Is the answer 17 (continued)?

All congruences will be modulo 289 (or, more generally, modulo the square of an odd prime).

Completing the square (or, equivalently, using the quadratic formula), we can reduce this to the problem of counting the solutions $y$ of $y^2\equiv a$ .

In the case $a\equiv 0$ , there are 17 solutions, namely, $y\equiv 17k$ for $1\leq k \leq 17$ .

We claim that for all other $a$ there are at most two solutions, so that the maximal number is $\boxed{17}$ .

We will show that if $y^2\equiv a$ has more than 2 solutions, then $a\equiv 0$ . Let $\pm p$ and $q$ be three pairwise non-congruent solutions, meaning that $p^2\equiv q^2 \equiv a$ . Then $p^2-q^2=(p-q)(p+q)\equiv 0$ . Since neither $p+q$ nor $p-q$ is divisible by 289, both must be divisible by 17, so that $p$ and $q$ are divisible by 17 as well. But this means that $p^2\equiv q^2 \equiv 0$ as claimed.

(The proof in the last paragraph addresses a very special case of Hensel's Lemma )