Proofs of Fermat's Theorem On Sums of Two Squares - Euler's Proof By Infinite Descent

Euler's Proof By Infinite Descent

Euler succeeded in proving Fermat's theorem on sums of two squares in 1749, when he was forty-two years old. He communicated this in a letter to Goldbach dated 12 April 1749. The proof relies on infinite descent, and is only briefly sketched in the letter. The full proof consists in five steps and is published in two papers. The four first steps are Propositions 1 to 4 of the first paper (and don't correspond exactly to the four steps below). The fifth step below is from the second paper.

1. The product of two numbers, each of which is a sum of two squares, is itself a sum of two squares.

This is a well known property, due to Diophantus of Alexandria.

2. If a number which is a sum of two squares is divisible by a prime which is a sum of two squares, then the quotient is a sum of two squares. (This is Euler's first Proposition).

Indeed, suppose for example that is divisible by and that this latter is a prime. Then divides

Since is a prime, it divides one of the two factors. Suppose that it divides . Since

(Diophantus identity) it follows that must divide . So the equation can be divided by the square of . Dividing the expression by yields:

and thus expresses the quotient as a sum of two squares, as claimed.

If divides, a similar argument holds by using

(Diophantus identity).

3. If a number which can be written as a sum of two squares is divisible by a number which is not a sum of two squares, then the quotient has a factor which is not a sum of two squares. (This is Euler's second Proposition).

Suppose divides and that the quotient, factored into its prime factors is . Then . If all factors can be written as sums of two squares, then we can divide successively by, etc., and applying the previous step we deduce that each quotient is a sum of two squares. This until we get to, concluding that would have to be the sum of two squares. So, by contraposition, if is not the sum of two squares, then at least one of the primes is not the sum of two squares.

4. If and are relatively prime then every factor of is a sum of two squares. (This is Euler's Proposition 4. The proof sketched below includes the proof of his Proposition 3).

This is the step that uses infinite descent. Let be a factor of . We can write

where and are at most half of in absolute value. This gives:

Therefore, must be divisible by, say . If and are not relatively prime, then their gcd cannot divide (if it did, then it would divide and which we assume are relatively prime). Thus the square of the gcd divides (as it divides ), giving us an expression of the form for relatively prime and, and with no more than half of, since

If and are relatively prime, then we can use them directly instead of switching to and .

If is not the sum of two squares, then by the third step there must be a factor of which is not the sum of two squares; call it . This gives an infinite descent, going from to a smaller number, both not the sums of two squares but dividing a sum of two squares. Since an infinite descent is impossible, we conclude that must be expressible as a sum of two squares, as claimed.

5. Every prime of the form is a sum of two squares. (This is the main result of Euler's second paper).

If, then by Fermat's Little Theorem each of the numbers is congruent to one modulo . The differences are therefore all divisible by . Each of these differences can be factored as

Since is prime, it must divide one of the two factors. If in any of the cases it divides the first factor, then by the previous step we conclude that is itself a sum of two squares (since and differ by, they are relatively prime). So it is enough to show that cannot always divide the second factor. If it divides all differences, then it would divide all differences of successive terms, all differences of the differences, and so forth. Since the th differences of the sequence are all equal to (Finite difference), the th differences would all be constant and equal to, which is certainly not divisible by . Therefore, cannot divide all the second factors which proves that is indeed the sum of two squares.