Is the answer 17?

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

By the quadratic formula we then have that

$a = \dfrac{3 \pm \sqrt{9 + 4*(19 + 289k)}}{2} = \dfrac{3 \pm \sqrt{85 + 1156k}}{2}.$

As we are looking for integer values for $a,$ we will require that

$85 + 1156k = 17(5 + 68k) = n^{2}$

for some integer $n.$ With $17$ being prime, since $17 | n^{2}$ we know that $17^{2} | n^{2},$ and so $5 + 68k$ would have to be divisible by $17.$ But as $5 + 68k \equiv 5 \pmod{17}$ this is not the case, and so $85 + 1156k$ cannot be a perfect square. Thus, there are $\boxed{0}$ integer solutions for $a.$

Hack: If there is any satisfying $a$ , then $a+289$ also satisfies the condition. Thus there can be no such $a$ , otherwise there would be infinitely many such $a$ , which doesn't fit to Brilliant's answer format.

Let's take $a^{2}-3a-19=(a-10)(a+7)+51$ .

Suppose $289$ divides $a^{2}-3a-19$ , $17$ must divide $(a-10)$ or $(a+7)$ .

If $17$ divides $a-10$ , $a \equiv 10 \pmod {17}$ . Hence, $a+7 \equiv 10+7 \equiv 17 \equiv 0 \pmod {17}$ .

If $17$ divides $a+7$ , $a \equiv -7 \equiv 10 \pmod {17}$ . Hence, $a-10 \equiv 10-10 \equiv 0 \pmod {17}$ .

Hence whenever $17$ divides one of $a-10$ and $a+7$ , it must divide the other also.

Thus, $17^{2}=289$ divides $(a-10)(a+7)$ .

Since $289$ divides $(a-10)(a+7)$ , $289$ must divide $51$ which is impossible.

Hence, there is no integer $a$ such that $289$ divides $a^{2}-3a-19$ .