{VERSION 6 0 "SUN SPARC SOLARIS" "6.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 276 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 281 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 282 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 283 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 285 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 286 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 287 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 288 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 290 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 291 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 292 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 293 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 294 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 295 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 296 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 297 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 298 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 299 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Outpu t" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 12 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 39 "Short vectors in a lattic e, an example." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 45 "The following is a result from number theory:" }}{PARA 0 "" 0 "" {TEXT -1 132 "Theorem. Let p be a prime number. Then there exi st integers a,b such that p = a^2+b^2 if and only if p is congruent to 1 or 2 mod 4." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "So now lets take such a prime number." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "restart; p:=nextprime(10^50); while irem(p, 4)<>1 do p:=nextprime(p) od;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"pG \"T^,+++++++++++++++++++++++\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\" pG\"TZ/+++++++++++++++++++++++\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% \"pG\"Tx0+++++++++++++++++++++++\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "How are we going to find a,b such that a^2+b^2 = p????" }} {PARA 0 "" 0 "" {TEXT -1 39 "After all, the theorem says they exist." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 116 "Lets l ook at this problem in a different way. Think of a,b as elements of F p. Then the equation a^2+b^2=p becomes:" }}{PARA 0 "" 0 "" {TEXT -1 16 " a^2 + b^2 = 0" }}{PARA 0 "" 0 "" {TEXT -1 72 "Suppose b is not \+ 0. Then we can divide the equation by b^2, and we get:" }}{PARA 0 "" 0 "" {TEXT -1 15 " (a/b)^2 = -1." }}{PARA 0 "" 0 "" {TEXT -1 95 "Deno te r = a/b. How can we find r? Well, r is a solution of the followin g polynomial in Fp[x]" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "f \+ := x^2 + 1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG,&*$)%\"xG\"\"#\" \"\"F*F*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "To find the soluti ons of this polynomial, all we have to do is to factor it. We now know how Maple does that." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "Fa ctor(f) mod p;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#*&,&%\"xG\"\"\"\"Sc' G/lS0j_l`(=aa>TRo8FSg_UvF&F&,&F%F&\"S@>d\\$f%ptWjC\"ea/)egJ'G(fRZdCF&F &" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "Lets take one of the two fac tors, and let r be the solution of that factor:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 56 "f1 := op(1,%): r := -coeff(f1,x,0)/coeff(f1,x ,1) mod p;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG\"S@>d\\$f%ptWjC\" ea/)egJ'G(fRZdC" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "Of course the other solution is -r. Because (-r)^2 = r^2, so if r^2 = -1 in Fp, t hen so is (-r)^2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "Do we now know a,b? No, we don't. All we know is that a/ b = r or a/b = -r." }}{PARA 0 "" 0 "" {TEXT -1 182 "We don't really care about minus signs, after all, we want the value of a^2+b^2 to be p, and when you take a square then the +/- sign doesn't matter. So we might as well assume that:" }}{PARA 0 "" 0 "" {TEXT -1 32 " a/b = r (this is in Fp)." }}{PARA 0 "" 0 "" {TEXT -1 15 "In other words: " }}{PARA 0 "" 0 "" {TEXT -1 37 " a/b congruent to r mod p. " }}{PARA 0 "" 0 "" {TEXT -1 15 "In other words:" }}{PARA 0 "" 0 "" {TEXT -1 37 " a - b*r congruent to 0 mod p." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 98 "If we took some arbitrary a,b such that a-b*r is 0 mod p, then we will have: irem(a^2+b^2,p) = 0." }}{PARA 0 "" 0 "" {TEXT -1 86 "But this does not imply that a^+b^ 2 = p, after all, it is possible that a^2+b^2 > p." }}{PARA 0 "" 0 " " {TEXT -1 62 "To make sure that a^2 + b^2 is not >p, what we need to \+ do is: " }}{PARA 0 "" 0 "" {TEXT -1 6 " " }{TEXT 257 66 "Take a s hortest nonzero vector (a,b) such that: a-b*r is 0 mod p." }}{PARA 0 "" 0 "" {TEXT -1 40 "That is an example of a lattice problem." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 11 "Definiti on:" }}{PARA 0 "" 0 "" {TEXT -1 17 "Consider the set " }{TEXT 259 1 "Z " }{TEXT -1 109 "^n, the set of all vectors with n integer entries. Th is is an Abelian group under addition. Any subgroup of " }{TEXT 260 1 "Z" }{TEXT -1 15 "^n is called a " }{TEXT 276 7 "lattice" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "S o a lattice is a subset L of " }{TEXT 261 1 "Z" }{TEXT -1 13 "^n such that:" }}{PARA 0 "" 0 "" {TEXT -1 20 "1) L is not empty." }}{PARA 0 "" 0 "" {TEXT -1 8 "2) If " }{TEXT 266 1 "u" }{TEXT -1 12 " in L t hen -" }{TEXT 267 1 "u" }{TEXT -1 14 " is also in L." }}{PARA 0 "" 0 " " {TEXT -1 8 "3) If " }{TEXT 268 3 "u,v" }{TEXT -1 11 " in L then " }{TEXT 269 3 "u+v" }{TEXT -1 6 " in L." }}{PARA 0 "" 0 "" {TEXT -1 53 "Note that these three conditions automatically imply:" }}{PARA 0 "" 0 "" {TEXT -1 6 "*) If " }{TEXT 263 1 "n" }{TEXT -1 4 " in " }{TEXT 262 2 "Z " }{TEXT -1 4 "and " }{TEXT 264 2 "v " }{TEXT -1 11 "in L the n " }{TEXT 265 3 "n v" }{TEXT -1 32 " is also in L. In other words: " }}{PARA 0 "" 0 "" {TEXT -1 10 "*) L is a " }{TEXT 270 1 "Z" }{TEXT -1 8 "-module." }}{PARA 0 "" 0 "" {TEXT -1 43 "Note that the phrases \+ \"abelian group\" and \"" }{TEXT 271 1 "Z" }{TEXT -1 31 "-module\" mea n exactly the same." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT 272 11 "Definition:" }}{PARA 0 "" 0 "" {TEXT -1 24 "For every \+ subgroup L of " }{TEXT 273 1 "Z" }{TEXT -1 77 "^n there exists a uniqu e integer r such that L is isomorphic (as a group) to " }{TEXT 274 1 " Z" }{TEXT -1 33 "^r. That number r is called the " }{TEXT 275 4 "rank " }{TEXT 277 1 " " }{TEXT -1 5 "of L." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 21 "If L is a lattice in " }{TEXT 279 1 " Z" }{TEXT -1 27 "^n then L is a subgroup of " }{TEXT 280 1 "Z" }{TEXT -1 108 "^n. Since these groups are commutative, it is also a normal s ubgroup. Hence we can form the quotient group:" }}{PARA 0 "" 0 "" {TEXT -1 4 " " }{TEXT 281 4 " Z" }{TEXT -1 7 "^n / L." }}{PARA 0 "" 0 "" {TEXT -1 31 "We have the following property:" }}{PARA 0 "" 0 " " {TEXT -1 6 " " }{TEXT 282 1 "Z" }{TEXT -1 42 "^n / L is finite <=====> rank(L) = n." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT 278 11 "Definition:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "If rank(L) = n then the " }{TEXT 284 5 "index" }{TEXT -1 17 " of lattice L in " }{TEXT 283 1 "Z" }{TEXT -1 32 "^n is the number of elements of " }{TEXT 285 1 "Z" }{TEXT -1 7 "^n / L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 183 "The way you can int erpret this number is as follows (although mathematically this interp retation is a little bit hazy because it is actually not possible to t ake a random element of " }{TEXT 290 1 "Z" }{TEXT -1 79 "^n with unifo rm distribution). Nevertheless, it does give the right intuition." }} {PARA 0 "" 0 "" {TEXT -1 32 "If you took a random element of " }{TEXT 289 3 " Z" }{TEXT -1 64 "^n, then the chance that element would be i n L is 1 / index(L)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT 286 11 "Definition:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "This number, the index of L in " }{TEXT 287 1 "Z" }{TEXT -1 26 "^n, is usually called the " }{TEXT 288 11 "determinant" }{TEXT -1 6 " of L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 2 " A " }{TEXT 292 5 "basis" }{TEXT -1 68 " of L is a set of linearly inde pendent vectors v1..vn for which L = " }{TEXT 293 1 "Z" }{TEXT -1 11 " v1 + .. + " }{TEXT 294 2 "Z " }{TEXT -1 3 "vn." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 291 12 "Proposition:" }}{PARA 0 "" 0 "" {TEXT -1 127 "If v1..vn is a basis for L, and we take the matr ix A = (v1 .. vn) whose columns are v1 .. vn, then the determinant of \+ L equals:" }}{PARA 0 "" 0 "" {TEXT -1 38 " det(L) = absolute value \+ of det(A)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 295 12 "Proposition:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 "If L1 and L2 are lattices of rank n. Then:" }}{PARA 0 "" 0 "" {TEXT -1 68 " L1 = L2 <======> L1 is subset of L2 and det(L1) = det(L2 )." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 191 "Now lets get back to our problem . I wrote that the problem we were looking at was a lattice problem. W hy was that? Well, look again at the problem: we are searching for a v ector v = (a,b) in " }{TEXT 296 1 "Z" }{TEXT -1 26 "^2 that has the p roperty:" }}{PARA 0 "" 0 "" {TEXT -1 24 " a*r - b is 0 mod p." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "What is t he lattice of this problem? Well, it's:" }}{PARA 0 "" 0 "" {TEXT -1 40 " L = \{ (a,b) | a*r - b is 0 mod p\}" }}{PARA 0 "" 0 "" {TEXT -1 117 "It is easy to verify that this is indeed a lattice (all \+ you have to check is the 3 properties 1)+2)+3) listed above)." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 "For every element (a,b) in L we have: a^2+b^2 is divisible by p." }}{PARA 0 "" 0 "" {TEXT -1 125 "We want it to be: equal to p, so: not 0, and not bi gger than p. Now a^2+b^2 is the square of the length of vector v = (a ,b)." }}{PARA 0 "" 0 "" {TEXT -1 76 "So to get a^2+b^2 not just divisi ble by p, but actually equal to p, we need:" }}{PARA 0 "" 0 "" {TEXT -1 45 " v = (a,b) a shortest nonzero vector in L." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 56 "But how do we compute i n L? How do you compute in any " }{TEXT 297 1 "Z" }{TEXT -1 44 "-mod ule? Well, we'll need a basis of course." }}{PARA 0 "" 0 "" {TEXT -1 10 "Lets take:" }}{PARA 0 "" 0 "" {TEXT -1 13 " v1 = (1, r)" }}{PARA 0 "" 0 "" {TEXT -1 13 " v2 = (0, p)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "It is clear that v1,v2 in L. So we hav e found a sublattice L' := " }{TEXT 298 2 "Z " }{TEXT -1 5 "v1 + " } {TEXT 299 1 "Z" }{TEXT -1 9 " v2 of L." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 80 "What is det(L')? Well, the matrix i s triangular, so that's easy: det(L') = p." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "What is det(L)? Look at it th is way: if you took random a,b, what is the chance that a*r-b is 0 mod p?" }}{PARA 0 "" 0 "" {TEXT -1 143 "I'd say that chance is 1/p. So d et(L)=p. But L' is a sublattice with the same determinant. Hence L' \+ = L, and hence v1, v2 form a basis of L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "v1 := [1,r]; v2:=[0 ,p]; basisL := [v1,v2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#v1G7$\" \"\"\"S@>d\\$f%ptWjC\"ea/)egJ'G(fRZdC" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#v2G7$\"\"!\"Tx0+++++++++++++++++++++++\"" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%'basisLG7$7$\"\"\"\"S@>d\\$f%ptWjC\"ea/)egJ'G(fRZdC7$ \"\"!\"Tx0+++++++++++++++++++++++\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "Now our basis elements v1, v2 are quite long, see:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "seq( evalf(sqrt(v[1]^2+v[2]^2)), v \+ = basisL);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"+gRZdC\"#S$\"+++++5\" #T" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "with(IntegerRelations ):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 183 "Note: If you use an older \+ version of Maple that does not contain the package \"IntegerRelations \", then instead of the command LLL use the command: lattice. That wil l do the same thing." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "new _basisL := LLL(basisL);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%+new_basi sLG7$7$!:,/XXUN3QMl5h(!:CaCU;#R(o!*oi['7$F(\":,/XXUN3QMl5h(" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "These basis elements are much shor ter:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "seq( evalf(sqrt(v[1 ]^2+v[2]^2)), v = new_basisL);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"+ ++++5\"#;F#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "short_vector := new_basisL[1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%-short_vectorG 7$!:,/XXUN3QMl5h(!:CaCU;#R(o!*oi['" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "a,b := op(short_vector);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6$%\"aG%\"bG6$!:,/XXUN3QMl5h(!:CaCU;#R(o!*oi['" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "a^2 + b^2 = p;" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#/\"Tx0+++++++++++++++++++++++\"F$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 1 0" 1 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }