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. 

Assuming $x$ is odd, $x\equiv1\pmod2$ so $2|x-1$ so $2|(x-1)+2=x+1,$
so $4|(x-1)(x+1)=x^2-1=x^2+2-3, $ so $ x^2+2\equiv3\pmod4$.
Let $p$ be a factor of $x^2+2$.  $p$ must be odd because $x$ and therefore $x^2+2$ is.
If all such factors were $\equiv1\pmod4$ then their product would be $\equiv1\pmod4$, a contradiction.
So $x^2+2$ has a prime factor $\equiv3\pmod4$.
