Around 1637 Pierre de Fermat noted in the margin of a book that the equation

$$x^n+y^n=z^n$$

has no solutions in positive whole numbers when $n$ is greater than 2, and that he had a marvellous proof which the margin was too narrow to contain. This is Fermat’s Last Theorem. Nobody ever found his proof, and the problem was only solved by Andrew Wiles in 1994.

Until 1800 the situation was modest. Fermat had proved the case $n=4$, Euler the case $n=3$. It was enough to study prime exponents $n=p$, but each prime needed its own trick, and nobody could get past 3. Sophie Germain was the first person to prove something for infinitely many exponents at once.

Two cases

Sophie split the problem in two. Suppose a solution $x^p+y^p=z^p$ existed. Either none of the three numbers $x$, $y$, $z$ is a multiple of $p$, which is now called the first case, or one of them is, the second case. Her theorem deals with the first.

Sophie Germain’s theorem. If $p$ is an odd prime and $2p+1$ is also prime, then there are no integers $x, y, z$, none of them a multiple of $p$, with $x^p+y^p=z^p$.

Primes $p$ with $2p+1$ also prime are now called Sophie Germain primes: $3, 5, 11, 23, 29, 41, 53, 83, 89, \ldots$

The idea, with p = 5

Let us look at the heart of the proof in the case $p=5$, where $2p+1=11$. Suppose $x^5+y^5=z^5$. Look at everything modulo 11. As we saw in the letters to Gauss, a fifth power modulo 11 can only be $0$, $1$ or $-1$.

If none of the three were a multiple of 11, the three fifth powers would be $\pm1$. But the sum of two numbers $\pm1$ gives $-2$, $0$ or $2$, and none of these is $\pm1$ modulo 11:

$$\pm1\pm1\in\{-2,\,0,\,2\}\not\ni\pm1\pmod{11}.$$

So one of the numbers $x$, $y$, $z$ must be a multiple of 11. That is the first step. In the second, Sophie uses the factorisation

$$z^5-y^5=(z-y)\left(z^4+z^3y+z^2y^2+zy^3+y^4\right)$$

and other congruences to show that if 11 divides one of the numbers, then 5 must also divide one of them. That contradicts the hypothesis of the first case, and the solution cannot exist.

The role of $2p+1$ is to be an auxiliary prime where $p$-th powers are so poor that the equation becomes impossible. Sophie realised that other auxiliary primes also worked and, with them, verified the first case for all primes below 100. Legendre extended the check up to 197.

After Sophie

The path she opened was followed by Dirichlet and Legendre, who proved the case $n=5$ in 1825, and by Lamé, who proved $n=7$ in 1839. In the mid-nineteenth century Ernst Kummer settled all the “regular” primes at once. Wiles’s complete proof in 1994 used entirely different ideas, but Sophie Germain’s theorem kept being cited as the first general result on the problem.

Exercise: use p = 3 and 2p + 1 = 7 to show that if x³ + y³ = z³, one of the numbers is a multiple of 7

Modulo 7, cubes can only be $0$, $1$ or $6\equiv-1$. If none of the numbers were a multiple of 7, we would have $\pm1\pm1\equiv\pm1\pmod 7$. But $\pm1\pm1$ only gives $-2\equiv5$, $0$ or $2$, and none is $1$ or $6$. So 7 divides $x$, $y$ or $z$.

Exercise: which primes below 100 are Sophie Germain primes?

Testing $2p+1$ for each prime $p$: $2\to5$, $3\to7$, $5\to11$, $11\to23$, $23\to47$, $29\to59$, $41\to83$, $53\to107$, $83\to167$ and $89\to179$ are all prime. The others fail, for example $7\to15$ and $13\to27$. There are 10 primes: $2, 3, 5, 11, 23, 29, 41, 53, 83, 89$.

Sources: Reinhard Laubenbacher and David Pengelley, “‘Voici ce que j’ai trouvé’: Sophie Germain’s grand plan to prove Fermat’s Last Theorem”, Historia Mathematica (2010) · biography of Sophie Germain, MacTutor · Harold M. Edwards, Fermat’s Last Theorem (1977).

All Sophie Germain topics · ← The letters of Monsieur Le Blanc: the arithmetic of remainders · Sophie Germain primes and cryptography →

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