There is a famous difficult problem:
For all natural numbers $a,b$ it's true that: $\displaystyle(ab+1)|(a^2+b^2)\implies \frac{a^2+b^2}{ab+1}$ is a perfect square.
I've noticed $(a=1,\dots 100)$ that for each $a$ there's a $b$ such that $(ab+1)|(a^2+b^2)$ and my question is if this can be proved generally? Induction seems to lead nowhere.
The first guaranteed value is $$ b = a^3 $$ which gives $$ \frac{a^2 + b^2}{ab+1} = \frac{a^2 + a^6}{a^4 + 1} = a^2 $$