If \(n\) were to be a sum of two integers, then there are three cases we need to consider: If both the integers are even, if both of them are odd and if one of them is odd while the other is even.:
$$n = a^2 + b^2 = \begin{cases} n = (2x)^2 + (2y)^2 \\ n = (2x+1)^2 + (2y+1)^2 \\ n = (2x)^2 + (2y+1)^2 \end{cases}$$
If \(n\) were to be a sum of two even square integers, then \(n = 4x^2 + 4y^2\), so \(n\) wold have been a multiple of 4 and \(n \not \equiv 3 \bmod 4\).
If \(n\) were to be a sum of two odd square integers, then \(n = 4x^2 + 4x + 4y^2 + 4y + 2\), so \(n\) wold have been a multiple of 2 and \(n \not \equiv 3 \bmod 4\).
If \(n\) were to be a sum of one odd square integer and one even square integer, then:
$$\begin{align} n &= 4x^2 + 4y^2 + 4y + 1 \\ n &= 4(x^2 + y^2 + y) + 1 \\ n - 1 &= 4k \end{align}$$
This means \(n ≡ 1 \bmod 4\), and therefore \(n \not \equiv 3 \bmod 4\). We concluded:
$$n = a^2 + b^2 \implies n \not \equiv 3 \bmod 4 $$
Using the contrapositive:
$$n \equiv 3 \bmod 4 \implies n \not = a^2 + b^2$$