This is an especially famous problem that you can check out an alternative asking and full solutions to here and, more formally, here. This post is not asking for solutions or full fledged proofs; rather, I am wondering whether an alternative motivation that I've thought of would be correct (if I were to write it out full-fledgedly and completely).
The problem asks:
Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Show that $$\frac{a^2 + b^2}{ab + 1}$$
is the square of an integer
My idea/motivation stems from the fact that a square of a number can only be $0$ or $1$ in mod(3). That means that $a^2 + b^2$ can equal either $0, 1$, or $2$ mod(3). If either $a$ or $b$ or both are equivalent to $0$ mod(3), that means that $ab + 1$ is 1, and the overall fraction becomes $0$ mod(3), which is a square of an integer. If $a$ and $b$ are both $1$, then that means that $\frac{a^2 + b^2}{ab + 1} = \frac{1^2 + 1^2}{1 \cdot 1 + 1} = 1$ in mod(3). Thus all cases are accounted for, and so the fraction in question is a perfect square.
Am I missing something (other than a formal proof)? Once again, my question differs from others in the idea as well as the topic that I am inquiring about. Thanks to all the people who help.
Best Answer
There is a logical flaw, we must have $\forall x \in \mathbb{Z} \implies x^2 \equiv 0 \pmod{3}$ or $x^2 \equiv 1 \pmod{3}$.
It doesn't mean that number that have remainder $0$ or $1$ are squares.
For example, any multiple of $3$ is congruent to $0$ $\pmod{3}$, but only some of them are squares.