4
$\begingroup$

Every positive integer can be written as the sum of 4 squares $n = a_1^2 + a_2^2 + a_3^2 + a_4^2$ however, if we only allow sum of 3 squares some numbers have to be left out:

$n = a^2 + b^2 + c^2$ $\longleftrightarrow$ $n \equiv 4^a (8k+7)$

Excuse me for using the same letter twice to mean different things...


I am trying trying to find a proof of this result from within Cassels' Rational Quadratic Forms, but it's not always the easiest to read. In that book Legendre's Theorem is stated in a very general way:

Let $g(x) = \color{lightgry}{a}_1 \,x_1^2 + a_2\, x_2^2 + a_3\, x_3^2$ with $a_1, a_2, a_3 \in \mathbb{Z}$ and $a_1 a_2 a_3$ is squarefree...

  • blah
  • blah
  • blah

Then where are $b_1, b_2, b_3 \in \mathbb{Z}$ not all of them zero with $a_1 \,b_1^2 + a_2\, b_2^2 + a_3\, b_3^2 = 0$

From what I gather, this Legendre theorem is strong form of the Hasse Principle and it can be proven using the geometry of numbers - which is the style I am endorsing today.

By the middle of Chapter 9 - Section 5 to be exact - we get the claim that proof of this theorem is complete. Because $x_1^2 + x_2^2 + x_3^2$ is the only quadratic form in it's genus. And that in indeed rested on the Hasse principle.


Come to think of it there is a rather serious question here. Mis-using the Hasse principle slightly we could way:

The only obstruction to having representation as sum of three squares $n = a^2 + b^2 + c^2 $ happens in the 2-adic numbers $\mathbb{Q}_2$.

How could a statement like this possible be proven geometrically? Minkowski's geometry of numbers is decidedly a Euclidean result and so the only possible completion that should be useful is $\mathbb{R}$.

In the case of sum of 2-squares the role of the Hasse principle is even more obvious since we can use multiplication identities like:

\begin{eqnarray} (a+bi)(c+id) &=& (ac-bd) + i(ad+bc) \\ (a^2 + b^2)(c^2 + d^2)&=& (ac-bd)^2 + (ad+bc)^2 \end{eqnarray} and check each time that $n = x_1^2 + x_2^2 \mod p$ and multiply the result. By Minkowski theorem this only works when $p = 4k+1$.


I am going to keep reading, but does anyone know what polyhedron or oval or other geometric shape leads to a proof of the sum of 3 squares theorem?

$\endgroup$
11
$\begingroup$

A proof of the three squares theorem by the geometry of numbers was given by Ankeny in 1957. His paper is available here.

P.S. Also, I think Legendre's proof was incomplete: he assumed Dirichlet's theorem about primes in arithmetic progressions, which was not known at that time. The first correct proof was given by Gauss. I admit I might be wrong here.

$\endgroup$
  • 1
    $\begingroup$ Wikipedia disagrees: wikiwand.com/en/Legendre's_three-square_theorem $\endgroup$ – Igor Rivin Nov 18 '15 at 21:31
  • $\begingroup$ thanks this is much shorter than what I was about to attempt $\endgroup$ – john mangual Nov 18 '15 at 21:32
  • 2
    $\begingroup$ @johnmangual, you give a link to a Pete L. Clark writeup in comment above. In that, he discusses a very pretty approach, which is also in Serre's little book: the sum of three squares is one of Pete's ADC forms, named after Aubry and Davenport-Cassels, and means that, if $x^2 + y^2 + z^2$ represents an integer with rational values for $x,y,z,$ then it also does so with integral values. Pete and I found all positive forms for which this holds, published an article. Turned out we were confirming (and slightly correcting) an existing list by G. Nebe. $\endgroup$ – Will Jagy Nov 18 '15 at 22:29
  • $\begingroup$ @IgorRivin: Yes, probably this is why I thought I was wrong here. At any rate, a modern exposition of Legendre's proof would be welcome (I recall Andre Weil wrote about this in his history book, but I am too lazy and busy to check). Thanks for your comment! $\endgroup$ – GH from MO Nov 18 '15 at 22:54
3
$\begingroup$

A complete proof seems to be given here: http://www.rsabey.pwp.blueyonder.co.uk/maths/sumof3squares.html

$\endgroup$
  • $\begingroup$ I notice Dirichlet's Theorem on prime numbers in arithmetic progressions is an input. So for example there are infinitely many prime numbers among $1,4,7,10,\dots$. And Quadratic Reciprocity of Legendre Symbols that $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}$ is also an input. $\endgroup$ – john mangual Nov 18 '15 at 21:40

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.