Diophantine Equation (2014 AMC 12A) There are exactly $N$ distinct rational numbers $k$ such that $|k| < 200$ and $$5x^2 + kx + 12 = 0  $$ has at least one integer solution for $x$. What is $N$?
(My idea was to consider the equation modulo 12, but it really hasn't gotten me anywhere). 
 A: Let the roots of the equations be $a, b$ where $a \in \mathbb{Z}$.
Then, 
$$\begin{align}5(x - a)(x - b) &= 0\\
5x^2 - 5(a + b)x + 5ab &= 0\end{align}$$
Hence, we have $k = -5(a + b)$, and $12 = 5ab \implies b = \frac{12}{5a}, a \not=0$.
From this, we have $k = -5(a + \frac{12}{5a}) = -5a - \frac{12}{a} \in \mathbb{Q}$ for all $a \not= 0$. Now, $|k| < 200$ for $-39 < a < 39, a \not=0$, giving us $39 \times 2 = 78$ solutions.
I went for the test as well, and I'm feeling rather flabergasted and pissed with myself now for not being able to solve this simple question during the test, perhaps because it was demoralizing to be placed quite far back. Oh well.
A: If $a$ is an integral solution to the equation, we have
$$(x-a)(5x-b) = 5x^2+kx+12$$
for some $b$. This gives us $5a+b=-k$ and $ab=12$. Substituting for $b$,
$$k = -5a - \frac{12}{a}$$
So for any non-zero integer $a$, $5x^2+kx+12$ has at least one integer solution for $x$ when $k = -5a - \frac{12}{a}$. Now we just have to count how many integers $a$ lead to values of $k \in (-200,200)$. We get $-39 \le a \le -1$ and $1 \le a \le 39$, for a total of 78 values.
