Prove that the equation $x^2 − y^2 = 2002$ has no integer solution.
If I had the equation $x^3 - 7y = 3$ I would easily conclude that for it to have any integer solution then $$x^2\equiv3\pmod 7$$ must be true.
Applying the same logic here, I know that $x² = y^2 +2002$ so I can conclude that for it to have any integer solution then $$x^2\equiv2002\pmod y$$ must be true.
Is this conclusion correct?
From that congruence I can prove that the equation hasn't any integer solution without much trouble but I am not sure if that congruence is valid.