A number theory problem by Anish Roy

$x^2 - 2y^2 = 1$ gives us $x^2 = 2y^2 + 1$ and hence $x$ must be an odd number. If $x = 2n + 1$ , then $x^2 = (2n + 1)^2 = 4n^2 + 4n + 1 = 2y^2 + 1$ .
This means that $y^2 = 2n(n + 1)$ . This means that $y^2$ is even and hence $y$ is an even integer.
Now, $y$ is also prime implies that $y = 2$ . This gives us $x = 3$ .
Thus the only solution is $x = 3 , y = 2$ .
$\therefore x + y = \boxed {5}$

If $y = 3$ then $x^2 = 19$ with no solutions and if $y \neq 3$ then $x^2 \equiv 2y^2 + 1 \equiv 0 \pmod{3} \implies x = 3 \implies y = 2$ .

Hence, $x + y = \boxed{5}$ .