2
$\begingroup$

I've been looking at Wikipedia's proof of the four-square theorem and trying to work out the details - I like that it doesn't need to separate the cases for $m$ even and odd, but there is one step that seems flawed if the $m$ even case is not dispensed with in advance.

Now let $m$ be the smallest positive integer such that $mp$ is the sum of four squares, $x_1^2+x_2^2+x_3^2+x_4^2$. We show by contradiction that $m$ equals $1$: supposing it is not the case, we prove the existence of a positive integer $r$ less than $m$, for which $rp$ is also the sum of four squares (this is the "infinite descent" method of Fermat).

For this purpose, we consider for each $x_i$ the $y_i$ which is in the same residue class modulo $m$ and between $-m/2+1$ and $m/2$ (included). It follows that $y_1^2+y_2^2+y_3^2+y_4^2=mr$, for some positive integer $r$ less than $m$.

What justifies the claim that $r<m$? From $y_i\in[-m/2+1,m/2]$ we get only $\sum y_i^2\le m^2$, and there still remains the case when each $y_i$ is equal to $m/2$.

Sorry to ask such a simple question, but there aren't any other proofs that I've seen that try to take this approach, and it would make my job easier (I'm writing a formal proof) if this approach is valid, since it saves me a few lemmas.

$\endgroup$
1
$\begingroup$

Assume $y_1=y_2=y_3=y_4=m/2$. Then $x_i=mk_i + m/2$, for some non-negative integers $k_1,...,k_4$ (because we can assume from the beginning of the problem that $x_1,...,x_4$ are non-negative). But this gives $mp=x_1^2+x_2^2+x_3^2+x_4^2=m^2(k_1^2+...+k_4^2+k_1+...+k_4+1)$, that is, $p=m(k_1^2+...+k_4^2+k_1+...+k_4+1)$, which is a contradiction since $p$ is prime.

Then $y_1,...,y_4$ cannot be all equal to $m/2$, and therefore $y_1^2+...+y_4^2<m^2$, which implies $r<m$.

$\endgroup$
  • $\begingroup$ Why the nonnegativity assumption? Seems it works fine without it. $\endgroup$ – Mario Carneiro Jul 16 '14 at 0:43
  • $\begingroup$ Well, I just wanted to avoid some troubles. What if $k_1^2+...+k_4^2+k_1+...+k_4 = 0$ and $m$ is prime? $\endgroup$ – Marco Flores Jul 16 '14 at 0:47
  • $\begingroup$ Oh, I see. But if that sum is zero, then $m=p$ and thus is odd ($p$ is an odd prime), in contradiction to $y_1=m/2$ being an integer. $\endgroup$ – Mario Carneiro Jul 16 '14 at 0:58
  • 1
    $\begingroup$ Hm, I got a bit farther in this problem and now I see that there is a similar problem in eliminating the case $r=0$, which happens when $y_i=0$. By a similar argument, you get that $m|p\implies m=p$, but now that's not a contradiction because $m$ could be odd in this case too. What gives? $\endgroup$ – Mario Carneiro Jul 16 '14 at 20:19
  • 1
    $\begingroup$ The previous paragraph at Wikipedia's proof shows the existence of an integer $n$, such that $np$ is the sum of four squares and $0<n<p$. Thus the $m$ may be choosen at the beginning satisfying $m<p$, and then $m=p$ is a contradiction. $\endgroup$ – Marco Flores Jul 16 '14 at 21:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.