Track / Stage 5 / 20 of 400 #620 of 1964

Find all integers $a, b$ such that $3 a^{2}=b^{2}+1$.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

I solved it I didn't Skip

Official solution

If we look modulo 3, we realize that a square can only be congruent to 0 or 1, so $b^{2}+1$ can never be divisible by 3: there are no solutions.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.