Does Diophantine equation $m^2+2010=n^2$ has a solution for $m,n\in\mathbb Z$?

Question

Number $2010$ is not a perfect square. We would have Pythagorean triple $(m,\sqrt{2010},n)$ for positive integers $m,n$?

Does this mean that equation $m^2+2010=n^2$ doesn't have a solution for any $m,n\in\mathbb Z$?

Answer 1

$$2010=(n-m)(n+m)$$

As $n-m+n+m=2n$ is even,

$n+m,n-m$ have the same parity

As $2010,n-m$ is even, so will be $n+m$

$\implies4|( n^2-m^2)$

But $2010\equiv?\pmod4\not\equiv0$