Is there any number in $\mathbb Z[\sqrt{5}]$ with norm equals 2?

Question

Is there any number $a+b\sqrt{5}$ with $a,b \in \mathbb{Z}$ with norm (defined by $|a^2−5b^2|$) equal 2?

Answer 1

No.

Since $a^2-5b^2\equiv a^2\pmod{5}$, the values of $a^2-5b^2$ must be congruent to either $0$, $1$, or $4$ modulo $5$, since those are the only squares modulo $5$. But neither $2$ nor $-2$ satisfy this condition, so it is impossible to have $a^2-5b^2=2$ or $a^2-5b^2=-2$ with $a,b\in\mathbb{Z}$.

Answer 2

Hint $\rm\,\ 2\:|\:a^2\!-\!5b^2\! = (a\!-\!b)(a\!+\!b)\!-\!4b^2\Rightarrow\, 2\:|\:a\pm b\:\Rightarrow\:2\:|\:a\mp b\:\Rightarrow\:4\:|\:a^2\!-\!5b^2$