Divisible quadratic?

Suppose $a^{2} - 3a - 19 = 289k \Longrightarrow a^{2} - 3a - (19 + 289k) = 0$ for some integer $k$ . Then

$a = \dfrac{3 \pm \sqrt{9 + 4(19 + 289k)}}{2}$ .

For $a$ to be an integer it is necessary, (although not sufficient), that the discriminant $9 + 4(19 + 289k)$ be a perfect square. Now

$9 + 4(19 + 289k) = 9 + 4*2 + 4(17 + 17^{2}k) = 17 + 4*17*(1 + 17k) = 17(5 + 4*17k)$ .

For this to be a perfect square, since $17$ is prime we will require that $5 + 4*17k$ be a multiple of $17$ , but as $5 + 4*17k \equiv 5 \pmod{17}$ for any integer $k$ , we can conclude that the discriminant can never be a perfect square, and thus that there are $\boxed{0}$ integer values of $a$ such that $289$ divided $a^{2} - 3a - 19$ .

If $a^2 - 3a - 19 \equiv 0 \pmod{289}$ then $4a^2 - 12a - 76 \equiv 0 \pmod{289}$ , and hence $(2a - 3)^2 \; \equiv \; (2a-3)^2 - 85 \; \equiv \; 4a^2 - 12a - 76 \; \equiv \; 0 \pmod{17}$ Thus $2a \equiv 3 \pmod{17}$ , and hence $a \equiv 10 \pmod{17}$ .

But then $a = 17b + 10$ for some integer $b$ , and hence $a^2 - 3a - 19 \; = \; 289b^2 + 289b + 51$ so that $a^2 - 3a - 19 \equiv 51 \not\equiv 0 \pmod{289}$ . There are $\boxed{0}$ positive integers that satisfy the desired condition.

A cheating way to solve this:

If a is a solution, then a+multiple of 289 also has to be a solution. So we either have 0 or infinite number of solutions. As we can't really type infinity as a solution, then 0 has to be the answer.