Dr. Sarah's Maple Demo on Sophie Germain's Modular Arithmetic Work on Fermat's Last Theorem

Based on Mathematical Expeditions by the Explorers by Reinhard Laubenbacher and David Pengelly.

Fermat said that + =

has no solutions in non-zero whole numbers for n>2.

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 solution for any p in N. But, if n>2, then we can never find non-zero integers x,y,z satisfying + =

Another way of saying this is: For all x,y,z in Z (the integers) x, y, and z non-zero implies that + is not equal to

**Germain's General Approach to Fermat's Last Theorem**
** **

Before Sophie Germain, this theorem had already been proven for n=3 by Euler, for n=4 by Fermat, and for n=5 by Legendre. These mathematicians

had completed proofs for specific cases of n. Germain was the first person to come up with a more generalized way to prove Fermat's Last Theorem, instead of looking at each power individually.

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 these.

Germain worked on the
**case of Fermat**
where
+
=

has no non-zero whole number solutions for an odd prime
**p which does not divide x,y, or z.**

She proved this case for p less than 100.

**Germain's Modular Arithmetic Approach to Fermat's Last Theorem **

In a letter to Gauss, Germain provides an explanation of her modular arithmetic approach to Fermat's Last Theorem :

"If the Fermat equation for the exponent p prime has a solution, and if t is a prime number with no nonzero consecutive pth powers modulo t,

then t must divide one of the numbers x, y, or z.

Recall that
For any integer a,
**a mod n**
, read "a mod n" or "a modulo n" is the
**remainder **
of
**a/n.**

**Proof of Germain's Modular Arithmetic Approach to Fermat's Last Theorem:**

Assume for contradiction that x,y and z are non-zero whole numbers satisfying + =

where p is a prime, and that t is a prime number with no nonzero consecutive pth powers modulo t,

and that t does not divide x, y or z.

Explain why this statement is the negation of Germain's theorem?

Notice that t does not divide x by assumption, and so x is not 0 mod t, by definition of modular

arithmetic. We'll use Fermat's Little Theorem to prove that x has a non-zero inverse (mod t)

**Fermat's Little Theorem**
: If k is a prime and k does not divide a then 1 mod k =

**Proof of Fermat's Little Theorem**
(From Burton's Elementary Number Theory):

We begin by considering the first k-1 positive multiples of a; that is, the integers a, 2a, 3a,..., (k-1)a

None of these numbers is congruent modular k to any other, nor is any congruent to zero. Indeed, if it happened that

ra = sa mod k where 1 less than or equal to r < s less than or equal to k-1,

then a could be canceled to give r=s mod k, which is impossible.

Explain why this is impossible?

Therefore, the above set of integers must be congruent mod k to 1,2,3,...,k-1, taken in some order (ie the order could be reversed or mixed up).

Multiplying all of these congruences together, we find that

a * 2a * 3a* ... * (k-1) a = 1 *2*3*...*(k-1) mod k and so

a^(k-1) * (k-1) ! = (k-1) ! mod k

Dividing both sides by (k-1)!, which we can do since k does not divide (k-1)! (ie neither side is 0 mod k), we see that

1 mod k =

as desired.

**Back to Proof of Germain's Modular Arithmetic Approach to Fermat's Last Theorem.**

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 detail).

Since x, y and z are whole number solutions to Fermat, we know that + =

and that t does not divide x,y or z, and so + = mod t

gives us a non-trivial equation (ie not 0=0).

Now multiply this equation by a^p to get + = mod t

which reduces to + = mod t since a is the inverse of x.

This reduces to 1 + = mod t.

Since t does not divide a, y or z, then

and must be consecutive non-zero numbers mod t. This contradicts the assumption that t has no consecutive non-zero pth powers.

Hence we have arrived at a contradiction to the fact that t does not divide x,y or z, and so t must divide

one of them, as desired.

Examples

**Example**
: To show that
**p=3**
and
**2p+1=7 **
satisfy the conditions in her Theorem, notice that 2*3+1=7 is prime, and we will show that

**7 has no consecutive 3rd powers mod 7:**

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

7n+1

7n+2

7n+3

7n+4

7n+5

or 7n+6

We will examine the 3rd power of each of these numbers modulo 7, and show that there are no consecutive (a difference of 1) pairs in the list:

`> `
**expand ((7*n+1)^3);**

`> `
**expand((7*n+1)^3) mod 7;**

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 expansion of

(7n+1)^3 are divisible by 7.

`> `
**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;**

Notice that the 3rd powers are all 1 or 6, so they are not consecutive, as desired.

`> `

**Non-example**
:
**p=7**
,
**2p+1 = 15**
, which is not prime.

Using Maple Statements, find 2 consecutive 7th powers mod 15 (ie two numbers a and b so that a^7+1=b^7 mod 15).

`> `

`> `

`> `

`> `

Now that we've seen an example and non-example, read thru the proof of her theorem again.

**Germain's Attempt at Using her Theorem to Prove Fermat for a Power p. **

Fix p. If one could find infinitely many primes t satisfying the condition that t has no non-zero consecutive pth powers modulo t, then, by her

theorem, each of these would have to divide one of x,y, and/or z, and thus one of these three numbers would be divisible by infinitely many primes, 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 proven for the exponent p.

**Why she never completely succeeded using this approach. **

Despite much effort, Germain never succeeded in proving Fermat for a single exponent p using this appoach, because she could not find 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 particular 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 size!

As she says in her letter to Gauss, "You can easily imagine, Monsieur, that I must have been able to prove that this equation is only possible for numbers who size frightens the imagination.... But all this is still nothing; it (proving Fermat has no non-zero whole number solutions) requires the infinite and not just the very large.

Germain never published her work on Fermat. One might speculate that her experience with the establishment at the Paris Academy of Sciences, and the fact that after several disputes relating to the publication of her prize-winning work on elasticity theory, she eneded up publishing that work at her own expense. A renewed battle over her number theory work migth have been two unpleasant for her to contemplate.

The only commonly known result of Germain's Fermat work appeared in 1825, as part of a supplement to the second edition of Legendre's Theory of Numbers.

In this supplement, Legendre presents his own proof (the first) for the p=5 case of Fermat, along with part of Germains' work, explicitly credited to her in a footnote.