Primes 3 mod 4 and sums of two squares

1. Suppose $p\mid a^2+b^2$ and $p\nmid b$. Let $x\equiv ab^{-1}\pmod p$. Then $x^2\equiv-1\pmod p$ and $x\not\equiv0$. By Fermat's little theorem,
$$1\equiv x^{p-1}=(x^2)^{(p-1)/2}\equiv(-1)^{(p-1)/2}\pmod p.$$
Since $p\equiv3\pmod4$, $\frac{p-1}2$ is odd, so the right side is $-1$. So $1\equiv-1$, i.e. $p=2$, contradiction. Hence $p\mid b$, and then $p\mid a^2$, so $p\mid a$.

2. By strong induction on $n$. Let $n=a^2+b^2$ and $p\equiv3\pmod4$ with $p\mid n$. By part 1, $p\mid a,b$, so $n=p^2(a_1^2+b_1^2)$ where $a=pa_1$, $b=pb_1$, and $n/p^2$ is again a sum of two squares, smaller than $n$. By induction, the exponent of $p$ in $n/p^2$ is even, so the exponent in $n$ is even too. (If $p\nmid n$ the exponent is $0$.)

3. Brahmagupta's identity:
$$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2.$$

Examples. $21=3\cdot7$ and $2021=43\cdot47$ are not sums of two squares (primes $\equiv3\pmod4$ with odd exponent); $45=9\cdot5=6^2+3^2$ is.

Lesson. To forbid a representation, find a prime for which the representation forces divisibility of both terms, then descend.