Sophie Germain’s Theorem

As early attempts to the proof of Fermat’s last theorem, many mathematicians solved the problem for small exponents. While these special cases are being studied, Sophie Germain, a French mathematician, came up with the following interesting result. (Look at https://www.agnesscott.edu/lriddle/women/germain.htm for her fascinating and revolutionary story)

Theorem 1: For any odd prime $p$ such that $2p+1$ is a prime, the equation $x^p+y^p=z^p$ has no solution in integers with $p\nmid xyz$ .

Proof : We prove by contradiction. Suppose there exists a solution $(x,y,z)$ with $p\nmid xyz$ . We can assume $x,y,z$ to be relatively prime for if they are common factors, they can be canceled out to get a primitive triple. Now, using the factorisation $x^p+y^p=(x+y)(x^{p-1}-x^{p-2}y...+y^{p-1})$ and the fact that $p\nmid xyz$ , we deduce that $x+y$ is a perfect $p^{\mbox{th}}$ power–Because if there was a prime factor $q$ common to both $x+y$ and $(x^{p-1}+x^{p-2}y...+y^{p-1})$ , then $x\equiv -y \mbox{ (mod }q)$ so that $(x^{p-1}-x^{p-2}y...+y^{p-1}) \equiv px^{p-1} \mbox{ (mod }q)$ –which is not possible for if

(1) $q$ divides $x$ it has to divide $y$ which can’t be true, and if

(2) $q$ divides $p$ , $p$ has to be $q$ and it has to divide $z^n$ and hence $z$ , which can’t be true.

So we have $x+y=m^p$ for some integer $m.$ Similarly we have $z-y=n^p$ and $z-x=l^p. .$ Now the fact that for every integer $\alpha$ , $\alpha^p\equiv \pm1 \mbox{ or }0 \mbox{ (mod }2p+1)$ implies that $2p+1$ divides one of $x,y,z .$ Lets us say it divides $x.$ (Similar arguments work for other cases also). Therefore $2x\equiv m^p+n^p-l^p\equiv 0 \mbox{ (mod } 2p+1).$ Using the above fact once again implies that $2p+1$ divides one of $l,m,n$ . It can’t divide $l,m$ for if it divides them it has to divides $y$ or $z$ along with $x$ which is not possible. Other case can also be dealt using same ideas above. Thus we will have a contradiction. $\Box$

Now the fact that the first case of FLT is true for the case $n=5$ follows from the above theorem by noting that $2.5+1=11$ is a prime. So this theorem proves the first case for a class of primes $p$ for which $2p+1$ is also a prime. However it is still not known whether there exist infinitely many such primes (now called Sophie Germain primes, by Hardy-Littlewood Heuristics there are about $2 \prod_{p>2} \frac{p(p-2)}{(p-1)^{2}} \frac{x}{(\ln x)^{2}}$ Sophie Germain primes less than $x$ ). They are further generalisations of this result about which I will talk sometime later!

Theorem 2 (Stronger):

If $p$ is an odd prime and there exists an “auxiliary prime” $q=2 n p+1$ satisfying: a) there are no consecutive $p^{\text {th }}$ power residues $\bmod ~q$ and (b) $p$ is not a $p^{\text {th }}$ power residue $\bmod ~q$
then in any solution to the Fermat equation $x^{p}+y^{p}=z^{p}, p^{2}$ must divide one of $x, y$ , or $z .$

The consecutive power residue condition can be rephrased equivalently as “ $x^p+y^p+z^p \equiv 0 \bmod q \implies x, y , \text{or} z \equiv 0 \bmod q$ .

Theorem 1 follows from Theorem 2 because the auxilary prime $2p+1$ satisfies the conditions of Theorem 2.

Second Condition: $x^p \equiv \pm 1 \bmod ~ q=2p+1$ and hence there is no solution to $x^p \equiv p \bmod ~ q$ .

First Condition: Also because $x^p \equiv \pm 1 \bmod ~ q$ , we have $x^p+y^p+z^p \equiv \pm 1 \pm 1\pm 1 \bmod ~ q.$ which cannot be $0 \mod q$

For $p=3$ , the only auxilary primes are $q=7, 13$ , every other $q$ doesn’t satisfy the consecutive power resides condition (I). Even for $p=5$ , the condition is false for $q=2np+1$ if $n$ is a multiple of $3$

The initial hope was that for every prime $p$ , one can find infinitely many auxiliary primes $q$ which implies that at least one of the elements $x, y, z$ has to be divisible by infinitely many of these auxiliary primes (using reformulation of the second condition given above), which implies FLT for $p.$ But for any prime $p$ , there are at most finitely many auxiliary primes and the strategy fails!

The proof of Theorem 2 is exactly similar to that of Theorem 1 if we just ask for divisibility of by $p$ . The divisibility by $p^2$ needs some slightly more analysis.

Sophie Germain and Legendre together founds auxiliary primes for $p <197$ with $n \le 10$ and hence solved case I of FLT for all these primes!

Some other methods to establish Case I of FLT is to show that a prime $p$ which fails Case I satisfies Wieferich prime conditions: $p^2$ divides $2^{p-1}-1$ . That is the Fermat quotient $\frac{2^{p-1}-1}{p}$ is divisible by $p$ . That allows us to computationally check. In fact, $p^2$ can be shown to divide $a^{p-1}-1$ for $a \le 113$ ,– so all these Wieferich conditions help you to computationally check that FLT case I has to be true for $p < 10^{14}.$ For instance primes less than $10^9$ that divide $\frac{2^{p-1}-1}{p}$ are only $p =1093, 3511$ and primes that divide $\frac{3^{p-1}-1}{p}$ are only $p =11, 1006002.$ Provin such criterion for bases $a \le 113$ allow to verify the range $p < 10^{14}.$