Integer as Sum of Two Squares

Theorem
Let $n$ be a positive integer.

Then:
 * $n$ can be expressed as the sum of two squares


 * each of its prime divisors of the form $4 k + 3$ (if any) occur to an even power.
 * each of its prime divisors of the form $4 k + 3$ (if any) occur to an even power.

Proof
Let us extract the largest square divisor of $n$, and write:
 * $n = m^2 r$

where $r$ is square-free.

Necessary Condition
Suppose $r$ has no prime divisor in the form $4 k + 3$.

If $r = 1$ then $n^2 = m^2 + 0^2$ and nothing needs to be proved.

If $r > 1$ then $r$ is a product of one or more primes, each of which is either $2$ or in the form $4 k + 1$.

Now from Fermat's Two Squares Theorem, each of these can be expressed as the sum of two squares.

By the extension to the Brahmagupta-Fibonacci Identity, the product of all these can itself be expressed as the sum of two squares.

So $r = a^2 + b^2$, say.

Then:
 * $n = m^2 \left({a^2 + b^2}\right) = \left({m a}\right)^2 + \left({m b}\right)^2$

So $n$ can be expressed as the sum of two squares.

Sufficient Condition
Now suppose that $n$ can be expressed as the sum of two squares, say:
 * $n = m^2 r = a^2 + b^2$

First, any common divisor of $a$ and $b$ may be cancelled as follows.

If $\gcd \left\{{a, b}\right\} = d$, then we can write:
 * $a = a' d, b = b' d$

where $\gcd \left\{{a', b'}\right\} = 1$ from Integers Divided by GCD are Coprime.

So:
 * $m^2 r = d^2 \left({a'^2 + b'^2}\right)$

As $r$ is square-free, $d \mathrel \backslash m$ and so, writing $m' = \dfrac m d$, we get:
 * $m'^2 r = a'^2 + b'^2$

We need to show that $r$ has no prime factor of the form $4 k + 3$.

To obtain a contradiction, suppose $p = 4 k + 3$ divides $r$.

Then:
 * $a'^2 + b'^2 \equiv 0 \implies a'^2 \equiv -b'^2 \pmod p$

If $p \mathrel \backslash a'$ we would have:
 * $p \mathrel \backslash b'$

which contradicts $\gcd \left\{{a', b'}\right\} = 1$.

So:
 * $\gcd \left\{{a', p}\right\} = \gcd \left\{{b', p}\right\} = 1$

and we can apply Fermat's Little Theorem:
 * $a'^{p-1} \equiv b'^{p-1} \equiv 1 \pmod p$

So, we put $p = 4 k + 3$:

$-1 \equiv 1 \pmod p$ is not possible for an odd prime.

So we have obtained our contradiction.

The result follows.