Units of $\mathbb{Z}[\sqrt {-2}]$, and are $3,5$ irreducible in $\mathbb{Z}[\sqrt {-2}]$? So for the first question I determined the units to be $+1$ and $-1$. I assigned $x = a + b \sqrt{-2}$ and proceeded to show $b = 0$ so $a^2 = 1$ or $-1$ so $x = 1$ or $-1$.
For the second part I am not sure what to do. Looking at the norm function $N$, we have $N(a + b\sqrt{-2}) = a^2 + 2b^2$ but I don't know how to go about with this information. I believe finding the norm would be important to determine the second part, right?
 A: If there exists a number $n=a+\sqrt{-2}b$ such that $n|3$ and $N(n)\neq 1$ (otherwise the number would be associated to $3$) it satisfies $N(n)|N(3)$ (since the norm is multiplicative). This means $a^2+2b^2|9$ and thus $N(n)=3$. But $a^2+2b^2=3$ means $(a,b)=(\pm1,\pm1)$. This means $\pm1\pm\sqrt{-2}$. A quick verification shows that $3=(1+\sqrt{-2})(1-\sqrt{-2})$. $3$ is reducible
For $5$ a similar reasoning shows $N(n)=5$, and this leads to $a^2+2b^2=5$. Reducing the equation $\equiv 8$ we see that there are no integer solutions (since $x^2\equiv_80,1,4$), and thus $5$ is irreducible. 
If you were to avoid the "modular" reasoning, a simple observation on $a^2\le 5;2b^2\le 5$ allows us to restrict to the values $a=1,2;b=1$: it is easy to verify by hand that none of these values satisfies the equation
A: Finding the norm is indeed important for determining the second part of this problem, whether or not $3$ and $5$ are both irreducible.
Note that $N(3) = 9$ and $N(5) = 25$. So, if there exist primes $p$ and $q$ in $\mathbb{Z}[\sqrt{-2}]$ such that $pq = 3$ and primes $r$ and $s$ such that $rs = 5$, they satisfy $N(p) = N(q) = 3$ and $N(r) = N(s) = 5$. We needn't worry about $N(p) = N(q) = -3$ or $N(r) = N(s) = -5$ because $a$ and $b$ are limited to $\mathbb{Z}$ and so $a^2 + 2b^2$ can't be negative.
Plugging $a = \pm 1$ and $b = \pm 1$ into the norm formula $N(a + b \sqrt{-2}) = a^2 + 2b^2$, we obtain $N(\pm 1 \pm 1 \sqrt{-2}) = 1^2 + 2 \times 1^2 = 1 + 2 = 3$. Bingo. This means $3$ is in fact not irreducible, that is to say, it is reducible.
Now, if for whatever reason you choose the "wrong" pair out of the six possibilities, such as $(-1 + \sqrt{-2})(1 + \sqrt{-2}) = -3$, you can bring your result to the "right" side of $0$ by multiplying by one of the units you have already discovered.
We have a rather different situation trying to solve $N(r) = 5$. There doesn't seem to be any solutions. We have $a^2 + 2b^2 = 5$. Subtracting $a^2$ from both sides gives us $2b^2 = 5 - a^2$. Then, halving both sides, we get $$b^2 = \frac{5 - a^2}{2}.$$ Obviously this has no solution in integers. Therefore $5$ is actually irreducible.
A: If you have no better ideas, you can almost always just do trial division. The two previous answerers did give better ideas, but I thought you might appreciate a different perspective to help confirm the better ideas.
If $5$ reducible in $\mathbb{Z}[\sqrt{-2}]$, that means there is some non-unit number (a number with norm not $\pm 1$) such that $5$ is divisible by it.
With trial division, we could start almost anywhere. I choose to start with $\sqrt{-2}$. We see that $$\frac{5}{\sqrt{-2}} = -\frac{5 \sqrt{-2}}{2} \not\in \mathbb{Z}[\sqrt{-2}].$$ Then I try $$\frac{5}{2 \sqrt{-2}}, \frac{5}{3 \sqrt{-2}}, \frac{5}{4 \sqrt{-2}}.$$ That last one is roughly $-0.884i$, so that tells me to stop on this line and move to the next line.
$5$ is trivially divisible by $1$, because of course $1$ is a unit. $5$ is not divisible by $1 + \sqrt{-2}$, nor by $1 + 2 \sqrt{-2}$. Long story short, by the time I get to $3 + 2 \sqrt{-2}$, I'm convinced $5$ is irreducible in this ring.
It plays out much differently with $3$. Clearly $3$'s not going to be divisible by $\sqrt{-2}$ any more than $5$ wasn't either. But soon I get to $1 + \sqrt{-2}$ and you how this turns out...
A: Here's an approach that doesn't involve calculation of the norm.
For $3$ it's directly checkable that $3 = (1 + \sqrt{-2})(1 - \sqrt{-2})$.

To show that $5$ is irreducible, it is enough to show that the quotient ring $\mathbb Z[\sqrt{-2}]/(5)$ is a field.
We then write $\mathbb Z[\sqrt{-2}]$ as $\mathbb Z[X]/(X^2 + 2)$ and get $\mathbb Z[\sqrt{-2}]/(5) \simeq \mathbb Z[X]/(5, X^2 + 2) \simeq \mathbb F_5[X]/(X^2 + 2)$.
Therefore it suffices to show that the polynomial $X^2 + 2$ is irreducible in $\mathbb F_5[X]$.
Being a polynomial of degree $2$, one only needs to check that it doesn't have any root in $\mathbb F_5$.
This is however the fact that $-2$ is a quadratic non-residue mod $5$.

This approach has the advantage that it easily generalizes to other primes $p$ with $\left(\frac{-2}{p}\right) = -1$. The ultimate answer of course is given by reciprocity law (+ the fact that $\mathbb Z[\sqrt {-2}]$ is a Euclidean domain).
