Show that $x^2+2 \equiv 3 \mod 4$ and deduce that there exists a prime $p$ with $p|x^2+2$ and $p \equiv 3 \mod 4$. Let $x$ be an odd natural number. Show that $x^2+2 \equiv 3 \mod 4$ and deduce that there exists a prime $p$ with $p|x^2+2$ and $p \equiv 3 \mod 4$.
For the first part I would assume 
$$x \equiv 1 \mod 2$$
$$x^2 \equiv 1^2 \mod 2^2$$
$$x^2 +2 \equiv 1^2+2 \mod 2^2$$
$$x^2 +2 \equiv 3 \mod 4$$
Can someone briefly tell me if this is correct and if it is a rule that squaring the number on the left means you square both numbers on the right?
Also I don't know how to start with the second part. 
Note: This question is to do with rings in general. 

For a more quick way: Notice, $x^2+2$ is always odd, as $x$ is odd. Therefore, all prime divisors of $x^2+2$ are either of form $4k+1$ or $4k+3$. Now, if all of them are of form $4k+1$, then simply check that $x^2+2\equiv 1\pmod{4}$. However, since $x$ is odd, $x^2+2\equiv 3\pmod{4}$, which is yielding a contradiction. 
