$p=4n+3$ never has a Decomposition into $2$ Squares, right? Primes of the form $p=4k+1\;$  have a unique decomposition as sum of squares $p=a^2+b^2$ with $0<a<b\;$, due to Thue's Lemma.
Is it correct to say that, primes of the form $p=4n+3$, never have a decomposition into $2$ squares, because sum of the quadratic residues $a^2+b^2$ with $a,b\in \Bbb{N}$
$$
a^2 \bmod 4 +b^2 \bmod 4 \le 2?
$$
If so, are there alternate ways to prove it?

Suppose $q$ is an odd sum of two squares.  Then one square is even and one is odd.  The odd square is, perforce, the square of an odd number and the even square is the square of
an even number. 
So we can write for some integers $m$ and $n$,
$$q = (2m)^ 2 + (2n + 1)^2 = 4(m^2 + n^2 + n) + 1.$$
We conclude that $q \equiv 1 \mod 4.$
This very naive derivation shows that if $q$ is any odd integer with $q\not\equiv 1\mod 4$, $q$ cannot be the sum of two squares.
