Number Theory : Solving $x^2$ $+$ $y^2$ $=$ $2^{10}$ - $1$

Question

I was working my way through some basic number theory problems and was all thumbs while solving this problem :

List all the pairs of integer solutions $(x, y)$ of the Diophantine equation : $x^2$ $+$ $y^2$ $=$ $2^{10}$$-$$1$ and show that the list is exhaustive

How can I go about it ? A hint would be a good way to start

Answer 1

It has no solutions, since $x^2+y^2\equiv -1\equiv 3\pmod{4}$, but $x^2$ can only be $0$ or $1$ modulo $4$.

$2^{10}-1\equiv -1\pmod{4}$, since $4\mid 2^{10}-1-(-1)=2^{10}=4\cdot 2^{8}$.