Track / Stage 5 / 47 of 400 #647 of 1964

Problem 647

AIME late

Number theory Difficulty 5.1 Prove it

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.

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 reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.