proof of Lagrange’s four-square theorem

The following proof is essentially Lagrange’s original, from around 1770. First, we need three lemmas.

Lemma 1.

For any integers $a\mathrm{,}b\mathrm{,}c\mathrm{,}d\mathrm{,}w\mathrm{,}x\mathrm{,}y\mathrm{,}z$,

$(a^2+b^2+c^2+d^2)(w^2+x^2+y^2+z^2)$$=$${(aw+bx+cy+dz)}^2$

$+$${(ax-bw-cz+dy)}^2$

$+$${(ay+bz-cw-dx)}^2$

$+$${(az-by+cx-dw)}^2.$

This is the Euler four-square identity, q.v., with different notation.

Lemma 2.

If $\mathrm{2}\mathit{}m$ is a sum of two squares, then so is $m$.

Proof.

Say $2m=x^2+y^2$. Then $x$ and $y$ are both even or both odd.Therefore, in the identity

$$m={\left(\frac{x-y}{2}\right)}^2+{\left(\frac{x+y}{2}\right)}^2,$$

both fractions on the right side are integers. ∎

Lemma 3.

If $p$ is an odd prime,then $a^{\mathrm{2}}\mathrm{+}{b}^{\mathrm{2}}\mathrm{+}\mathrm{1}\mathrm{=}k\mathit{}p$ for some integers $a\mathrm{,}b\mathrm{,}k$ with .

Proof.

Let $p=2n+1$. Consider the sets

$$A:=\{a^2\mid a=0,1,\mathrm{\dots },n\}\mathit{\hspace{1em}}\text{ and }\mathit{\hspace{1em}}B:=\{-b^2-1\mid b=0,1,\mathrm{\dots },n\}.$$

We have the following facts:

1. No two elements in $A$ are congruent mod $p$, for if $a^2\equiv c^2(modp)$, then either $p\mid (a-c)$ or $p\mid (a+c)$ by unique factorization of primes. Since , and $0\le a,c$, we must have $a=c$.

2. Similarly, no two elements in $B$ are congruent mod $p$.

3. Furthermore, $A\cap B=\mathrm{\varnothing }$ since elements of $A$ are all non-negative, while elements of $B$ are all negative.

4. Therefore, $C:=A\cup B$ has $2n+2$, or $p+1$ elements.

Therefore, by the pigeonhole principle, two elements in $C$ must be congruent mod $p$. In addition, by the first two facts, the two elements must come from different sets. As a result, we have the following equation:

$$a^2+b^2+1=kp$$

for some $k$. Clearly $k$ is positive. Also, $p^2={(2n+1)}^2>2n^2+1\ge a^2+b^2+1=kp$, so $p>k$.∎

Basically, Lemma 3 says that for any prime $p$, some multiple of $p$ is a sum of four squares, since $a^2+b^2+1=a^2+b^2+1^2+0^2$.

Proof of Theorem.

By Lemma 1 we need only show that an arbitrary prime $p$ is a sum offour squares. Since that is trivial for $p=2$, suppose $p$ is odd. By Lemma 3, we know

$$mp=a^2+b^2+c^2+d^2$$

for some $m,a,b,c,d$ with . If $m=1$, then we are done. To complete the proof, we will show that if $m>1$ then $np$ is a sum of four squares for some $n$ with .

If $m$ is even, then none, two, or all four of $a,b,c,d$ are even; in any of those cases, we may break up $a,b,c,d$ into two groups, each group containing elements of the same parity. Then Lemma 2 allows us to take $n=m/2$.

Now assume $m$ is odd but $>1$. Write

$w$$\equiv$$a(modm)$

$x$$\equiv$$b(modm)$

$y$$\equiv$$c(modm)$

$z$$\equiv$$d(modm)$

where $w,x,y,z$ are all in the interval $(-m/2,m/2)$. We have

$$w^2+x^2+y^2+z^2\equiv 0(modm).$$

So $w^2+x^2+y^2+z^2=nm$ for some integer non-negative $n$. Since , . In addition, if $n=0$, then $w=x=y=z=0$, so that $a\equiv b\equiv c\equiv d\equiv 0(modm)$, which implies $mp=a^2+b^2+c^2+d^2=m^{2}q$, or that $m|p$. But $p$ is prime, forcing $m=p$, and contradicting . So . Look at the product $(a^2+b^2+c^2+d^2)(w^2+x^2+y^2+z^2)$ and examine Lemma 1. On the left is $n{m}^{2}p$. One the right, we have a sum of four squares. Evidently three of them

$$ax-bw-cz+dy=(ax-bw)+(dy-cz)$$

$$ay+bz-cw-dx=(ay-cw)+(bz-dx)$$

$$az-by+cx-dw=(az-dw)+(cx-by)$$

are multiples of $m$. The same is true of the other sum on theright in Lemma 1:

$$aw+bx+cy+dz\equiv w^2+x^2+y^2+z^2\equiv 0(modm).$$

The equation in Lemma 1 can therefore be divided through by $m^2$.The result is an expression for $np$ as a sum of four squares.Since , the proof is complete.∎

Remark: Lemma 3 can be improved: it is enough for$p$ to be an odd number, not necessarily prime.But that stronger statement requires a longer proof.