Number Theory Simple Proof

Question

I am looking at a solution for a problem where the following line is stated but not explained, and I can not seem to make sense of it:

If a prime $p\equiv 3\pmod 4$ then why is $\frac{p(p+1)}{2}$ always even?

Answer 1

If $p\equiv3\mod 4$ then $p=4m+3$ for some $m\in\Bbb Z$ then $$\frac{p(p+1)}{2}=2(m+1)(4m+3),$$ hence your claim.

Answer 2

If $p\equiv 3\pmod 4$ then $p+1\equiv 0\pmod 4$, so that $p+1$ is divisible by $4$ $\implies\ \displaystyle\frac{p+1}2$ is even.

Answer 3

$n=\frac{p^2+p}{2}\ mod\ 4 = \frac{(p^2\ mod\ 4) + (p\ mod\ 4)}{(2\ mod\ 4)} = \frac{(p\ mod\ 4)^2 + (p\ mod\ 4)}{(2\ mod\ 4)} = \frac{3^2+3}{2} = \frac{9+3}{2} = \frac{12}{2} = 6 \equiv 2\ (mod\ 4)$

Thus, (assuming $n\in\Bbb N$) $n$ would be a $n=4k+2\ ,\ k\in\Bbb N$ number, so it is even, always positioned at $+2$ of a multiple of $4$.