Inspired by Anish Harsha

Not sure if I solved this in the most efficient way. I used to love these kind of problems in Number Theory class.

$a^2+6a+7\equiv0\pmod{289}$

$a^2+6a\equiv-7\pmod{289}$

$a^2+6a+9\equiv2\pmod{289}$

$(a+3)^2\equiv2\pmod{289}$

Let $x=a+3$

In order for $2$ to be a quadratic residue modulo $289$ , it must be a quadratic residue modulo $17$ . A quick search yields $x^2\equiv2\pmod{17}$ only when $x\equiv6\pmod{17}$ or $x\equiv11\pmod{17}$ .

Case 1: $x=17k+6$ , $k\in\mathbb{Z}$

$(17k+6)^2\equiv2\pmod{289}$

$289k^2+204k+36\equiv2\pmod{289}$

$204k\equiv-34\pmod{289}$

$12k\equiv-2\pmod{17}$

$6k\equiv-1\pmod{17}$

$6k\equiv-18\pmod{17}$

$k\equiv-3\pmod{17}$

$k\equiv14\pmod{17}$

$a=17k+3$

$a$ must be a positive integer less than $289$ , and $14$ is the only possible value for $k$ that would put $a$ in this range. $a=17(14)+3=241$ .

Case 2: $x=17k+11$ , $k\in\mathbb{Z}$

$(17k+11)^2\equiv2\pmod{289}$

$289k^2+374k+121\equiv2\pmod{289}$

$85k\equiv-119\pmod{289}$

$5k\equiv-7\pmod{17}$

$5k\equiv10\pmod{17}$

$k\equiv2\pmod{17}$

$a=17k+8$

$a$ must be a positive integer less than $289$ , and $2$ is the only possible value for $k$ that would put $a$ in this range. $a=17(2)+8=42$ .

There are only $\boxed{2}$ solutions for $a$ that are positive integers less than $289$ : $a=42$ and $a=241$ .

Andy has written a fine solution... let me suggest an alternative, for the sake of variety. All congruences will be modulo 289.

Like Aareyan, I'm writing $(a+3)^2\equiv 2$ . If we let $a+3=b$ , we can write $b^2\equiv 2$ ... we have to count the "square roots" of 2 modulo 289. We quickly see that $b=\pm 45$ are solutions, but the key step is to show that there are no other solutions. We will argue by contradiction, assuming that there exists an $x$ that is congruent to neither 45 nor -45 such that $45^2\equiv x^2 \equiv 2$ . Then $(45^2-x^2)=(45-x)(45+x)\equiv 0$ , meaning that both $45+x$ and $45-x$ are divisible by 17. This implies that 90 is divisible by 17, a contradiction.

Thus our congruence has only $\boxed{2}$ solutions.