Problem 3

Fix a prime $p$ and positive integers $a,b$ with $g(a)\equiv g(b)\pmod p$. Choose a sufficiently large positive integer $M$ so that both $v_p(M+g(a))$ and $v_p(M+g(b))$ are odd; this can be done by choosing a suitable residue of $M$ modulo a high enough power of $p$: if the two constants differ by at least two powers of $p$ (or are equal), both valuations can be made $1$, while if they differ by exactly one power of $p$, they can be made $1$ and $3$. Since $P(M,a)=(g(M)+a)(g(a)+M)$ is a square, the odd valuation of $M+g(a)$ forces $a+g(M)$ to have odd p-adic valuation, hence $p\mid a+g(M)$. The same argument with $b$ gives $p\mid b+g(M)$. Therefore $a\equiv b\pmod p$.