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{SECT 0 {EXCHG {PARA 257 "" 0 "" {TEXT -1 87 "Maple worksheet to perfo
rm computations for RSA algorithm             MME 529   6/15/99" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }{XPPEDIT 18 0 "" "6#%#%?G" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }{MPLTEXT 1 0 19 "p:=13;q:=17;n:=p*q;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"pG\"#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"qG
\"#<" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"nG\"$@#" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 15 "k:=(p-1)*(q-1);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"kG\"$#>" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "no
w I need a number which is not a divisor of k. To make life easy, I'll
 peek at its divisors and then avoid them." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 16 "with(numtheory):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "divisors(k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<0\"
\"\"\"\"#\"\"$\"\"%\"\"'\"\")\"#7\"#;\"#C\"#K\"#[\"#k\"#'*\"$#>" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "d:=65;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"dG\"#l" }}}{EXCHG {PARA 256 "" 1 "" {TEXT -1 99 "Ne
xt, I need the multiplicative inverse of d, mod k; that is, the soluti
on to  65x = 1(mod 192). I " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "ca
n solve 65x + 192y=1 using the Maple function \"" }{TEXT 264 7 "isolve
\"" }{TEXT -1 45 " instead of using Euclid's algorithm by hand!" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "isolve(d*x+k*y=1);" }{TEXT 
-1 51 "(in what follows, N1 is arbitrary. if we set it to " }{TEXT 
266 3 "-1," }{TEXT -1 13 " we get that " }{TEXT 265 7 "e is 65" }
{TEXT -1 72 " also - a chance event - a number being its own multiplic
ative inverse!)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<$/%\"xG,&!$F\"\"\"
\"%$_N1G!$#>/%\"yG,&\"#VF(F)\"#l" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
24 "the \"public key\" is now " }{TEXT 267 11 "e=65, n=221" }{TEXT -1 
27 ".  Let's encode the letter " }{TEXT 257 1 "B" }{TEXT -1 30 " which
 we'll give a value of  " }{TEXT 256 2 "74" }{TEXT -1 4 " to:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "e:=65;B:=74;" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"eG\"#l" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"
BG\"#u" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 15 "C:=(B^e)mod(n);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"CG\"$f\"" }}}{PARA 0 "" 0 "" {TEXT -1 7 "So the " }{TEXT 
271 7 "encoded" }{TEXT -1 26 " version of the character " }{TEXT 268 
2 "74" }{TEXT -1 4 " is " }{TEXT 269 3 "159" }{TEXT -1 57 ". To check \+
that things work alright, let's see if we can " }{TEXT 270 6 "decode" 
}{TEXT -1 8 " it back" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "(C^
d)mod(n);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#u" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 78 "good! we got the same thing back. This does have i
ts limitations - if you use " }{TEXT 262 5 "B > n" }{TEXT -1 55 ", you
 will only get a number back which is equal to B, " }{TEXT 263 5 "mod \+
n" }{TEXT -1 270 ", not the original B - try it! At this point, I have
 the machinery set up to code and decode numbers using the public key \+
of e=65 and n = 221.  This is a terribly poor key because n is so smal
l, but it does serve to illustrate the commands needed to do the calcu
lations." }}}}{MARK "12" 0 }{VIEWOPTS 1 1 0 1 1 1803 }