" }{TEXT 293 2 "Z " }{TEXT -1 5 "w1 + " }{TEXT 294 2 "Z " }{TEXT -1 
603 "w2). For each point P in the connected component that contains th
e base-point, we can compute an integral that is a real number because
 we can find a path through real points from the base-point to P and t
he integrand (1/y)*dx is also real, so the resulting integral is real.
 If there is a second connected component, and if P is in that compone
nt, then P+P is in the first component again. So the integral to the p
oint P is a complex number, that when you multiply it by two then modu
lo w1 and w2 it must again be real. We may assume that w1 is real and \+
w2 is not real, so then the set of integrals is " }{TEXT 338 1 "R" }
{TEXT -1 9 " (modulo " }{TEXT 339 2 "Z " }{TEXT -1 5 "w1 + " }{TEXT 
340 2 "Z " }{TEXT -1 39 "w2) for the first component, and it is " }
{TEXT 343 1 "R" }{TEXT -1 18 "+(1/2)*w2 (modulo " }{TEXT 341 2 "Z " }
{TEXT -1 5 "w1 + " }{TEXT 342 2 "Z " }{TEXT -1 62 "w2) for the second \+
component (if there is a second component)." }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 42 "Elliptic curves over t
he rational numbers." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 195 "Since the rational numbers are a subset of the real nu
mbers, the group of points over the rational numbers (i.e. the group o
f those points that have rational numbers as entries) is a subgroup of
:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 295 3 "R/Z" }}{PARA 0 "" 0 "
" {TEXT -1 6 "or of:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 296 9 "R/
Z * C2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
35 "A point P on the curve is called a " }{TEXT 297 13 "torsion point
" }{TEXT -1 138 " if there exists a positive integer n such that n*P i
s zero in the group (so n*P is the point at infinity). n*P is short fo
r: P+P+P+...+P." }}{PARA 0 "" 0 "" {TEXT -1 164 "Since the group is co
mmutative, it is clear that the set of torsion points forms a subgroup
. The set of torsion points with real coefficients must be isomorphic \+
to:" }}{PARA 0 "" 0 "" {TEXT 298 3 "Q/Z" }}{PARA 0 "" 0 "" {TEXT -1 3 
"or:" }}{PARA 0 "" 0 "" {TEXT 299 8 "Q/Z * C2" }}{PARA 0 "" 0 "" 
{TEXT -1 6 "where " }{TEXT 300 1 "Q" }{TEXT -1 65 " is the set of rati
onal numbers. Why, well if A is an element of " }{TEXT 301 4 "R/Z," }
{TEXT -1 272 " so A can be represented by a real number and we compute
 modulo integers (so adding or substracting integers has no effect) th
en take the denominator of A, which is an integer, and if we multiply \+
A by that integer then the product is an integer (the numerator), so m
odulo " }{TEXT 302 2 "Z " }{TEXT -1 104 "the product is zero. So there
 is a positive integer such that A times that integer is zero in the g
roup." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "
A point P is an " }{TEXT 313 16 "inflection point" }{TEXT -1 57 " when
 3*P is zero. Over the complex numbers the group is " }{TEXT 314 1 "C
" }{TEXT -1 2 "/(" }{TEXT 315 2 "Z " }{TEXT -1 5 "w1 + " }{TEXT 316 2 
"Z " }{TEXT -1 72 "w2) so it is clear that there are 9 inflection poin
ts, corresponding to:" }}{PARA 0 "" 0 "" {TEXT -1 21 "  i/3 * w1 + j/3
 * w2" }}{PARA 0 "" 0 "" {TEXT -1 20 "for i,j from 0 to 2." }}{PARA 0 
"" 0 "" {TEXT -1 35 "Over the real numbers the group is " }{TEXT 317 
8 "R/Z * C2" }{TEXT -1 4 " or " }{TEXT 318 3 "R/Z" }{TEXT -1 342 " and
 so the number of inflection points is always 3 (the point at infinity
 and two finite points). More generally, the number of points P for wh
ich n*P is zero is n^2 over the complex numbers. Over the real numbers
 the number is n when n is odd, and it is either n or 2*n (depending o
n whether there is 1 or 2 real components) when n is even." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "The " }{TEXT 303 
13 "torsion index" }{TEXT -1 18 ", also called the " }{TEXT 319 5 "ord
er" }{TEXT -1 362 " of a torsion point P is the smallest positive inte
ger n such that n*P is zero. We saw that over the real numbers the ord
er of points in an elliptic curve C can be arbitrarily high. However, \+
if C is a rational elliptic curve (so we restrict to rational points) \+
then the order can not be arbitrarily high, which implies that the gro
up of torsion points is finite." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 659 "Why? The reason is the following (I won'
t explain what \"good reduction\" precisely means). There are infinite
ly many prime numbers p such that the point P and the curve have \"goo
d reduction\" mod p. If P is a torsion point, and Q is the point on th
e reduced curve mod p, then the order of P (assuming that P is a torsi
on point) equals some power of p times the order of Q. Apply good redu
ction modulo 2 distinct primes and you get a bound for the order. Note
 that also works in the case of higher genus curves, reduction mod two
 primes gives you a way to bound the torsion index of a divisor (under
 the assumption that that divisor is indeed a torsion divisor)." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "Conclusio
n: The set of " }{TEXT 306 43 "torsion points on a rational elliptic c
urve" }{TEXT -1 34 " is a group that is isomorphic to:" }}{PARA 0 "" 
0 "" {TEXT 304 2 "Cn" }}{PARA 0 "" 0 "" {TEXT -1 2 "or" }}{PARA 0 "" 
0 "" {TEXT 305 5 "Cn*C2" }}{PARA 0 "" 0 "" {TEXT -1 129 "where Cn is t
he cyclic group with n elements. The number n is the highest torsion i
ndex that occurs. Note that in the second case" }{TEXT 307 6 " Cn*C2" 
}{TEXT -1 56 " we may assume that n is even, because if n is odd then \+
" }{TEXT 308 5 "Cn*C2" }{TEXT -1 18 " is isomorphic to " }{TEXT 309 2 
"Cm" }{TEXT -1 49 " where m=2*n, so that falls under the first case." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 138 "We rea
ched this conclusion relatively easily. However, the precise statement
 (which n can occur and which can't) is in fact hard to prove:" }}
{PARA 0 "" 0 "" {TEXT 310 161 "Mazur's Theorem: the group of torsion p
oints on an elliptic curve over the rational numbers is isomorphic to \+
one of the following groups, all of which can occur:" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }{TEXT 311 33 "Cn,   where 1 <= n <= 10 or n=12." }}
{PARA 0 "" 0 "" {TEXT -1 2 "or" }}{PARA 0 "" 0 "" {TEXT 312 32 "Cn*C2 \+
 where n is 2, 4, 6, or 8." }}{PARA 0 "" 0 "" {TEXT -1 121 "That all o
f these groups occur can be shown by examples, but that other groups s
uch as C11 do not occur is hard to prove." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "The theorem makes it very easy to \+
check if a given point P is a torsion point or not. Just compute:" }}
{PARA 0 "" 0 "" {TEXT -1 57 "P, 2*P, 3*P, 4*P, 5*P, 6*P, 7*P, 8*P, 9*P
, 10*P, and 12*P" }}{PARA 0 "" 0 "" {TEXT -1 50 "and check if one of t
hem is the point at infinity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 299 "n_times_point:=proc(n,P,f,x
,y)\n   if n=0 then\n      [0,1,0]\n   elif n<0 then\n      n_times_po
int(-n, [P[1],-P[2]], f,x,y)\n   elif n=1 then\n      P\n   elif type(
n,even) then\n      n_times_point(n/2, addPoints(P,P,f,x,y), f,x,y)\n \+
  else\n      addPoints(P, n_times_point(n-1,P,f,x,y), f,x,y)\n   fi\n
end;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%.n_times_pointGR6'%\"nG%\"PG
%\"fG%\"xG%\"yG6\"F,F,@+/9$\"\"!7%F0\"\"\"F02F/F0-F$6',$F/!\"\"7$&9%6#
F2,$&F:6#\"\"#F79&9'9(/F/F2F:-%%typeG6$F/%%evenG-F$6',$F/#F2F?-%*addPo
intsG6'F:F:F@FAFBF@FAFB-FM6'F:-F$6',&F/F2F7F2F:F@FAFBF@FAFBF,F,F," }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "f;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,,*$)%\"yG\"\"#\"\"\"F(*$)%\"xG\"\"$F(!\"%*$)F+F'F(\"\"
)F+F-!\"\"F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "P1;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#7$\"\"!\"\"\"" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 29 "Q:=n_times_point(1,P1,f,x,y);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"QG7$\"\"!\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 29 "Q:=n_times_point(2,P1,f,x,y);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"QG7$\"\"$!\"(" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 29 "Q:=n_times_point(3,P1,f,x,y);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"QG7$#\"\"(\"\"*#\"#H\"#F" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 15 "map(ifactor,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#7$*&-%!G6#\"\"(\"\"\"*$)-F&6#\"\"$\"\"#F)!\"\"*&-F&6#\"#HF)*$)F,F.F)F
0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "If P1 were a torsion point, \+
then Q should also be a torsion point." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 216 "However, there is a theorem that say
s that if the polynomial f=y^2-fx and fx has integer coefficients, the
n every torsion point has integer coefficients as well. So Q can not b
e a torsion point, hence neither can P1." }}{PARA 0 "" 0 "" {TEXT -1 
66 "A part of the proof of the theorem consists of the following fact:
" }}{PARA 0 "" 0 "" {TEXT -1 36 "If fx has integer coefficients, and \+
" }{TEXT 320 123 "if a prime number p divides the denominator of the y
-coordinate of Q, then p also divides the denominator of 2*Q, 3*Q, 4*Q
," }{TEXT -1 135 " etc. Note that to prove the theorem we need to do m
ore than just prove this fact, but we won't do that. Lets check these \+
denominators:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "n_times_po
int(2,Q,f,x,y);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$#\"%SR\"%pv#!'P2!
)\"'.&e'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "map(ifactor,%);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$*&*()-%!G6#\"\"#F*\"\"\"-F(6#\"
\"&F+-F(6#\"$(>F+F+*&)-F(6#\"\"$F*F+)-F(6#\"#HF*F+!\"\",$*&*(-F(6#\"\"
(F+-F(6#\"#tF+-F(6#\"%n:F+F+*&)F4F6F+)F8F6F+F;F;" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 25 "n_times_point(3,Q,f,x,y);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#7$#\"*![b**p\")\"G<T$#!/\"G<*)GK_$\"-@$Q!z#*>" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "map(ifactor,%);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#7$*&*.)-%!G6#\"\"#\"\"$\"\"\"-F(6#\"\"&F,-F(6#
\"\"(F,-F(6#\"#hF,-F(6#\"$2\"F,-F(6#\"$$QF,F,*()-F(6#F+\"\"%F,)-F(6#\"
#6F*F,)-F(6#\"#fF*F,!\"\",$*&*&-F(6#\"#HF,-F(6#\".*Q9l!\\@\"F,F,*()F>
\"\"'F,)FBF+F,)FFF+F,FIFI" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
25 "n_times_point(4,Q,f,x,y);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$#\"
1V7gr6n\"*[\"1hh/w%*3`[#\"9ZK!e=V#\\ZFH\"Q$\"9frR4u#px-i3Q$" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "map(ifactor,%);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#7$*&*0-%!G6#\"#>\"\"\"-F'6#\"#VF*-F'6#\"#ZF*-F'
6#\"#zF*-F'6#\"$\">F*-F'6#\"%R9F*-F'6#\"%neF*F**,)-F'6#\"\"$\"\"#F*)-F
'6#\"\"(FBF*)-F'6#\"#HFBF*)-F'6#\"#tFBF*)-F'6#\"%n:FBF*!\"\"*&*,-F'6#
\"$$QF*-F'6#\"(`I(>F*-F'6#\"%.DF*-F'6#\"(\"G4[F*-F'6#\"&rr$F*F**,)F?FA
F*)FDFAF*)FHFAF*)FLFAF*)FPFAF*FS" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
213 "Indeed, we see that we're not getting rid of the factor 3 in the \+
denominator. Why is it so that if we have a prime p (in the example p=
3) in the denominator of Q, that then all multiples of Q also have tha
t prime?" }}{PARA 0 "" 0 "" {TEXT -1 603 "There is a simple explanatio
n: If Q has 3 in the denominator, then \"modulo 3\" the point Q is inf
inity. Just like if you have a function with x in the denominator, the
n modulo x (that means: take x=0) the function goes to infinity. So mo
dulo 3, Q is infinity, which is the zero element of the group. In othe
r words, when we compute Q, 2*Q, 3*Q, ... then modulo 3 we are just co
mputing the zero element of the group, 2 * the zero element, 3 * the z
ero element, and of course this stays the zero element, i.e. the point
 at infinity, so modulo 3 it stays infinity, in other words 3 stays in
 the denominator." }}{PARA 0 "" 0 "" {TEXT -1 212 "This makes the abov
e given fact plausible but it is not an actual proof because we ignore
d if the point and curve had \"good reduction modulo p\", and we'll no
t complete the rest of the proof of the theorem either." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "fx;" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#,**$)%\"xG\"\"$\"\"\"\"\"%*$)F&\"\"#F(!\")F&F)F(F(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "discrim(%,x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#!$)o" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "ifac
tor(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&)-%!G6#\"\"#\"\"%\"\"\"
-F'6#\"#VF+!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 486 "There exists
 another theorem that says that the square of the y-coordinate of a to
rsion point is either 0 or it must divide the discriminant. So the y-c
oordinate in this example must be either 0, -2 or 2, and one quickly f
inds this way that there are no torsion points. For example, the point
 [2,3] does not satisfy this condition and hence it can not be a torsi
on point. By taking a multiple of [2,3] that has non-integer entries w
e can also show that [2,3] can not be a torsion point." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "P:=[2,3];" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"PG7$\"\"#\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "n_times_point(2,P,f,x,y);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#7$#\"\"(\"\"*#\"#H\"#F" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 137 "If P is a torsion point then so is 2*P, but 2*P is not a
 torsion point because the entries are not integers. So P is not a tor
sion point." }}{PARA 0 "" 0 "" {TEXT -1 60 "So this curve has no non-t
rivial torsion points. That means:" }}{PARA 0 "" 0 "" {TEXT -1 73 "For
 every rational point P on the curve (except P=infinity) we have that:
" }}{PARA 0 "" 0 "" {TEXT -1 16 "P, 2*P, 3*P, ..." }}{PARA 0 "" 0 "" 
{TEXT -1 46 "are all distinct and none of them is infinity." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 344 22 "Mordell-Weil T
heorem: " }{TEXT -1 72 "The group of rational points on an elliptic cu
rve is finitely generated." }}{PARA 0 "" 0 "" {TEXT -1 79 "Since the g
roup is commutative this means that the group is isomorphic to:\nF * \+
" }{TEXT 321 2 "Z " }{TEXT -1 2 "* " }{TEXT 322 1 "Z" }{TEXT -1 3 " * \+
" }{TEXT 323 1 "Z" }{TEXT -1 10 " * .... * " }{TEXT 324 3 "Z =" }
{TEXT -1 5 " F * " }{TEXT 326 1 "Z" }{TEXT -1 2 "^s" }{TEXT 325 1 " " 
}}{PARA 0 "" 0 "" {TEXT -1 68 "where F is a finite group (the group of
 torsion points). The number " }{TEXT 327 3 "s, " }{TEXT -1 66 "the nu
mber of independent points of infinite order, is called the " }{TEXT 
345 4 "rank" }{TEXT -1 228 " of the elliptic curve. In our example F =
 \{1\}, and s is at least 1 because we already found points with infin
ite order. We can't be sure if the rank is 1 or higher than 1 because \+
all the points we've found so far are dependent:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 9 "P:=[1,1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%\"PG7$\"\"\"F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "seq(n_ti
mes_point(i,P,f,x,y),i=0..10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6-7%\"
\"!\"\"\"F$7$F%F%7$F$!\"\"7$\"\"#!\"$7$\"\"$\"\"(7$#F%\"\"%#\"\"&F17$#
F.\"\"*#!#H\"#F7$\"#A!$(>7$#\"#g\"#\\#\"$$Q\"$V$7$#!#>\"$@\"#!$J&\"%J8
7$#\"$4'\"$+%#!%Fl\"%+S" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "Lets f
irst see if we can find more integer points:" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 77 "for i from -1000 to 1000 do if not irreduc(subs(
x=i,f)) then lprint(i) fi od;" }}{PARA 6 "" 1 "" {TEXT -1 1 "0" }}
{PARA 6 "" 1 "" {TEXT -1 1 "1" }}{PARA 6 "" 1 "" {TEXT -1 1 "2" }}
{PARA 6 "" 1 "" {TEXT -1 1 "3" }}{PARA 6 "" 1 "" {TEXT -1 2 "22" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "OK, so if there are any more integ
er points, the x-coordinate needs to have absolute value > 1000." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "How about
 rational points? Well, we can search but we don't know if we find eve
rything:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 120 "S:=\{\}; for j
 from 1 to 11 do for i from -j^2 to 500 do if not irreduc(subs(x=i/j^2
,f)) then S:=S union \{i/j^2\} fi od od;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"SG<\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "S;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#<+#\"\"\"\"\"%\"\"!F%\"\"#\"\"$\"#A#\"
#g\"#\\#\"\"(\"\"*#!#>\"$@\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 195 "
We've searched denominators 1^2, 2^2, 3^2,...,11^2 with some pretty la
rge numerators and still didn't find any new points. So if there are m
ore rational points (other than the ones n*P with n in " }{TEXT 328 1 
"Z" }{TEXT -1 139 ") then they can't be very easy points, they must ei
ther have a not-so-small denominator or a not-so-small numerator (in o
ther words: their " }{TEXT 329 6 "height" }{TEXT -1 15 " is not small)
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "In g
eneral, given a positive number B, " }{TEXT 346 63 "there are only fin
itely many rational numbers with  height <= B" }{TEXT -1 179 ", so if \+
we check for all of those rational numbers if they are the x-coordinat
e of a rational point then we can decide for sure if we have found all
 points with height <=B or not." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 370 "So: if we could prove some bound on the \+
height, a bound that tells us how far we need to search in order to be
 sure that we have found a basis for all rational points, then the pro
blem of computing the rank would be decidable. Computing all torsion p
oints is very easy, but for the non-torsion points it would be nice to
 have an algorithm that can compute the following:" }}{PARA 0 "" 0 "" 
{TEXT -1 36 "1) a basis of all non-torsion points" }}{PARA 0 "" 0 "" 
{TEXT -1 37 "2) a proof that there aren't any more" }}{PARA 0 "" 0 "" 
{TEXT -1 198 "A complete algorithm that always gives you 1 and 2 does \+
not exist (so the bound we would like to have does not exist either). \+
So we can not for every elliptic curve decide for sure what its rank i
s." }}{PARA 0 "" 0 "" {TEXT -1 477 "However, there are very clever alg
orithms that work in most cases. That these algorithms do not always w
ork is because the computation of a proof that there there are no addi
tional (not-yet found) rational points does not always need to termina
te. If it terminates, then we can be sure what the rank is, but if it \+
doesn't terminate then we may have found some points but we only have \+
obtained some lower bound for the rank, and can't prove that the lower
 bound equals the rank." }}{PARA 0 "" 0 "" {TEXT -1 376 "Most likely t
hose algorithms would have no troubles with the curve we were computin
g with, and they would give a basis for the rational points and constr
uct a proof that the rank is not higher than the number of elements in
 this basis. I have not actually ran the algorithm but I think it is l
ikely that the rank of the above curve is 1, and that a basis is \{P\}
 where P=[1,1]." }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 35 "Elliptic cur
ves over finite fields." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 28 "Let p be a prime number and " }{TEXT 347 2 "Fp" }
{TEXT -1 27 " the field with p elements." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "For those of you not familiar wit
h finite fields, here is a table for \"+\" and \"*\" in the finite fie
ld " }{TEXT 348 2 "F7" }{TEXT -1 201 " with 7 elements. The elements o
f Fp are represented by 0,1,...,p-1, and we compute modulo p, so that \+
0 is the same as p, and 1 is the same as 1+p, and A is the same as A+B
*p for all integers A,B, etc." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 5 "p:=7;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"pG\"\"(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "for j from 0 to p-1 do print(seq(ca
t(i,\"+\",j,\"=\",i+j mod p,\"   \"),i=0..p-1)) od;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6)%)0+0=0~~~G%)1+0=1~~~G%)2+0=2~~~G%)3+0=3~~~G%)4+0=4~~
~G%)5+0=5~~~G%)6+0=6~~~G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6)%)0+1=1~~~
G%)1+1=2~~~G%)2+1=3~~~G%)3+1=4~~~G%)4+1=5~~~G%)5+1=6~~~G%)6+1=0~~~G" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6)%)0+2=2~~~G%)1+2=3~~~G%)2+2=4~~~G%)3+
2=5~~~G%)4+2=6~~~G%)5+2=0~~~G%)6+2=1~~~G" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6)%)0+3=3~~~G%)1+3=4~~~G%)2+3=5~~~G%)3+3=6~~~G%)4+3=0~~~G%)5+3=1~~
~G%)6+3=2~~~G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6)%)0+4=4~~~G%)1+4=5~~~
G%)2+4=6~~~G%)3+4=0~~~G%)4+4=1~~~G%)5+4=2~~~G%)6+4=3~~~G" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6)%)0+5=5~~~G%)1+5=6~~~G%)2+5=0~~~G%)3+5=1~~~G%)4+
5=2~~~G%)5+5=3~~~G%)6+5=4~~~G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6)%)0+6
=6~~~G%)1+6=0~~~G%)2+6=1~~~G%)3+6=2~~~G%)4+6=3~~~G%)5+6=4~~~G%)6+6=5~~
~G" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "for j from 0 to p-1 d
o print(seq(cat(i,\"*\",j,\"=\",i*j mod p,\"   \"),i=0..p-1)) od;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6)%)0*0=0~~~G%)1*0=0~~~G%)2*0=0~~~G%)3*0
=0~~~G%)4*0=0~~~G%)5*0=0~~~G%)6*0=0~~~G" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6)%)0*1=0~~~G%)1*1=1~~~G%)2*1=2~~~G%)3*1=3~~~G%)4*1=4~~~G%)5*1=5~~
~G%)6*1=6~~~G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6)%)0*2=0~~~G%)1*2=2~~~
G%)2*2=4~~~G%)3*2=6~~~G%)4*2=1~~~G%)5*2=3~~~G%)6*2=5~~~G" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6)%)0*3=0~~~G%)1*3=3~~~G%)2*3=6~~~G%)3*3=2~~~G%)4*
3=5~~~G%)5*3=1~~~G%)6*3=4~~~G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6)%)0*4
=0~~~G%)1*4=4~~~G%)2*4=1~~~G%)3*4=5~~~G%)4*4=2~~~G%)5*4=6~~~G%)6*4=3~~
~G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6)%)0*5=0~~~G%)1*5=5~~~G%)2*5=3~~~
G%)3*5=1~~~G%)4*5=6~~~G%)5*5=4~~~G%)6*5=2~~~G" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6)%)0*6=0~~~G%)1*6=6~~~G%)2*6=5~~~G%)3*6=4~~~G%)4*6=3~~~G
%)5*6=2~~~G%)6*6=1~~~G" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "The tab
les illustrate how to add and multiply in F7." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "Now lets consider an elli
ptic curve over F7." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "f:=y
^2-x*(x-1)*(x-2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG,&*$)%\"yG
\"\"#\"\"\"F**(%\"xGF*,&F,F*!\"\"F*F*,&F,F*!\"#F*F*F." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 1600 "addPoints_Fp:=proc(P1,P2,f,x,y,p)
\n  local su1,su2,L,a,b,c,i,P3;\n  if P1=[0,1,0] then RETURN(P2) fi;\n
  if P2=[0,1,0] then RETURN(P1) fi;\n  if P1[1]=P2[1] and P1[2]=-P2[2]
 mod p then\n      # P1 and P2 are on a vertical line, sum=0\n      RE
TURN([0,1,0])\n  fi;\n  su1:=x=P1[1], y=P1[2]:\n  su2:=x=P2[1], y=P2[2
]:\n  if P1=P2 then\n      # If P1=P2 then we need to take a tangent l
ine.\n      # The following line has the correct angle:\n      L:=subs
(su1,diff(f,x))*x + subs(su1,diff(f,y))*y + c mod p;\n      # And now \+
it also goes through the point P1:\n      L:=subs(msolve(subs(su1,L)=0
,p),L);\n      # Now compute the intersection points, and\n      # thr
ow away P1:\n      P3:=\{seq(subs(i,[x,y]), i=\{msolve(\{L,f\},p)\})\}
;\n      if nops(P3)=1 then\n         # there is only 1 intersection p
oint with the line\n         # and the curve, so that must be an infle
ction point.\n         P3:=P1\n      else\n         P3:=op(P3 minus \{
P1\})\n      fi\n  else\n      L:=a*x+b*y+c;\n      L:=subs(msolve(\{s
ubs(su1,L),subs(su2,L)\},p),L);\n      L:=subs(_NN1=1,L) mod p;\n     \+
 P3:=\{seq(subs(i,[x,y]), i=\{msolve(\{L,f\},p)\})\};\n      if nops(P
3)=2 then\n         # there are only 2 points on this line, so one\n  \+
       # of the two must be a tangent point.\n         L:=Normal(subs(
solve(\{L\},\{y\}),f)/((x-P1[1])*(x-P2[1]))) mod p;\n         if subs(
x=P1[1],L) mod p=0 then P3:=P1\n         elif subs(x=P2[1],L) mod p=0 \+
then P3:=P2 else ERROR()\n         fi\n      else\n         # there ar
e 3 points on the line:\n         P3:=op(P3 minus \{P1,P2\});\n      f
i\n  fi;\n      # Now put in the -sign at the y-coordinate:\n      [P3
[1],-P3[2]] mod p;\nend:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
340 "n_times_point_Fp:=proc(n,P,f,x,y,p)\n   if n=0 then\n      [0,1,0
]\n   elif n<0 then\n      n_times_point_P(-n, [P[1],-P[2]] mod p, f,x
,y,p)\n   elif n=1 then\n      P mod p\n   elif type(n,even) then\n   \+
   n_times_point_Fp(n/2, addPoints_Fp(P,P,f,x,y,p), f,x,y,p)\n   else
\n      addPoints_Fp(P, n_times_point_Fp(n-1,P,f,x,y,p), f,x,y,p)\n   \+
fi\nend:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "Now lets compute all \+
points over Fp on the curve f." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 80 "points:=[seq(seq(subs(j,[i,y]),j=[msolve(subs(x=i,f),p)]),i=0.
.p-1),  [0,1,0] ];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'pointsG7*7$\"
\"!F'7$\"\"\"F'7$\"\"#F'7$\"\"&F-7$F-F+7$\"\"'F07$F0F)7%F'F)F'" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 153 "for i from 1 to nops(points
) do for j from i to nops(points) do print(cat(points[i],\"+\",points[
j],\"=\",addPoints_Fp(points[i],points[j],f,x,y,p))) od; od;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"!F*Q\"+6\"F)Q\"=F,7%F*\"\"\"F*
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"!F*Q\"+6\"7$\"\"\"F*
Q\"=F,7$\"\"#F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"!F*Q
\"+6\"7$\"\"#F*Q\"=F,7$\"\"\"F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((
((%!G7$\"\"!F*Q\"+6\"7$\"\"&F.Q\"=F,7$\"\"'\"\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#(((((%!G7$\"\"!F*Q\"+6\"7$\"\"&\"\"#Q\"=F,7$\"\"'F2" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"!F*Q\"+6\"7$\"\"'F.Q\"=F
,7$\"\"&\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"!F*Q\"+
6\"7$\"\"'\"\"\"Q\"=F,7$\"\"&F2" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((
((%!G7$\"\"!F*Q\"+6\"7%F*\"\"\"F*Q\"=F,F)" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#(((((%!G7$\"\"\"\"\"!Q\"+6\"F)Q\"=F-7%F+F*F+" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"\"\"\"!Q\"+6\"7$\"\"#F+Q\"=F-7
$F+F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"\"\"\"!Q\"+6\"7
$\"\"&F/Q\"=F-7$\"\"'F2" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$
\"\"\"\"\"!Q\"+6\"7$\"\"&\"\"#Q\"=F-7$\"\"'F*" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#(((((%!G7$\"\"\"\"\"!Q\"+6\"7$\"\"'F/Q\"=F-7$\"\"&F2" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"\"\"\"!Q\"+6\"7$\"\"'F*
Q\"=F-7$\"\"&\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"\"
\"\"!Q\"+6\"7%F+F*F+Q\"=F-F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%
!G7$\"\"#\"\"!Q\"+6\"F)Q\"=F-7%F+\"\"\"F+" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#(((((%!G7$\"\"#\"\"!Q\"+6\"7$\"\"&F/Q\"=F-7$F/F*" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"#\"\"!Q\"+6\"7$\"\"&F*Q
\"=F-7$F/F/" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"#\"\"!Q\"
+6\"7$\"\"'F/Q\"=F-7$F/\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((
%!G7$\"\"#\"\"!Q\"+6\"7$\"\"'\"\"\"Q\"=F-7$F/F/" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#(((((%!G7$\"\"#\"\"!Q\"+6\"7%F+\"\"\"F+Q\"=F-F)" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"&F*Q\"+6\"F)Q\"=F,7$\"\"
#\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"&F*Q\"+6\"7$F*
\"\"#Q\"=F,7%\"\"!\"\"\"F1" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G
7$\"\"&F*Q\"+6\"7$\"\"'F.Q\"=F,7$\"\"!F1" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#(((((%!G7$\"\"&F*Q\"+6\"7$\"\"'\"\"\"Q\"=F,7$F/\"\"!" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"&F*Q\"+6\"7%\"\"!\"\"\"F.Q\"=F
,F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"&\"\"#Q\"+6\"F)Q
\"=F-7$F+\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"&\"\"#
Q\"+6\"7$\"\"'F/Q\"=F-7$\"\"\"\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#(((((%!G7$\"\"&\"\"#Q\"+6\"7$\"\"'\"\"\"Q\"=F-7$\"\"!F3" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"&\"\"#Q\"+6\"7%\"\"!\"\"\"F/Q\"=F-
F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"'F*Q\"+6\"F)Q\"=F,
7$\"\"#\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"'F*Q\"+6
\"7$F*\"\"\"Q\"=F,7%\"\"!F.F1" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((
%!G7$\"\"'F*Q\"+6\"7%\"\"!\"\"\"F.Q\"=F,F)" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#(((((%!G7$\"\"'\"\"\"Q\"+6\"F)Q\"=F-7$\"\"#\"\"!" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7$\"\"'\"\"\"Q\"+6\"7%\"\"!F+F
/Q\"=F-F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((((%!G7%\"\"!\"\"\"F*Q
\"+6\"F)Q\"=F-F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 131 "There are 4 \+
points of order 4, 3 points of order 2, and one point of order 1, so t
he group of points must be isomorphic to C2 * C4." }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 7 "p:=101;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
\"pG\"$,\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "points:=[seq(
seq(subs(j,[i,y]),j=[msolve(subs(x=i,f),p)]),i=0..p-1),  [0,1,0] ];" }
}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%'pointsG7dq7$\"\"!F'7$\"\"\"F'7$\"
\"#F'7$\"\"$\"#R7$F-\"#i7$\"\"%\"#B7$F2\"#y7$\"\"'\"#D7$F7\"#w7$\"\")
\"#s7$F<\"#H7$\"\"*\"#57$FA\"#\"*7$FB\"#m7$FB\"#N7$\"#6FA7$FJ\"##*7$\"
#8FB7$FNFD7$\"#@\"$+\"7$FQF)7$\"#A\"\"(7$FU\"#%*7$F3\"#$)7$F3\"#=7$\"#
CF37$FhnF57$F8F<7$F8\"#$*7$\"#F\"#t7$F^o\"#G7$\"#IF87$FcoF:7$\"#MFV7$F
foFX7$FH\"#`7$FH\"#[7$\"#OF27$F]p\"#(*7$F.\"#r7$F.Fco7$\"#VFF7$FdpFH7$
\"#W\"#_7$Fgp\"#\\7$\"#Y\"#o7$F\\q\"#L7$F[pFV7$F[pFX7$\"#]F77$Fcq\"#&*
7$\"#^FF7$FgqFH7$Fhp\"#a7$Fhp\"#Z7$Fio\"#T7$Fio\"#g7$\"#b\"#q7$Fbr\"#J
7$\"#d\"#u7$FgrF^o7$\"#f\"#')7$F[s\"#:7$F`rFjq7$F`rF\\r7$\"#k\"#)*7$Fb
sF-7$\"#n\"#S7$Ffs\"#h7$F]qF87$F]qF:7$\"#pFcr7$F]tFer7$F_oFio7$F_oF[p7
$F:F37$F:F57$F5FQ7$F5\"#!)7$\"#zF_o7$FgtFao7$FetFU7$FetFgt7$\"#\")Fcr7
$F\\uFer7$\"##)FB7$F_uFD7$\"#!*FR7$FbuF)7$FLFJ7$FLFbu7$F\\oFjq7$F\\oF
\\r7$FXFR7$FXF)7$Feq\"#))7$FeqFN7$F_pFio7$F_pF[p7$\"#**F_o7$F`vFao7$FR
\"#()7$FR\"#97%F'F)F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "no
ps(points);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"$/\"" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 87 "Now we have a commutative group with 104 \+
elements. There are only a few of such groups." }}{PARA 0 "" 0 "" 
{TEXT -1 254 "Like before, the group can not be cyclic because we have
 4 elements of order <=2 (3 elements of order 2 and one of order 1). S
o the next guess would be that the group is C2 * C52. If there exists \+
an element of order 52 then that must be indeed the group." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "ifactor(52);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#*&)-%!G6#\"\"#F(\"\"\"-F&6#\"#8F)" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 13 "P:=points[4];" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"PG7$\"\"$\"#R" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "
n_times_point_Fp(52/13,P,f,x,y,p);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
7$\"#C\"#y" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "n_times_point
_Fp(52/2,P,f,x,y,p);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$\"\"\"\"\"!
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "So 52/2 * P is not zero, nor \+
is 52/13, so P must have order 52, so the group is C2*C52." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "fx := (x^3-x+2);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%#fxG,(*$)%\"xG\"\"$\"\"\"F*F(!\"\"\"\"#F*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "f:=y^2 - fx;" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"fG,**$)%\"yG\"\"#\"\"\"F**$)%\"xG\"\"$F*!\"\"F
-F*!\"#F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "p:=71;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"pG\"#r" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 80 "points:=[seq(seq(subs(j,[i,y]),j=[msolve(subs(x=i,f
),p)]),i=0..p-1),  [0,1,0] ];" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%'po
intsG7[o7$\"\"!\"#f7$F'\"#77$\"\"\"F(7$F,F*7$\"\"#\"#C7$F/\"#Z7$\"\"(
\"#d7$F4\"#97$\"\")\"#o7$F9\"\"$7$\"\"*\"#c7$F>\"#:7$FA\"#m7$FA\"\"&7$
\"#?\"#K7$FG\"#R7$\"#@F?7$FLFA7$\"#AF47$FO\"#k7$\"#B\"#a7$FS\"#<7$\"#E
FL7$FX\"#]7$\"#GFO7$Ffn\"#\\7$\"#HF:7$FjnF<7$\"#MF:7$F]oF<7$\"#N\"#57$
F`o\"#h7$\"#QFeo7$Feo\"#L7$\"#TF?7$FioFA7$\"#X\"#U7$F\\pFjn7$\"#^F`p7$
F`pFG7$\"#_F47$FcpFQ7$\"#`F>7$Ffp\"#i7$FTFT7$FTFV7$\"#bFfp7$F\\q\"#=7$
F?\"#67$F?\"#g7$\"#e\"#W7$Fdq\"#F7$Fco\"#>7$FcoFcp7$\"#lFT7$F\\rFV7$FC
Feo7$FCFgo7$F:F47$F:FQ7$\"#qF(7$FcrF*7%F'F,F'" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 13 "nops(points);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#\"#j" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "ifactor(%);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#*&)-%!G6#\"\"$\"\"#\"\"\"-F&6#\"\"(F*
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "P:=points[9];" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"PG7$\"\")\"#o" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 33 "n_times_point_Fp(63/3,P,f,x,y,p);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#7$\"\"#\"#C" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 33 "n_times_point_Fp(63/7,P,f,x,y,p);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#7$\"#q\"#f" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 269 "So \+
P is a point in a group with 63 elements, so the order of P divides 63
, however, 63/3 * P is not zero, nor is 63/7 * P, so the order of P do
es not divide 63/3 nor 63/7, so the order of P can not be smaller than
 63. Hence the order of P is 63. So the group is cyclic." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "How many points do
es an elliptic curve over Fp have? So far we've seen examples with:" }
}{PARA 0 "" 0 "" {TEXT -1 17 "p=7 and 8 points." }}{PARA 0 "" 0 "" 
{TEXT -1 21 "p=101 and 104 points." }}{PARA 0 "" 0 "" {TEXT -1 19 "p=7
1 and 63 points." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 104 "So the number of points seems pretty close to p. Lets es
timate the number of points in a simplistic way:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 3 "fx;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*$)%
\"xG\"\"$\"\"\"F(F&!\"\"\"\"#F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 18 "i:=rand(0..p-1)();" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"iG\"
#J" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "K := subs(x=i,fx) mod
 p;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"KG\"#8" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 367 "For this random chosen x-value  x=rand(0..p-1)(),
 how many y-values do we expect? Well, f = y^2-fx, so y^2 should be eq
ual to K. Consider the following two cases:\nCase 1. K=0 in Fp. Then w
e get 1 solution for y.\nCase 2. K is not 0 in Fp.\n   Case 2a. K is a
 square in Fp  ->  we find 2 solutions for y.\n   Case 2b. K is not a \+
square in Fp -> we find 0 solutions for y." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 349 "If p is odd, then precisely 50
% of the non-zero elements of Fp are squares, and 50% are not squares.
 So in Case 2 it is reasonable to say that with 50% probability we get
 2 points and with 50% probability we get 1 point. So the expectation \+
value for the number of points in Case 2 equals 1 point, and also in C
ase 1 the expectation value is 1 point." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 53 "So the expectation value for the num
ber of points is:" }}{PARA 0 "" 0 "" {TEXT -1 65 "1 point for each val
ue of x in Fp, plus 1 more point at infinity." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "So, " }{TEXT 349 86 "the e
xpectation value for the number of points on an elliptic curve over Fp
 equals p+1" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 320 "But, as we have seen in the examples, it doesn
't always have to be equal to p+1. We could think that we essentially \+
throw a coin p+1 times, and each time we get either 2 or 0 points (or \+
sometimes 1 point, but that can happen at most 4 times, for the roots \+
of fx (fx has 0, 1 or 3 roots in Fp) and the point at infinity)." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 253 "If you t
hrow a coin 10000 times, and you get 2 dollars every time it falls on \+
head, 0 dollars for tail, your expectation value is: 10000. Furthermor
e, it is not likely that the actual amount is much different from 1000
0, here is a list of probabilities:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 263 "abs(actual amount minus 10000) <= 100 (
about 68.75% chance)\nabs(actual amount minus 10000) <= 200 (about 95.
56% chance)\nabs(actual amount minus 10000) <= 300 (about 99.74% chanc
e)\nabs(actual amount minus 10000) <= 1000 (about 99.99999999999999999
9998733% chance)\n" }}{PARA 0 "" 0 "" {TEXT -1 236 "That is why the fo
llowing theorem is at first sight not surprising, but at second sight \+
very surprising. As always, the elliptic curve is given by a polynomia
l \nf=y^2-fx where fx is a polynomial in x of degree 3 with no multipl
e roots.\n" }}{PARA 258 "" 0 "" {TEXT -1 151 "Theorem (Hasse, Weil): I
f p is the number of elements of a finite field F, and the number of p
oints of the curve is N, then: abs(p+1 - N) <= 2*sqrt(p)." }}{PARA 0 "
" 0 "" {TEXT -1 94 "Note: for arbitrary non-singular curves, the state
ment is <=2*g*sqrt(p), where g is the genus." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 253 "The theorem says that the numb
er of points is not too far away from the expectation value p+1, at fi
rst sight that's what we expected so that's not surprising. However, w
ith the coins game it is \"only\" probable to be close to the expectat
ion value, and " }{TEXT 350 11 "not certain" }{TEXT -1 473 ". With a v
ery small, but non-zero, probability we can be far away from the expec
tation value with the coins game; we are off by more than 1000$ in the
 coins game with 0.1267*10^(-20) procent probability, which is small b
ut greater than zero. However, the theorem says that for elliptic curv
es we can never be far away from the expectation value, not even with \+
0.1267*10^(-20) procent probability, and certainly not with the 100-95
.56=4.44% probability as in the coins game." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 245 "The probility distributions (a
lthough they clearly must be different because with the curves the pro
bability drops to zero if you stray too far from q+1) look a bit alike
, like a bell shaped curve. Closer to q+1 is more likely then further \+
away." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 257 
"Note that if you have one curve: y^2-fx that has q+1+k points, so abs
(k)<=2*sqrt(q), then you can easily make another curve with q+1-k poin
ts. How? Just take an element c in the field that is not a square, and
 consider the curve: y^2-c*fx. This is called a " }{TEXT 351 5 "twist
" }{TEXT -1 142 " of the curve. It is a curve with the same j-invarian
t, so over the complex numbers it is the same curve, but over Fp it's \+
not the same curve." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "fx;" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*$)%\"xG\"\"$\"\"\"F(F&!\"\"\"\"#F
(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "f;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,**$)%\"yG\"\"#\"\"\"F(*$)%\"xG\"\"$F(!\"\"F+F(!\"#F(" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "p;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"#r" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "k:=n
ops(points) - (p+1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG!\"*" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "So we have a curve with p+1+k=71+
1+(-9)=63 points. Lets now make a curve with p+1-k=81 points." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "Factor(y^2+1) mod p;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*$)%\"yG\"\"#\"\"\"F(F(F(" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 97 "So -1 is not a square in Fp (p sho
uld be congruent to 1 modulo 4 in order for -1 to be a square)." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "f:=y^2 - (-1)*fx mod p;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG,**$)%\"yG\"\"#\"\"\"F**$)%\"xG
\"\"$F*F*F-\"#qF)F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "poin
ts:=[seq(seq(subs(j,[i,y]),j=[msolve(subs(x=i,f),p)]),i=0..p-1),  [0,1
,0] ];" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%'pointsG7]p7$\"\"$\"#^7$F'
\"#?7$\"\"%\"#o7$F,F'7$\"\"&\"#P7$F0\"#M7$\"\"'\"#q7$F5\"\"\"7$\"#5\"#
f7$F:\"#77$\"#6\"#e7$F?\"#87$F=\"#D7$F=\"#Y7$FB\"#@7$FB\"#]7$\"#9\"#E7
$FL\"#X7$\"#;F57$FQ\"#l7$\"#<F67$FUF87$\"#=\"#>7$FX\"#_7$FYF(7$FYF*7$
\"#C\"#b7$FinFQ7$FD\"#O7$FD\"#N7$\"#FF-7$FaoF'7$\"#IFin7$Fdo\"#Z7$\"#J
\"#A7$Fho\"#\\7$\"#K\"\")7$F]p\"#j7$\"#LF17$FbpF37$F]o\"#m7$F]oF07$F1
\"#a7$F1FU7$\"#R\"#V7$F[q\"#G7$\"#SF-7$F`qF'7$\"#UFhp7$FcqFU7$F\\q\"#d
7$F\\qFL7$\"#WFio7$FiqF[p7$FF\"\"(7$FF\"#k7$Ffo\"#Q7$FfoFbp7$\"#[F67$F
crF87$F[pF(7$F[pF*7$FJFin7$FJFfo7$FfqF]p7$FfqF[q7$F;\"\"*7$F;\"#i7$\"#
g\"#`7$F`sFX7$F^sFin7$F^sFfo7$F`pFhp7$F`pFU7$F^rF?7$F^rF`s7$\"#nFio7$F
jsF[p7$\"#pF]t7$F]t\"\"#7%\"\"!F8Fat" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 13 "nops(points);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#
\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "ifactor(%);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#*$)-%!G6#\"\"$\"\"%\"\"\"" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 64 "Lets see, is this a cyclic group and if s
o, what is a generator?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "
P:=points[1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG7$\"\"$\"#^" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "n_times_point_Fp(81/3,P,f,x
,y,p);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$\"#Y\"#k" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 67 "OK, not the zero element, so the group is gener
ated by the point P." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
}}}{MARK "1 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 
1 2 33 1 1 }