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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 91 "Dr. Sarah's Maple Demo \+
on Sophie Germain's Modular Arithmetic Work on Fermat's Last Theorem"
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 95 "Based on Mathematical Expeditions by the Explorers by Rei
nhard Laubenbacher and David Pengelly." }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "Fermat said that \+
" }{XPPEDIT 19 1 "x^n;" "6#)%\"xG%\"nG" }{TEXT -1 2 "+ " }{XPPEDIT
19 1 "y^n;" "6#)%\"yG%\"nG" }{TEXT -1 2 "= " }{XPPEDIT 19 1 "z^n;" "6#
)%\"zG%\"nG" }}{PARA 0 "" 0 "" {TEXT -1 51 "has no solutions in non-ze
ro whole numbers for n>2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 283 "For n=2, we know that we can find many solutions:
the sides of any right triangle give us a solution (example 3,4,5). \+
And (as in the test solutions), we can prove that (3p,4p,5p) is a sol
ution for any p in N. But, if n>2, then we can never find non-zero in
tegers x,y,z satisfying " }{XPPEDIT 19 1 "x^n;" "6#)%\"xG%\"nG" }
{TEXT -1 2 "+ " }{XPPEDIT 19 1 "y^n;" "6#)%\"yG%\"nG" }{TEXT -1 2 "= \+
" }{XPPEDIT 19 1 "z^n;" "6#)%\"zG%\"nG" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "Another way of s
aying this is: For all x,y,z in Z (the integers) x, y, and z non-zero
implies that " }{XPPEDIT 19 1 "x^n;" "6#)%\"xG%\"nG" }{TEXT -1 3 " + \+
" }{XPPEDIT 19 1 "y^n;" "6#)%\"yG%\"nG" }{TEXT -1 17 " is not equal to
" }{XPPEDIT 19 1 "z^n;" "6#)%\"zG%\"nG" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 259 "" 0 "" {TEXT 267 51 "Germain's General Appr
oach to Fermat's Last Theorem" }{TEXT 268 1 " " }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 143 "Before Sophie Germain, t
his theorem had already been proven for n=3 by Euler, for n=4 by Ferma
t, and for n=5 by Legendre. These mathematicians" }}{PARA 0 "" 0 ""
{TEXT -1 193 "had completed proofs for specific cases of n. Germain w
as the first person to come up with a more generalized way to prove Fe
rmat's Last Theorem, instead of looking at each power individually." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 150 "People \+
knew that it was only necessary to prove Fermat for prime powers (ie n
=p, a prime) since all other cases of powers can be broken up into the
se." }}{PARA 0 "" 0 "" {TEXT -1 22 "Germain worked on the " }{TEXT
257 14 "case of Fermat" }{TEXT -1 7 " where " }{XPPEDIT 19 1 "x^p;" "6
#)%\"xG%\"pG" }{TEXT -1 2 "+ " }{XPPEDIT 19 1 "y^p;" "6#)%\"yG%\"pG" }
{TEXT -1 2 "= " }{XPPEDIT 19 1 "z^p;" "6#)%\"zG%\"pG" }}{PARA 0 "" 0 "
" {TEXT -1 56 "has no non-zero whole number solutions for an odd prime
" }{TEXT 258 34 "p which does not divide x,y, or z." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "She proved this case fo
r p less than 100." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 258 "" 0 "" {TEXT 260 63 "Germain's Modular Arithmetic
Approach to Fermat's Last Theorem " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 115 "In a letter to Gauss, Germain provides \+
an explanation of her modular arithmetic approach to Fermat's Last The
orem :" }}{PARA 0 "" 0 "" {TEXT -1 140 "\"If the Fermat equation for t
he exponent p prime has a solution, and if t is a prime number with no
nonzero consecutive pth powers modulo t," }}{PARA 0 "" 0 "" {TEXT -1
49 "then t must divide one of the numbers x, y, or z." }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "Recall that " }{TEXT
275 19 "For any integer a, " }{TEXT 272 7 "a mod n" }{TEXT 276 40 ", r
ead \"a mod n\" or \"a modulo n\" is the " }{TEXT 273 10 "remainder "
}{TEXT 277 3 "of " }{TEXT 274 4 "a/n." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 257 "" 0 "" {TEXT -1 72 "Proof of Germain's Modular Arithmet
ic Approach to Fermat's Last Theorem:" }}{PARA 0 "" 0 "" {TEXT -1 85 "
Assume for contradiction that x,y and z are non-zero whole number
s satisfying " }{XPPEDIT 19 1 "x^p;" "6#)%\"xG%\"pG" }{TEXT -1 3 " +
" }{XPPEDIT 19 1 "y^p;" "6#)%\"yG%\"pG" }{TEXT -1 2 "= " }{XPPEDIT
19 1 "z^p;" "6#)%\"zG%\"pG" }}{PARA 0 "" 0 "" {TEXT -1 98 "where p is \+
a prime, and that t is a prime number with no nonzero consecutive pth \+
powers modulo t, " }}{PARA 0 "" 0 "" {TEXT -1 37 "and that t does not \+
divide x, y or z." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 259 64 "Expl
ain why this statement is the negation of Germain's theorem?" }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 96 "Notice that t does not divide x by assumption, and s
o x is not 0 mod t, by definition of modular" }}{PARA 0 "" 0 "" {TEXT
-1 93 "arithmetic. We'll use Fermat's Little Theorem to prove that x \+
has a non-zero inverse (mod t)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 262
23 "Fermat's Little Theorem" }{TEXT -1 71 ": If k is a prime and k do
es not divide a then 1 mod k = " }{XPPEDIT 19 1 "a^(k-1)
" "6#)%\"aG,&%\"kG\"\"\"\"\"\"!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "
" {TEXT 263 32 "Proof of Fermat's Little Theorem" }{TEXT -1 44 " (From
Burton's Elementary Number Theory): " }}{PARA 0 "" 0 "" {TEXT -1
118 " We begin by considering the first k-1 positive multiples o
f a; that is, the integers a, 2a, 3a,..., (k-1)a" }}{PARA 0 "" 0
"" {TEXT -1 117 "None of these numbers is congruent modular k to any o
ther, nor is any congruent to zero. Indeed, if it happened that" }}
{PARA 0 "" 0 "" {TEXT -1 76 "ra = sa mod k where 1 less than or equal \+
to r < s less than or equal to k-1," }}{PARA 0 "" 0 "" {TEXT -1 64 "th
en a could be canceled to give r=s mod k, which is impossible." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 261 31 "Explain why this is impos
sible?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 146 "Therefore, the above set of integers
must be congruent mod k to 1,2,3,...,k-1, taken in some order (ie t
he order could be reversed or mixed up)." }}{PARA 0 "" 0 "" {TEXT -1
59 "Multiplying all of these congruences together, we find that" }}
{PARA 0 "" 0 "" {TEXT -1 64 "a * 2a * 3a* ... * (k-1) a = 1 *2*3*...*
(k-1) mod k and so" }}{PARA 0 "" 0 "" {TEXT -1 35 "a^(k-1) * (k-1
) ! = (k-1) ! mod k " }}{PARA 0 "" 0 "" {TEXT -1 120 "Dividing both s
ides by (k-1)!, which we can do since k does not divide (k-1)! (ie nei
ther side is 0 mod k), we see that " }}{PARA 0 "" 0 "" {TEXT -1 10 "1 \+
mod k = " }{XPPEDIT 19 1 "a^(k-1)" "6#)%\"aG,&%\"kG\"\"\"\"\"\"!\"\""
}}{PARA 0 "" 0 "" {TEXT -1 11 "as desired." }}}{EXCHG {PARA 0 "" 0 ""
{TEXT 264 80 "Back to Proof of Germain's Modular Arithmetic Approach t
o Fermat's Last Theorem." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }
{TEXT 265 132 "Explain why x has a non-zero inverse (mod t), call it a
, by applying Fermat's Little Thorem to find the inverse (explain in d
etail)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 266
0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "Since x, y and z are whole number \+
solutions to Fermat, we know that " }{XPPEDIT 19 1 "x^p;" "6#)%\"xG%\"
pG" }{TEXT -1 4 " + " }{XPPEDIT 19 1 "y^p;" "6#)%\"yG%\"pG" }{TEXT
-1 2 "= " }{XPPEDIT 19 1 "z^p;" "6#)%\"zG%\"pG" }}{PARA 0 "" 0 ""
{TEXT -1 43 "and that t does not divide x,y or z, and so" }{XPPEDIT
19 1 "x^p;" "6#)%\"xG%\"pG" }{TEXT -1 5 " + " }{XPPEDIT 19 1 "y^p;"
"6#)%\"yG%\"pG" }{TEXT -1 2 "= " }{XPPEDIT 19 1 "z^p;" "6#)%\"zG%\"pG
" }{TEXT -1 5 "mod t" }}{PARA 0 "" 0 "" {TEXT -1 45 "gives us a non-tr
ivial equation (ie not 0=0)." }}{PARA 0 "" 0 "" {TEXT -1 41 "Now multi
ply this equation by a^p to get " }{XPPEDIT 19 1 "ax^p;" "6#)%#axG%\"p
G" }{TEXT -1 5 " + " }{XPPEDIT 19 1 "ay^p;" "6#)%#ayG%\"pG" }{TEXT
-1 2 "= " }{XPPEDIT 19 1 "az^p;" "6#)%#azG%\"pG" }{TEXT -1 5 "mod t" }
}{PARA 0 "" 0 "" {TEXT -1 17 "which reduces to " }{XPPEDIT 19 1 "1^p;
" "6#)\"\"\"%\"pG" }{TEXT -1 5 " + " }{XPPEDIT 19 1 "ay^p;" "6#)%#ay
G%\"pG" }{TEXT -1 2 "= " }{XPPEDIT 19 1 "az^p;" "6#)%#azG%\"pG" }
{TEXT -1 34 "mod t since a is the inverse of x." }}{PARA 0 "" 0 ""
{TEXT -1 21 "This reduces to 1 + " }{XPPEDIT 19 1 "ay^p;" "6#)%#ayG%
\"pG" }{TEXT -1 2 "= " }{XPPEDIT 19 1 "az^p;" "6#)%#azG%\"pG" }{TEXT
-1 8 "mod t. " }}{PARA 0 "" 0 "" {TEXT -1 40 "Since t does not divide
a, y or z, then " }{XPPEDIT 19 1 "ay^p" "6#)%#ayG%\"pG" }}{PARA 0 ""
0 "" {TEXT -1 5 "and " }{XPPEDIT 19 1 "az^p;" "6#)%#azG%\"pG" }{TEXT
-1 123 "must be consecutive non-zero numbers mod t. This contradicts \+
the assumption that t has no consecutive non-zero pth powers." }}
{PARA 0 "" 0 "" {TEXT -1 106 "Hence we have arrived at a contradiction
to the fact that t does not divide x,y or z, and so t must divide" }}
{PARA 0 "" 0 "" {TEXT -1 24 "one of them, as desired." }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}}{EXCHG {PARA 262 "" 0 "" {TEXT -1 8 "Examples" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT 284 7 "Example" }{TEXT -1 16 ": To show \+
that " }{TEXT 278 3 "p=3" }{TEXT -1 5 " and " }{TEXT 279 7 "2p+1=7 " }
{TEXT -1 91 "satisfy the conditions in her Theorem, notice that 2*3+1=
7 is prime, and we will show that " }}{PARA 260 "" 0 "" {TEXT -1 38 "7
has no consecutive 3rd powers mod 7:" }}{PARA 0 "" 0 "" {TEXT -1 158
"Proof: Let's examine all the possible 3rd powers of numbers modulo 7
. So, let m be any number that is not divisible by 7. Then m can be \+
written in the form" }}{PARA 0 "" 0 "" {TEXT -1 8 " 7n+1" }}{PARA
0 "" 0 "" {TEXT -1 8 " 7n+2" }}{PARA 0 "" 0 "" {TEXT -1 8 " 7n+3
" }}{PARA 0 "" 0 "" {TEXT -1 9 " 7n+4 " }}{PARA 0 "" 0 "" {TEXT -1
8 " 7n+5" }}{PARA 0 "" 0 "" {TEXT -1 7 "or 7n+6" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 142 "We will examine the 3rd \+
power of each of these numbers modulo 7, and show that there are no co
nsecutive (a difference of 1) pairs in the list:" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 19 "expand ((7*n+1)^3);" }}{PARA 11 "" 1 ""
{XPPMATH 20 "6#,**$)%\"nG\"\"$\"\"\"\"$V$*$)F&\"\"#F(\"$Z\"F&\"#@\"\"
\"F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "expand((7*n+1)^3) m
od 7;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 ""
0 "" {TEXT -1 126 "This is equivalent to 1 mod 7. In fact notice that
all we needed to look at is 1^3, since the other terms of the expansi
on of" }}{PARA 0 "" 0 "" {TEXT -1 31 "(7n+1)^3 are divisible by 7. \+
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "expand((7*n+2)^3) mod \+
7; expand((7*n+3)^3) mod 7; expand((7*n+4)^3) mod 7; expand((7*n+5)^3)
mod 7; expand((7*n+6)^3) mod 7;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"
\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"'" }}{PARA 11 "" 1 ""
{XPPMATH 20 "6#\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"'" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"'" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 83 "Notice that the 3rd powers are all 1 or 6, so they are no
t consecutive, as desired." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 285 11 "Non-example" }{TEXT -1 3
": " }{TEXT 280 3 "p=7" }{TEXT -1 2 ", " }{TEXT 281 9 "2p+1 = 15" }
{TEXT -1 23 ", which is not prime. " }}{PARA 0 "" 0 "" {TEXT -1 112 "
Using Maple Statements, find 2 consecutive 7th powers mod 15 (ie two n
umbers a and b so that a^7+1=b^7 mod 15). " }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT
271 89 "Now that we've seen an example and non-example, read thru the \+
proof of her theorem again." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT 269 70 "Germain's Attempt at Using her Theorem \+
to Prove Fermat for a Power p. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 144 "Fix p. If one could find infinitely man
y primes t satisfying the condition that t has no non-zero consecutive
pth powers modulo t, then, by her" }}{PARA 0 "" 0 "" {TEXT -1 449 "th
eorem, each of these would have to divide one of x,y, and/or z, and th
us one of these three numbers would be divisible by infinitely many pr
imes, which is absurd since x, y and z are just non-zero whole numbers
, and so they only have a finite number of primes dividing into them. \+
Hence, we would arrive at a contradiction, and so the Fermat equation
could not have a solution for the exponent p. Thus Fermat would be p
roven for the exponent p. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 270 56 "Why she never \+
completely succeeded using this approach. " }}{PARA 0 "" 0 "" {TEXT
-1 625 "Despite much effort, Germain never succeeded in proving Fermat
for a single exponent p using this appoach, because she could not fin
d infinitely many primes t satisfying the condition that t has no non-
zero consecutive pth powers modulo t. Hower, she did invent a method \+
for producing many primes t satisfying the above condition. For any p
articular exponent, her method could show that any solutions to Fermat
would have to be quite large. She made many such applications of her
mthod in her manuscripts. For instance, for p=5, she showed that any
solutions to Fermat would have to be at least 30 decimal digits in si
ze!" }}{PARA 0 "" 0 "" {TEXT -1 334 "As she says in her letter to Gaus
s, \"You can easily imagine, Monsieur, that I must have been able to p
rove that this equation is only possible for numbers who size frighten
s the imagination.... But all this is still nothing; it (proving Ferma
t has no non-zero whole number solutions) requires the infinite and no
t just the very large." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 416 "Germain never published her work on Ferm
at. One might speculate that her experience with the establishment at
the Paris Academy of Sciences, and the fact that after several disput
es relating to the publication of her prize-winning work on elasticity
theory, she eneded up publishing that work at her own expense. A ren
ewed battle over her number theory work migth have been two unpleasant
for her to contemplate. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 152 "The only commonly known result of Germain's Ferm
at work appeared in 1825, as part of a supplement to the second editio
n of Legendre's Theory of Numbers." }}{PARA 0 "" 0 "" {TEXT -1 170 "In
this supplement, Legendre presents his own proof (the first) for the \+
p=5 case of Fermat, along with part of Germains' work, explicitly cred
ited to her in a footnote. " }}}}{MARK "1 0 0" 0 }{VIEWOPTS 1 1 0 1
1 1803 }