Geometry of Numbers, Lecture 4: Lagrange’s Four Squares Theorem (Mike)

We aim to give a proof of the following theorem, by using Minkowski’s First Theorem.

Theorem (Lagrange)

Every positive integer is the sum of four squares.

To establish this theorem, we shall require 3 lemmata.

Lemma 1

Let $m$ be an odd positive integer, then there exist $a,b \in \mathbb{Z}$ s.t.

$a^{2}+b^{2}+1 \equiv 0 \pmod{m}$ .

Proof
We split into 3 cases.
(i) $m = p$ , an odd prime.

Let

$A = \{a^{2}, a=0,1,\ldots,(p-1)/2\},$
$B = \{-b^{2}-1, b=0,1,\ldots,(p-1)/2\}.$

Clearly $|A| = |B| = (p+1)/2$ and the elements of $A$ (resp. $B$ ) are pairwise incogruent modulo $p$ . To see this,
assume $a, a' \in A$ satisfy $a^{2} \equiv a'^{2} \pmod{p}$ . Clearly $a' \equiv 0 \Rightarrow a \equiv 0$ . Assume
$a' \not\equiv 0 \pmod{p}$ . Then $(aa'^{-1})^{2} \equiv 1 \pmod{p}$ . By Lagrange’s theorem,

$x^{2} \equiv 1 \pmod{p}$

has only 2 solutions, and these are $-1$ and $1$ . If $aa'^{-1} \equiv -1 \pmod{p}$ , then $a \equiv -a' \pmod{p}$ , which is a contradiction to how $A$ was defined.

So $aa'^{-1} \equiv 1 \pmod{p}$ , and hence $a \equiv a' \pmod{p}$ . Similarly for $B$ .

So by the pigeonhole priniciple, the 2 sets can’t be distinct, and it follows that there exist integers $a,b$ such that $a^{2} + b^{2} + 1 \equiv 0 \pmod{p}$ .

(ii) $m = p^{k}, k\geq 1$ , $p$ odd prime.

We proceed by induction. We have the case $k=1$ from before. For some $k \geq 1$ , assume there exist integers $a,b$ such that

$a^{2} + b^{2} +1 \equiv 0 \pmod{p^{k}}$ .

Then there exists some integer $s$ such that

$a^{2} + b^{2} +1 = sp^{k}$ .

Clearly $p$ cant divide both $a$ and $b$ . Assume $a \not\equiv 0 \pmod{p}$ . Then we have $(2a,p)=1$ , and so there exists some integer $t$ such that $s+2at \equiv 0 \pmod{p}$ . This can be found by solving the equation.

Let $a_{1} = a + tp^{k}$ .

Then we have

$a_{1}^{2} = a^{2} + 2atp^{k} +t^{2}p^{2k}$
$=-b^{2} -1 +sp^{k}+2atp^{k} +t^{2}p^{2k}$
$\equiv -b^{2} -1 + (s+2at)p^{k} \pmod{p^{k+1}}$
$\equiv -b^{2} -1 \pmod{p^{k+1}}.$

Hence this case follows by induction.

(iii) $m$ is an odd positive integer

The case $m=1$ is trivial. So let

$m = \prod_{i=1}^{r}p_{i}^{k_{i}},$

where the $p_{i}$ are odd primes, and $k_{i} \geq 1$ are integers.

Then for each $i$ , we have integers $a_{i},b_{i}$ such that

$a_{i}^{2} + b_{i}^{2} +1 \equiv 0 \pmod{p_{i}^{k_{i}}}.$

We can then use the Chinese remainder theorem to find integers $a,b$ such that

$a \equiv a_{i} \pmod{p_{i}^{k_{i}}}, b \equiv b_{i} \pmod{p_{i}^{k_{i}}}$

for each $1 \leq i \leq r$ . It follows that

$a^{2} + b^{2} + 1 \equiv 0 \pmod{m}.$

Lemma 2
If every odd positive integer is the sum of four squares, then every positive integer is the sum of four squares.

Proof
If some integer $n$ is the sum of four squares, say

$n= a^{2} + b^{2} + c^{2} +d^{2}.$

Then we have

$2n = (a+b)^{2} + (a-b)^{2} + (c+d)^{2} + (c-d)^{2}.$

We can continue like this to show that $2^{k}n$ is the sum of four squares for any $k \geq 0$ .
As any positive integer is of the form $2^{k}n$ , for some $k \geq 0$ , and some odd integer $n$ , the lemma follows.

Lemma 3
Let $B_{r}(\textbf{0})$ be the ball of radius $r$ in $\mathbb{R}^{4}$ . Then

$\text{vol}(B_{r}(\textbf{0}))= \pi^{2}r^{4}/2.$

Proof
It is not hard to see that the volume of this ball is equal to

$\int_{-r}^{r}\frac{4}{3}\pi(r^{2}-z^{2})^{3/2}\text{d}z.$

Solve this by using the substitution $z = r\sin{\theta}$ .

We are now ready to prove the main theorem
Proof
By lemma 2, it suffices to prove for odd positive integers. Let $m$ be an odd positive integer. By lemma 1, there exist integers $a,b$ such that

$a^{2} + b^{2} + 1 \equiv 0 \pmod{m}.$

Let $\Gamma \subset \mathbb{Z}^{4}$ be the lattice with basis vectors

$\textbf{a}_{1} = (m,0,0,0)$
$\textbf{a}_{2} = (0,m,0,0)$
$\textbf{a}_{3} = (a,b,1,0)$
$\textbf{a}_{4} = (b,-a,0,1).$

We have $\text{det}(\Gamma) = m^{2}$ , and $u \in \Gamma$ can be written as

$\textbf{u} = u_{1}\textbf{a}_{1} +u_{2}\textbf{a}_{2} +u_{3}\textbf{a}_{3} +u_{4}\textbf{a}_{4}$
$(u_{1}m+u_{3}a+u_{4}b)\textbf{e}_{1}+(u_{2}m+u_{3}b-u_{4}a)\textbf{e}_{2}+u_{3}\textbf{e}_{3}+u_{4}\textbf{e}_{4},$

with $u_{1},\ldots,u_{4} \in \mathbb{Z}$ .
So

$|\textbf{u}|^{2} = (u_{1}m+u_{3}a+u_{4}b)^{2} + (u_{2}m+u_{3}b-u_{4}a)^{2} + u_{3}^{2} + u_{4}^{2}$
$\equiv u_{3}(a^{2} + b^{2} + 1) + u_{4}(a^{2} + b^{2} + 1) \pmod{m}$
$\equiv 0 \pmod{m},$

for all $\textbf{u} \in \Gamma$ .

Let $K = B_{\sqrt{2m}}(\textbf{0})$ be the ball of radius $2m$ in $\mathbb{R}^{4}$ . It is clear this is a symmetric convex body, amd has volume

$2\pi^{2}m^{2} > 16m^{2} = 2^{4}\text{det}(\Gamma)$ .

Hence we can apply Minkowski’s first theorem, and we have $K$ contains some non-trivial lattice point

$\textbf{u} = u_{1}\textbf{a}_{1} +u_{2}\textbf{a}_{2} +u_{3}\textbf{a}_{3} +u_{4}\textbf{a}_{4}$
$= v_{1}\textbf{e}_{1} + v_{2}\textbf{e}_{2} + v_{3}\textbf{e}_{3} + v_{4}\textbf{e}_{4},$