Solving equation over $\mathbb{Z}$ involving squares 
Solve the equation $$a^2 - 10b^2 = 2$$ for $a,b \in \mathbb{Z}$.

I tried to consider the equation as a polynomial in the indeterminate $a$ and what I get is $$a_{1,2} = \pm \sqrt{10b^2 + 2}$$ which does not really help (I would have to find $b \in \mathbb{Z}$ such that $10b^2 + 2 = n^2$ for some natural number $n$). So I looked at the solutions where it is written: obviously, this equation is not solvable over $\mathbb{Z}$. For me, it is not that obvious. Could anyone please show me how to see, that this equation is not solvable over $\mathbb{Z}$?

If you consider the equation modulo $5$, you get $$a^2\equiv2\pmod5.$$ Checking the squares modulo $5$, we see that $2$ isn't a square, hence this equation has no solutions in the integers.
