<s>[INST] 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.
 [/INST] 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.
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