I wanted to present a proof of Lagrange’s four square theorem that I had seen a few years ago that I really like. I think there’s also some approaches using analytic number theory and algebraic number theory, but this one uses convex geometry (!) First, let me state the theorem (my convention here is that denotes the nonnegative integers):

Theorem (Lagrange). Every nonnegative integer can be written as a sum of four squares, i.e., the function given by is surjective.



In order to make this (reasonably) self-contained, let’s give some definitions we’ll use.

A lattice is a discrete subgroup of which spans in the sense of vector spaces. Basically, a lattice will be some subgroup of which is isomorphic to . If is a basis for , define to be the determinant of . Define the fundamental parallelepiped of to be the set .

Now we just need some convex geometry. The main tool will be Minkowski’s theorem. The proof is short, so I’ll include it.

Lemma (Blichfeldt). Let be a lattice and be a measurable set. If , then there exist distinct such that .

Proof. Let be the fundamental parallelepiped of . Then (disjoint union), and hence . Define . Then

.

Since each , there must exist distinct such that . Take and set and .

Lemma (Minkowski). Let be a lattice and be a centrally symmetric (i.e., implies ) convex set such that . Then contains a nonzero element of .

Proof. Set , so . Then there exist distinct such that . Since , we have .

Now we can write down a proof of Lagrange’s four square theorem.

Proof. Pick . We break the proof up into three steps.

Step 1: Reduce to the case that is prime. The sum of four squares is the square of the norm of a quaternion . Since norm is multiplicative, the product of two sums of four squares is itself a sum of four squares. This can be checked directly, but it’s not really that interesting.

Step 2: Find such that . If , take , . Otherwise, assume is odd. Define

. Choose with . Then , and since , which implies that , and hence . So has elements.

Similarly, define . Pick such that . Then as before, , so also has elements. Hence cannot be empty by the pigeonhole principle, so we can find and such that .

Step 3: Construct a suitable lattice and centrally symmetric convex body such that a lattice point in will correspond to an expression of as a sum of four squares. Define . Being a subgroup of , it is clear that is discrete. Also, the set surjects onto under the projection, so has finite index in , and hence has full rank and . Define . Since

,

we can apply Minkowski’s theorem to find such that . Now implies , and hence is a multiple of . From , this multiple must be 1.

And there you have it!

-Steven