Prove that if $a$ divides both $2$ and $\sqrt{10}$ in $\mathbb{Z}[\sqrt{10}]$, then $a$ is a unit Prove that if $a$ divides both $2$ and $\sqrt{10}$  in $\mathbb{Z}[\sqrt{10}]$, then $a$ is a unit. 
Further, show that you can't express $a$ as  $a = 2b + \sqrt{10} c$ where $b, c \in \mathbb{Z}[\sqrt{10}]$.
The only hint provided is that you should use the fact that  $A^{2} + 10B^{2} = 2, -2$ is insoluble.
 A: We prove the first assertion, and give a hint for the second.
The norm of $a$ divides the norm of $2$ and the norm of $\sqrt{10}$. So the norm of $a$ divides $4$ and $-10$, and therefore divides $2$.
We want to show that the norm of $a$ cannot be $2$ or $-2$. From that it follows that $a$ must have norm $1$ or $-1$ and therefore be a unit.
Let $a=x+y\sqrt{10}$. Then $a$ has norm $x^2-10y^2$. We show this can be neither $2$ nor $-2$. Suppose to the contrary that $x^2-10y^2=2$. Then $x^2\equiv 2\pmod{5}$, which is impossible. The same argument works for $-2$.
For the second problem, show that the norm of $2b+\sqrt{10}\,c$ is even.
A: Remember that the norm is multiplicative: $$N((\alpha + \beta \sqrt{10})(\gamma + \delta \sqrt{10})) = N(\alpha + \beta \sqrt{10})N(\gamma + \delta \sqrt{10}).$$ If $a$ divides two numbers with different norms, the absolute value of its own norm can't be greater than the greatest common divisor of the two numbers' norms.
$2$ has a norm of $4$ and $\sqrt{10}$ has a norm of $-10$. Since $a$ divides both $2$ and $\sqrt{10}$ and $\gcd(4, -10) = 2$, its norm must be $-2$, $-1$, $1$ or $2$. But no number in this domain can have a norm of $-2$ or $2$. That leaves $-1$ and $1$ as possible norms for $a$, hence $a$ is a unit.
For example, let's say $a = 3 + \sqrt{10}$. Then $$\frac{2}{a} = -6 + 2 \sqrt{10}$$ and $$\frac{\sqrt{10}}{a} = 10 - 3 \sqrt{10}.$$
For the second part of your question, I wondered if you meant $b, c \in \mathbb{Z}$ rather than $b, c \in \mathbb{Z}[\sqrt{10}]$. But as it turns out, $a = 2b + c \sqrt{10}$ has no solution either way. $2b$ has a norm divisible by $4$, while $c \sqrt{10}$ has a norm divisible by $-10$. From this it follows that $N(2b + c \sqrt{10})$ must be even, contradicting what was just proven about $a$ being a unit.
A: Somewhere in the book it should say something along the lines of: 

The norm function is defined thus: $N(A + B \sqrt{10}) = A^2 - 10B^2$.

It should also say something like:

This function is multiplicative.

These are worth reviewing; the book might even provide concrete examples that make the exercises much easier to do.
Check that $N(2) = 4$ and $N(\sqrt{10}) = -10$.
If $a|2$ and $a|\sqrt{10}$, then $2 = ak$ and $\sqrt{10} = aj$; we don't really care what $k$ and $j$ are but we do care that $k, j \in \mathbb{Z}[\sqrt{10}]$.
If $a$ is not a unit, then $|N(a)| \neq 1$ (the bars represent the absolute value function, in case you're wondering). Therefore $N(a) = 2$ and $N(k) = 2$ also ($N(a) = -2$ and $N(k) = -2$ works, too). But like the hint said, $|A^2 - 10B^2| = 2$ has no solutions, meaning that no number $x \in \mathbb{Z}[\sqrt{10}]$ satisfies $|N(x)| = 2$.
But since $a$ is not a unit, this means that $N(a) = 4$ and $N(k) = 1$ (so $k$ must be the unit). But now we have a much bigger problem when we try to make sense of $\sqrt{10} = aj$, since the only possible solution to $-10 = 4N(j)$ is $N(j) = -\frac{5}{2}$, but that takes us out of $\mathbb{Z}[\sqrt{10}]$ to the "wider world" of $\mathbb{Q}(\sqrt{10})$. Maybe $N(j) = 5$ and $N(a) = 2$, but, wait a minute, no, that doesn't work either.
Denying that $a$ is a unit has prevented us from figuring out $2 = ak$ and $\sqrt{10} = aj$. When you have eliminated the impossible... $|N(a)| = 1$ and there are infinitely many values of $a$ that also satisfy both $a|2$ and $a|\sqrt{10}$.
