Three Squares Theorem by Geometry of Numbers

Three Squares Theorem: if ${m}$ is a positive integer not of the form ${4^{a}(8 n+7)}$ , then ${m}$ is the sum of three squares.

We present a proof using geometry of numbers due to Ankeny.

We prove it for the case ${m\equiv 3(\bmod 8)}$ , and squarefree. Let ${m=p_{1} p_{2} \cdots p_{r}}$

Find a prime ${q}$ such that ${\begin{aligned} \left(\frac{-2 q }{ p_{i}}\right) =+1, q \equiv 1(\bmod 4) \end{aligned}}$ for all ${i}$ .

$\displaystyle \left(\frac{-m}{ q}\right) =\left(\frac{-2}{ m}\right) \left(\frac{-m}{ q}\right)$

$\displaystyle =\left(\frac{-2}{ m}\right) \left(\frac{m}{ q}\right)$

$\displaystyle =\left(\frac{-2}{ m}\right)\prod_{i} \left(\frac{p_i}{ q}\right)$

$\displaystyle =\left(\frac{-2}{ m}\right)\prod_{i} \left(\frac{q}{ p_i}\right) =\prod_{i} \left(\frac{-2q}{ p_i}\right)=1$
Therefore we find solutions to ${b^{2}-4 q h=-m.}$

${\left(\frac{-2 q }{ p_{i}}\right) =+1}$ implies ${t^{2} \equiv\frac{-1 }{ 2 q}(\bmod m)}$ for some ${t.}$

$\displaystyle R=2tqX+t b Y+m Z$

$\displaystyle S=(2q)^{1/2}X+ \frac{b}{(2q)^{1/2}}Y$

$\displaystyle T=\frac{m^{1 / 2} }{(2 q)^{1 / 2}} Y$

Now consider the ${R^{2}+S^{2}+T^{2}<2 m}$ which represent a symmetric convex region of area ${\frac{2^{7 / 2} \pi}{3} >8.}$
Hence by Minkowski’s theorem contains a non-zero integers point ${(X,Y,Z).}$

Note that ${S,T}$ are not integers.

$\displaystyle \begin{aligned} R^{2}+S^{2}+T^{2}=&\left(2tqX+t b Y+m Z\right)^{2} \\ &+\left( (2q)^{1/2}X+ \frac{b}{(2q)^{1/2}}Y\right)^{2}+\left( \frac{m^{1 / 2} }{(2 q)^{1 / 2}} Y\right)^{2} \\ \equiv & ~~t^{2}\left(2 qX+bY\right)^{2}+\frac{1}{2 q}\left(2 qX+bY\right)^{2} \\ \equiv &~~ 0~~(\bmod m) \end{aligned}$

$\displaystyle R^{2}+S^{2}+T^{2}=R^{2}+2\left(qX^{2}+b XY+hY^{2}\right)$ ${R^{2}+S^{2}+T^{2}<2 m}$ implies that ${R^{2}+2\left(qX^{2}+b XY+hY^{2}\right) =m}$

If we show that ${v=2\left(qX^{2}+b XY+hY^{2}\right) }$ is a sum of two squares we are done.

Let ${p}$ an odd prime dividing ${v}$ with odd multiplicity.

${2q v=\left(2 qX+bY\right)^{2}+m Y^{2}}$

Assume ${p}$ doesnt divide ${m}$ – we see that ${\left(\frac{m}{ p}\right)=1}$
If ${p}$ divides ${q}$ , ${b^{2}-4 q h=-m}$ gives ${\left(\frac{-m}{ p}\right)=1}$
if ${p}$ doesnt divide ${q}$ , ${p^{2 n+1} || \left(2 qX+bY\right)^{2}+m Y^{2}}$ , so ${\left(\frac{-m}{ p}\right)=1}$
Thus we have ${\left(\frac{m}{ p}\right)=1}$ and ${\left(\frac{-m}{ p}\right)=1}$ which implies ${\left(\frac{-1}{ p}\right)=1}$

If ${p}$ divides ${m}$ , then ${\frac{1}{2 q} \frac{m}{p}Y^{2}=\frac{m}{p}(\bmod p)}$

${Y^{2} \equiv 2 q~(\bmod p) \implies \quad\left(\frac{2 q}{p}\right)=+1}$ But ${\left(\frac{-2 q }{ p_{i}}\right) =+1}$ , for ${p_i}$ dividing ${m}$ . Hence ${\left(\frac{-}{p}\right)=1}$

In all the cases, we get ${p \equiv 1(\bmod 4),}$ thereby establishing that all odd factors of ${v}$ with odd multiplicity are ${1}$ modulo ${4}$ which implies that ${v}$ can be written as sum of squares.