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. 

First, you can square both sides in the congruence relation but you must not square the number that you are "moding".
If $x$ is odd, then $x\equiv1$ or $3$ (mod $4$). So, by some calculations, 
$$x^2\equiv1(\mod 4)$$
$$x^2+2=3(\mod 4)$$
Then, for the next part, we will use prove by contradiction.
First, we can see that all of the factors of $x^2+2$ is odd because $x^2+2$ is odd.
If $x^2+2$ has no prime factor which has a remainder of $3$ when divided by $4$, then $x^2+2$ should have only prime factors which has a remainder of $1$ when divided by $4$. As the product of any number of numbers that leaves a remainder of $1$ when divided by $4$ also has a remainder of $1$ when divided by $4$, so it contradicts that $x^2+2\equiv3 (\mod 4)$
$\therefore x^2+2 $ has at least one factor that leaves a remainder of $3$ when divided by $4$.
