## Euler and the Four-Squares Theorem

There are several questions in the Euler-Goldbach correspondence that I am unable to answer. Sometimes it does not take very much: in his letter to Goldbach dated June 9th, 1750, Euler conjectured that every odd number can be written as a sum of four squares in such a way that $n = a^2 + b^2 + c^2 + d^2$ and $a+b+c+d = 1$. I was just about to post this to MO when I saw that Euler's conjecture can be reduced to the Three-Squares Theorem in one line (am I supposed to spoil this right away?). Here's another one where I haven't found a proof yet.

In his letter to Goldbach dated Apr.15, 1747, Euler wrote:

The theorem Any number can be split into four squares'' depends on this:Any number of the form $4m+2$ can always be split into two parts such as $4x+1$ and $4y+1$, none of which has any divisor of the form $4p-1$'' (which does not appear difficult, although I cannot yet prove it).

Later, Euler attributed to Goldbach the much stronger claim that the two summands can be chosen to be prime, which is a strong form of the Goldbach conjecture.

Euler's intention was proving the Four-Squares Theorem (which he almost did. Assuming this result, write $4m+2 = a^2+b^2+c^2+d^2$; then congruences modulo $8$ show that two numbers on the right hand side, say $a$ and $b$, are even, and the other two are odd. Now $a^2 + c^2 = 4x+1$ and $b^2 + d^2 = 4y+1$ satisfy Euler's conditions except when $a$ and $c$ (or $b$ and $d$) have a common prime factor of the form $4n-1$. Can this be excluded somehow?

Hermite [Oeuvres I, p. 259] considered a similar problem:

Tout nombre impair est decomposable en quatre carres et, parmi ces decompositions, il en existe toujours de telles que la somme de deux carrees soit sans diviseurs communs avec la somme de deux autres. (Every odd number can be decomposed into four squares, and among these decompositions, there always exist some for which the sum of two squares is coprime to the sum of the other two.)

Hermite's proof contains a gap. Can Hermite's claim be proved somehow?

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Euler's intention was proving the Four-Squares Theorem, which is trivial for sums of two squares. But 18 = 3^2 + 3^2 is not what Euler was looking for. – Franz Lemmermeyer Sep 1 2010 at 9:51
As for the claim that for every positive odd integer $n$ there is always an integral point on the intersection of the hypersphere $2n+1 = a^2 + b^2 + c^2 + d^2$ and the hyperplane $a+b+c+d=1$: solve the last equation for $d$, plug it into the first, and multiply through by $4$ and subtract $1$. Then $8m+3$ becomes a sum of three squares, so Euler's result is equivalent to this special case of the Three-Squares Theorem, or, if you happen to know it, to Gauss's theorem that every number is the sum of three triangular numbers. – Franz Lemmermeyer Sep 1 2010 at 9:54