Prove that $x^2-3y^2=2729$ cannot have an integral solution

Question

Do I need to first break it down in $a^2-b^2$ and then proceed ?

Answer 1

All squares are $0$ or $1$ mod $3$.

We have $x^2-3y^2 \equiv x^2 \mod 3$.

We also have $2729 \equiv 2 \mod 3$.

Therefore, we have $x^2 \equiv 2\mod 3$.

Referring to the first equation, this can't be possible.

Therefore, this equation has no solutions.