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,

(a2+b2+c2+d2)(w2+x2+y2+z2) = (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 2m is a sum of two squares, then so is m.


Say 2m=x2+y2. Then x and y are both even or both odd. Therefore, in the identity


both fractions on the right side are integers. ∎

Lemma 3.

If p is an odd prime, then a2+b2+1=kp for some integers a,b,k with 0<k<p.


Let p=2n+1. Consider the sets

A:={a2a=0,1,,n} and B:={-b2-1b=0,1,,n}.

We have the following facts:

  1. 1.

    No two elements in A are congruentMathworldPlanetmath mod p, for if a2c2(modp), then either p(a-c) or p(a+c) by unique factorization of primes. Since a-c,a+c2n<p, and 0a,c, we must have a=c.

  2. 2.

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

  3. 3.

    Furthermore, AB= since elements of A are all non-negative, while elements of B are all negative.

  4. 4.

    Therefore, C:=AB has 2n+2, or p+1 elements.

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


for some k. Clearly k is positive. Also, p2=(2n+1)2>2n2+1a2+b2+1=kp, so p>k. ∎

Basically, Lemma 3 says that for any prime p, some multipleMathworldPlanetmath 0<m<p of p is a sum of four squares, since a2+b2+1=a2+b2+12+02.

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


for some m,a,b,c,d with 0<m<p. If m=1, then we are done. To completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof, we will show that if m>1 then np is a sum of four squares for some n with 1n<m.

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 a(modm)
x b(modm)
y c(modm)
z d(modm)

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


So w2+x2+y2+z2=nm for some integer non-negative n. Since w2+x2+y2+z2<m2, n<m. In addition, if n=0, then w=x=y=z=0, so that abcd0(modm), which implies mp=a2+b2+c2+d2=m2q, or that m|p. But p is prime, forcingMathworldPlanetmath m=p, and contradicting m<p. So 0<n<m. Look at the product (a2+b2+c2+d2)(w2+x2+y2+z2) and examine Lemma 1. On the left is nm2p. One the right, we have a sum of four squares. Evidently three of them


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


The equation in Lemma 1 can therefore be divided through by m2. The result is an expression for np as a sum of four squares. Since 0<n<m, the proof is complete. ∎

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

Title proof of Lagrange’s four-square theorem
Canonical name ProofOfLagrangesFoursquareTheorem
Date of creation 2013-03-22 13:21:07
Last modified on 2013-03-22 13:21:07
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 13
Author CWoo (3771)
Entry type Proof
Classification msc 11P05