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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 257 0 "" }{TEXT 258 62 "Chapter
 17: Using Maple to Solve kth Root Modulo m Congruences" }}{PARA 0 "" 
0 "" {TEXT 256 0 "" }}{PARA 0 "" 0 "" {TEXT -1 114 "Maple can be use t
o save a tremendous amount of time in solving root congruences. We fir
st unassign all variables." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "Maple has a packag
e, " }{TEXT 259 9 "numtheory" }{TEXT -1 126 ", that is designed to do \+
many different types of number theory computations. We enter this pack
age into our package using the " }{TEXT 260 4 "with" }{TEXT -1 11 " st
atement." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "with(numtheory)
;" }}{PARA 7 "" 1 "" {TEXT -1 69 "Warning, the protected name order ha
s been redefined and unprotected\n" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#
7Q%&GIgcdG%)bigomegaG%&cfracG%)cfracpolG%+cyclotomicG%)divisorsG%)fact
orEQG%*factorsetG%'fermatG%)imagunitG%&indexG%/integral_basisG%)invcfr
acG%'invphiG%*issqrfreeG%'jacobiG%*kroneckerG%'lambdaG%)legendreG%)mco
mbineG%)mersenneG%(migcdexG%*minkowskiG%(mipolysG%%mlogG%'mobiusG%&mro
otG%&msqrtG%)nearestpG%*nthconverG%)nthdenomG%)nthnumerG%'nthpowG%&ord
erG%)pdexpandG%$phiG%#piG%*pprimrootG%)primrootG%(quadresG%+rootsunity
G%*safeprimeG%&sigmaG%*sq2factorG%(sum2sqrG%$tauG%%thueG" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 42 "Suppose we desire to solve the congruence
:" }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "x^20075;" "6#*$%\"xG\"&v+#" }
{TEXT -1 23 "= 13973 (mod 43279809)." }}{PARA 0 "" 0 "" {TEXT -1 51 "W
e enter and store the parameters for the exponent " }{TEXT 261 1 "k" }
{TEXT -1 9 ", number " }{TEXT 262 1 "b" }{TEXT -1 14 ", and modulus " 
}{TEXT 263 1 "m" }{TEXT -1 2 ". " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 38 "k := 20075; b := 13973; m := 43279809;" }{TEXT -1 0 "
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"&v+#" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"bG\"&tR\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"mG
\")4)zK%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 112 "Step 1 of the soluti
on process requires us the compute the Euler phi function of the given
 modulus. The command " }{TEXT 264 4 "phi " }{TEXT -1 7 "in the " }
{TEXT 265 9 "numtheory" }{TEXT -1 125 " package computes the Euler phi
 function for a given number. To enter the number of integers relative
ly prime to the modulus " }{TEXT 266 1 "m" }{TEXT -1 18 " = 43279809, \+
enter" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "phi(m);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#\");#R#G" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 179 "The next command uses the gcd command to check that the greate
st common divisor of the necessary parameters in our problem is indeed
 1 to guarantee that the solution method works." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 24 "gcd(b,m); gcd(k,phi(m));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "Step 2 of the solution process req
uires to us to solve " }{TEXT 267 3 "ku " }{TEXT -1 2 "+ " }{XPPEDIT 
18 0 "phi;" "6#%$phiG" }{TEXT -1 1 "(" }{TEXT 269 1 "m" }{TEXT -1 1 ")
" }{TEXT 268 1 "v" }{TEXT -1 12 " = 1, where " }{TEXT 270 1 "u" }
{TEXT -1 5 " and " }{TEXT 271 1 "v" }{TEXT -1 37 " are positive soluti
ons. The command " }{TEXT 273 6 "igcdex" }{TEXT -1 41 " will perform t
he Euclidean algorithm on " }{TEXT 290 1 "k" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 1 "(" }{TEXT 272 1 "m" }
{TEXT -1 65 ") to start the solution process. This command has the par
ameters " }{TEXT 278 6 "igcdex" }{TEXT -1 1 "(" }{TEXT 274 1 "a" }
{TEXT -1 2 ", " }{TEXT 275 1 "b" }{TEXT -1 2 ", " }{TEXT 276 1 "x" }
{TEXT -1 2 ", " }{TEXT 277 1 "y" }{TEXT -1 9 "), where " }{TEXT 285 1 
"a" }{TEXT -1 5 " and " }{TEXT 286 1 "b" }{TEXT -1 86 " are the two nu
mbers we will compute the greatest common divisor of, that is, the gcd
(" }{TEXT 283 1 "a" }{TEXT -1 1 "," }{TEXT 284 1 "b" }{TEXT -1 17 "). \+
The variables " }{TEXT 281 1 "x" }{TEXT -1 5 " and " }{TEXT 282 1 "y" 
}{TEXT -1 135 " are optional parameters (usually stored in single quot
es to ensure variable unassignment) that represents the coefficient so
lutions in" }{TEXT 279 3 " ax" }{TEXT -1 3 " + " }{TEXT 280 2 "by" }
{TEXT -1 21 " = 1. We first solve " }{TEXT 287 3 "ku " }{TEXT -1 2 "+ \+
" }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 1 "(" }{TEXT 289 1 "m" }
{TEXT -1 1 ")" }{TEXT 288 1 "v" }{TEXT -1 16 " = 1 by entering" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "igcdex(k,phi(m),'u','v'); " 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 13 "Only the gcd(" }{TEXT 292 1 "k" }{TEXT -1 2 ", " }
{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 1 "(" }{TEXT 291 1 "m" }
{TEXT -1 41 ")) is output in the last command. To see " }{TEXT 293 1 "
u" }{TEXT -1 5 " and " }{TEXT 294 1 "v" }{TEXT -1 6 ", type" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "u; v;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#!(X)GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"%6?" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "The next command can be used to ch
eck that " }{TEXT 311 2 "u " }{TEXT -1 4 "and " }{TEXT 312 2 "v " }
{TEXT -1 11 "are correct" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 
"k*u + phi(m)*v = 1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/\"\"\"F$" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "However, recall that to be able to
 used " }{TEXT 313 1 "u" }{TEXT -1 41 " as an exponent in the solution
 process, " }{TEXT 314 2 "u " }{TEXT -1 46 "must be represented as a p
ositve solution mod " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 1 "(
" }{TEXT 315 1 "m" }{TEXT -1 69 ") in its congruence class. The next c
ommand performs this conversion." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "u := u mod phi(m);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%\"uG\")r.TD" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "Note that u can
 be found also by entering the following command:" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 24 "u := k^(-1) mod phi(m); " }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"uG\")r.TD" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 
"These are the values for " }{TEXT 295 1 "m" }{TEXT -1 5 " and " }
{TEXT 296 1 "b" }{TEXT -1 18 " we entered above." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 5 "m; b;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\")4)
zK%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"&tR\"" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 30 "Step 3 requires us to compute " }{XPPEDIT 18 0 "b^u;
" "6#)%\"bG%\"uG" }{TEXT -1 93 " to find the solution. To perform modu
lar exponentiation quickly and efficiently, we use the " }{TEXT 297 2 
"&^" }{TEXT -1 53 " operator. The next command will compute solution t
o " }{XPPEDIT 18 0 "x^20075;" "6#*$%\"xG\"&v+#" }{TEXT -1 27 "= 13973 \+
(mod 43279809) for " }{TEXT 298 1 "x" }{TEXT -1 1 "." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 18 "x := b &^ u mod m;" }{TEXT -1 0 "" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"xG\")Sgm;" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 61 "The next command checks this answer to see if it is co
rrect. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "x &^ k mod m = b
 mod m;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/\"&tR\"F$" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 142 "It is in
teresting to see if a solution can be produced using this method if th
e initial assumptions are not satisfied, that is, if the    gcd(" }
{TEXT 300 1 "k" }{TEXT -1 1 "," }{XPPEDIT 18 0 "phi;" "6#%$phiG" }
{TEXT -1 1 "(" }{TEXT 299 1 "m" }{TEXT -1 9 "))or gcd(" }{TEXT 301 1 "
b" }{TEXT -1 1 "," }{TEXT 302 1 "m" }{TEXT -1 36 ") is not 1. Suppose \+
we want to solve" }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "x^132234;" "6#*$%
\"xG\"'MA8" }{TEXT -1 23 "=  12312 (mod 34244221)" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 39 "k := 132234; b := 12312; m := 34244221;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"'MA8" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"bG\"&7B\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"mG
\")@UCM" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "After performing step \+
1 of the process," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "phi(m);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\")+68J" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 32 "the next commands show that gcd(" }{TEXT 304 1 "k" }
{TEXT -1 1 "," }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 1 "(" }
{TEXT 303 1 "m" }{TEXT -1 12 ")) is not 1." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 25 "gcd(k, phi(m)); gcd(b,m);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "The next commands demonstrate step
 2 of the process. Note that u is positive when solving " }{TEXT 305 
3 "ku " }{TEXT -1 2 "+ " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 1 
"(" }{TEXT 307 1 "m" }{TEXT -1 1 ")" }{TEXT 306 1 "v" }{TEXT -1 7 " = \+
gcd(" }{TEXT 309 1 "k" }{TEXT -1 1 "," }{XPPEDIT 18 0 "phi;" "6#%$phiG
" }{TEXT -1 1 "(" }{TEXT 308 1 "m" }{TEXT -1 31 ")) and no adjustment \+
is needed." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "igcdex(k, phi
(m), 'u', 'v');" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"#" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "u; v;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#\"(.BY#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#!&f/\"" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "k*u + phi(m)*v = 2;" }{TEXT -1 0 "
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/\"\"#F$" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 39 "We next attempt to find a solution for " }{TEXT 310 1 "
x" }{TEXT -1 25 " using the given process." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 18 "x := b &^ u mod m;" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%\"xG\")AdYB" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "However, the \+
next command shows that this is clearly not a solution." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "x &^ k mod m = b mod m;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#/\")g%3Y\"\"&7B\"" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 83 "Actually, the inverse of k mod phi(m) does not exist as
 the next command indicates." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 18 "k^(-1) mod phi(m);" }}{PARA 8 "" 1 "" {TEXT -1 42 "Error, the mo
dular inverse does not exist\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}}{MARK "0 0 1" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }