# 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 $a,b,c,d,w,x,y,z$,

 $\displaystyle(a^{2}+b^{2}+c^{2}+d^{2})(w^{2}+x^{2}+y^{2}+z^{2})$ $\displaystyle=$ $\displaystyle(aw+bx+cy+dz)^{2}$ $\displaystyle+$ $\displaystyle(ax-bw-cz+dy)^{2}$ $\displaystyle+$ $\displaystyle(ay+bz-cw-dx)^{2}$ $\displaystyle+$ $\displaystyle(az-by+cx-dw)^{2}.$

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

###### Lemma 2.

If $2m$ 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

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

both fractions on the right side are integers. ∎

###### Lemma 3.

If $p$ is an odd prime, then $a^{2}+b^{2}+1=kp$ for some integers $a,b,k$ with $0.

###### Proof.

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

 $A:=\{a^{2}\mid a=0,1,...,n\}\quad\mbox{ and }\quad B:=\{-b^{2}-1\mid b=0,1,...% ,n\}.$

We have the following facts:

1. 1.

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

2. 2.

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

3. 3.

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

4. 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, $p^{2}=(2n+1)^{2}>2n^{2}+1\geq a^{2}+b^{2}+1=kp$, so $p>k$. ∎

Basically, Lemma 3 says that for any prime $p$, some multiple  $0 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 of four 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 $0. 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 $1\leq n.

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

 $\displaystyle w$ $\displaystyle\equiv$ $\displaystyle a\pmod{m}$ $\displaystyle x$ $\displaystyle\equiv$ $\displaystyle b\pmod{m}$ $\displaystyle y$ $\displaystyle\equiv$ $\displaystyle c\pmod{m}$ $\displaystyle z$ $\displaystyle\equiv$ $\displaystyle d\pmod{m}$

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

 $w^{2}+x^{2}+y^{2}+z^{2}<4\cdot\frac{m^{2}}{4}=m^{2}$
 $w^{2}+x^{2}+y^{2}+z^{2}\equiv 0\pmod{m}.$

So $w^{2}+x^{2}+y^{2}+z^{2}=nm$ for some integer non-negative $n$. Since $w^{2}+x^{2}+y^{2}+z^{2}, $n. In addition, if $n=0$, then $w=x=y=z=0$, so that $a\equiv b\equiv c\equiv d\equiv 0\pmod{m}$, 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 $m. So $0. 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 $nm^{2}p$. 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 the right in Lemma 1:

 $aw+bx+cy+dz\equiv w^{2}+x^{2}+y^{2}+z^{2}\equiv 0\pmod{m}.$

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 $0, 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.

Title proof of Lagrange’s four-square theorem ProofOfLagrangesFoursquareTheorem 2013-03-22 13:21:07 2013-03-22 13:21:07 CWoo (3771) CWoo (3771) 13 CWoo (3771) Proof msc 11P05