Can an integer of the form $4n+3$ written as a sum of two squares? Let $u$ be an integer of the form $4n+3$, where $n$ is a positive integer. Can we find integers $a$ and $b$ such that $u = a^2 + b^2$? If not, how to establish this for a fact? 

Lemma 1: $a$ is odd $\Longrightarrow$ $a^2\equiv 1(\operatorname{mod} 4)$.
Proof: $a^2-1=(a-1)(a+1)$. Since $a$ is odd, both $a-1$ and $a+1$ are even, so that $a^2-1$ is divisible by $4$. $\blacksquare$
Lemma 2: $a$ is even $\Longrightarrow$ $a^2\equiv 0(\operatorname{mod} 4)$.
Proof: Trivial. $\blacksquare$
Now, suppose that $u=a^2+b^2$.
(1) If both $a$ and $b$ are even, then $u$ is divisible by four by lemma 2.
(2) If both $a$ and $b$ are odd, then $u\equiv 2(\operatorname{mod}4)$ by lemma 1.
(3) If $a$ is even and $b$ is odd (wlog), then $u\equiv1(\operatorname{mod}4)$ by lemmas 1 and 2.
That is, it is never the case that $u\equiv 3(\operatorname{mod} 4)$.
