Fermat's Two Squares Theorem

Theorem
Let $p$ be a prime number.

Then $p$ can be expressed as the sum of two squares either:
 * $p = 2$

or:
 * $p \equiv 1 \pmod 4$

The expression of a prime of the form $4 k + 1$ as the sum of two squares is unique except for the order of the two summands.

Proof of Existence
There are three possibilities for a prime:
 * $(1): \quad p = 2$

or:
 * $(2): \quad p \equiv 1 \pmod 4$

or:
 * $(3): \quad p \equiv 3 \pmod 4$

Necessary Condition
Suppose $p$ can be expressed as the sum of two squares.

First we note that $2 = 1^2 + 1^2$, which is the sum of two squares.

This disposes of the case where $p = 2$.

Let $p = a^2 + b^2$.

From Sum of Two Squares not Congruent to 3 modulo 4, $p \not \equiv 3 \pmod 4$ whatever $a$ and $b$ are.

So either $p = 2$, or $p \equiv 1 \pmod 4$.

Sufficient Condition
We have already noted that $2 = 1^2 + 1^2$, which is the sum of two squares.

Let $p$ be a prime number of the form $p \equiv 1 \pmod 4$.

Suppose $m p = x^2 + y^2$ has a solution such that $1 < m < p$.

Let $u, v$ be the least absolute residues modulo $m$ of $x$ and $y$ respectively.

That is:
 * $u \equiv x, v \equiv y \pmod m: \dfrac {-m} 2 < u, v \le \dfrac m 2$

Then:
 * $u^2 + v^2 \equiv x^2 + y^2 \pmod m$

Thus:
 * $\exists r \in \Z, r \ge 0: u^2 + v^2 = m r$

We are going to establish a descent step.

That is, we aim to show that $r p$ is the sum of two squares with $1 \le r < m$.

First we show that $r$ does lie in this range.

If $r = 0$ then $u = v = 0$ and so $m$ divides both $x$ and $y$.

But then from $m p = x^2 + y^2$ we have that $m \divides p$.

This cannot happen as $p$ is prime.

So:
 * $1 \le r = \dfrac {u^2 + v^2} m \le \dfrac 1 m \times \paren {\dfrac {m^2} 4 + \dfrac {n^2} 4} = \dfrac m 2 < m$

So $1 \le r < m$.

Now we show that $r p$ is the sum of two squares.

Multiplying $m p = x^2 + y^2$ and $m r = u^2 + v^2$:

Now:
 * $x u + y v \equiv x^2 + y^2 \equiv 0 \pmod m$, so $m \divides x u + y v$
 * $x v - y u \equiv x y - x y \equiv 0 \pmod m$, so $m \divides x v - y u$

So, putting $m X = x u + y v, m Y = x v - y u$, we get:


 * $m^2 r p = m^2 X^2 + m^2 Y^2$

That is:
 * $r p = X^2 + Y^2$

Hence the descent step is established.

Next we need to show that $m p = x^2 + y^2$ has a solution for some $m$ with $1 \le m < p$.

From First Supplement to Law of Quadratic Reciprocity, we have that $-1$ is a quadratic residue for each prime $p \equiv 1 \pmod 4$.

Hence the congruence $x^2 + 1 \equiv 0 \pmod p$ has a least positive solution $x_1$ such that $1 \le x_1 \le p - 1$.

So there exists a positive integer $m$ such that $m p = x_1^2 + 1^2$.

This is just what we want, because:
 * $m = \dfrac {x_1^2 + 1^2} p \le \dfrac {\paren {p - 1}^2 + 1} p = \dfrac {p^2 - 2 \paren {p - 1}^2} p < p$

If this solution has $m > 1$, then our descent step (above) guarantees a solution for a smaller positive value of $m$.

Eventually we will reach a solution with $m = 1$, that is:
 * $p = x^2 + y^2$

Also see

 * Definition:Pythagorean Prime