Problem - 121
Show that, if $a,b$ are positive integers satisfying $4(ab-1)\mid (4a^2-1)$, then $a=b$
Because $(4ab-1)\mid (4a^2-1) \implies (4ab-1)\mid 4a(a-b)$ and $(4a,4ab-1) = 1$, we have $(4ab-1)\mid (a-b)$.
Meanwhile, because $|a-b| < 4ab-1$, therefore $|a-b|=0 \implies a=b$, as desired.