In this note, we are interested in the following well-known result:

**Theorem:** A prime can be written as the sum of two squares if and only if .

There exist many different proofs of this theorem, but a surprising one is exposed by D. Zagier in his article *A one-sentence proof that every prime is a sum of two squares*. Our note is dedicated to this proof.

First of all, it is easy to prove that the sum of two squares is either congruent to , or modulo . Thus, because is necessarily odd, if it can be written as the sum of two squares then .

Conversely, let be a prime satisfying , and define the set

.

Notice that is nonempty since : there exists a such that

.

Now, let be the map defined by

It is not difficult to show that is well-defined on , because for all . Furthermore, an easy calculation proves that the image of is included into .

Now, we want to prove that is an involution with exactly one fixed point.

**The map is an involution.** Let . If , then implies

;

if , then implies

;

if , then implies

.

Therefore, ie. is an involution.

**The map has only one fixed point.** Clearly, if then is a fixed point of . Conversely, let be a fixed point of . If then is a solution of

,

hence , which is impossible. If then is a solution of

,

hence . Thus, implies . Because is prime, necessarily , and we deduce that .

Finally, if then is a solution of

,

hence , which is impossible. Therefore, has exactly one fixed point: .

In order to conclude the proof of our theorem, we need the following lemma:

**Lemma:** Let be a set and an involution. Then .

**Proof.** Let be a maximal subset of satisfying . Then

,

which implies . This proves the lemma.

Now we are able to conclude the proof of our theorem. Let be a new map defined by . Clearly, is a well-defined involution of . By applying our previous lemma twice, we deduce that

.

In particular, is nonempty. Let be one of its points. Then

.

This proves that can be written as the sum of two squares, and the proof is complete.