Fermat's Two Squares Theorem

Theorem
Let $$p$$ be a prime number.

Then $$p$$ can be expressed as the sum of two squares iff:
 * $$p = 2$$, 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) $$p = 2$$, or
 * 2) $$p \equiv 1 \pmod 4$$, or
 * 3) $$p \equiv 3 \pmod 4$$.

Sufficient 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 Square Modulo 4, either $$a^2 \equiv 0$$ or $$a^2 \equiv 1 \pmod 4$$. Similarly for $$b^2$$.

So $$a^2 + b^2 \not \equiv 3 \pmod 4$$ whatever $$a$$ and $$b$$ are.

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

Necessary 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:
 * $$u \equiv x, v \equiv y \pmod m: \frac {-m} 2 < u, v \le \frac m 2$$.

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 $$m \backslash p$$ which can't happen as $$p$$ is prime.

So:
 * $$1 \le r = \frac {u^2 + v^2} m \le \frac 1 m \times \left({\frac {m^2} 4 + \frac {n^2} 4}\right) = \frac m 2 < m$$.

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 $$m \backslash x u + y v$$.
 * $$x v - y u \equiv x y - x y \equiv 0 \pmod m$$, so $$m \backslash x v - y u$$.

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 the First Supplement to the 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:
 * $$m = \frac {x_1^2 + 1^2} p \le \frac {\left({p-1}\right)^2 + 1} p = \frac {p^2 - 2\left({p-1}\right)^2} p < p$$.

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:
 * $$p = x^2 + y^2$$.

Proof of Uniqueness
Let $$p$$ be prime such that $$p \equiv 1 \pmod 4$$.

Suppose $$p = a^2 + b^2 = c^2 + d^2$$ where $$a > b > 0$$ and $$c > d > 0$$.

We are going to show that $$a = c$$ and $$b = d$$.

From the two expressions for $$p$$, we have:

$$ $$ $$ $$ $$

So we have $$\left({a d - b c}\right) \left({a d + b c}\right) \equiv 0 \pmod p$$.

From Euclid's Lemma, that means $$p \backslash \left({a d - b c}\right)$$ or $$p \backslash \left({a d + b c}\right)$$.

So, suppose $$p \backslash \left({a d + b c}\right)$$.

Now, we have that each of $$a^2, b^2, c^2, d^2$$ must be less than $$p$$.

Hence $$0 < a, b, c, d < \sqrt p$$.

This implies $$0 < a d + b c < 2p$$.

That must mean that $$a d + b c = p$$.

But then:

$$ $$ $$

That means $$a c + b d = 0$$

But since $$a > b$$ and $$c > d$$ we have $$a c > b d$$.

This contradiction shows that $$a d + b c$$ can not be divisible by $$p$$.

So this means $$p \backslash \left({a d - b c}\right)$$.

Similarly, because $$0 < a, b, c, d < \sqrt p$$ we have $$-p < a d - b c < p$$

This means $$a d = b c$$.

So $$a \backslash b c$$.

But $$a \perp b$$ otherwise $$a^2 + b^2$$ has a common divisor greater than $$1$$ and less than $$p$$ and it can't because $$p$$ is prime.

So by Euclid's Lemma $$a \backslash c$$.

So we can put $$c = k a$$ and so $$a d = b c$$ becomes $$d = k b$$.

Hence:
 * $$p = c^2 + d^2 = k^2 \left({a^2 + b^2}\right) = k^2 p$$.

This means $$k = 1$$ which means $$a = c$$ and $$b = d$$ as we wanted to show.

Historical Note
This theorem was initially stated without proof by Albert Girard in 1632.

Fermat announced its proof in a letter to Marin Mersenne dated December 25, 1640.

For this reason it is sometimes referred to as Fermat's Christmas Theorem.