# Classify all integers which are the sum of two squares but can't be written as the sum of two nonzero squares

So immediately I just want to say that all these integers n must themselves be squares. Then, I know the following from class:

1. primes that are the sums of two squares are 1 (mod 4)
2. n, an integer, is the sum of two squares if it's prime factorization has primes congruent congruent to 3 (mod 4) occurring only with even powers
3. products of sums of two squares are also the sums of two squares
• You want $n=x^2+y^2$ and $n=z^2$, right? So we have all Pythagorean triples. – Dietrich Burde Nov 16 '19 at 20:05
• wrong @Dieterich we want $$n=x^2+y^2\land (x=0\lor y=0)$$ – user645636 Nov 16 '19 at 22:28

Let $$n=x^{2}+y^{2}$$ be such an integer. if $$x \neq 0$$ and $$y \neq 0$$ we have a contradiction. So we may assume that $$y = 0$$.

Therefore $$n = x^{2}$$.$$\$$ Hence $$n$$ is a square.

Assume $$n>1$$, since $$n$$ can't be written as as the sum of two nonzero squares, all of its odd prime factors must be $$3$$ mod $$4$$.

If not, assume $$p | n$$ and $$p = 1$$ mod $$4$$

Then we may write:$$\ \ \ p=a^{2}+b^{2} \Rightarrow p^2 = (a^{2}-b^{2})^{2}+(2ab)^{2}.$$

$$n = \frac{x^{2}}{p^{2}} p^{2} = [\frac{x}{p}(a^{2}-b^{2})]^{2}+[\frac{x}{p}(2ab)]^{2}$$ which is the sum of 2 non-zero squares.

Therefore $$n$$ is either $$0, 1$$ or a number of the form: $$\ \ \ \ 2^{2k}\prod_{ì=1}^{s}p_{i}^{2k_{i}},$$

Where each $$p_{i}$$ is $$3$$ mod $$4$$.

On the other hand, if $$n$$ can be written in such a way, and $$\ n = x^{2}+y^{2}$$,

$$\Rightarrow n = (x + yi)(x-yi)$$, By unique factorization of $$\mathbb{Z}[i]$$, there are 2 cases to consider:

Either one of the $$p_{i}$$ would factor non trivially in $$\mathbb{Z}[i]$$, but it doesn't as $$p_{i}$$ is $$3$$ mod $$4$$.

The only other possibility is that $$\ \ 2^{2k}=a^{2}+b^{2}, \ a \neq 0, \ b \neq 0$$

but you can check that this is not possible by taking congruences mod $$4$$.