Elementary number theory and quadratic Find the value(s) of $a$ for which the equation 
$ax^2-4x+9=0$ 
has integral roots(i.e. $x$ is an integer).
My attempt: 
By hit and trial I am getting answer as $a=\frac{1}{3}$. No information about nature of $a$ is given.
 A: If $n\in\mathbb{Z}\setminus\{0\}$ is a root then $a=(4n-9)/n^2$. It follows that the roots are $n$ and 
$$\frac{9n}{4n-9}=2+\frac{n+18}{4n-9}.$$
Now if $(n+18)/(4n-9)$ is an integer then $-2\leq n\leq 9$ (because $|n+18|$ has to be greater or equal to $|4n-9|$. By checking, we find that it is an integer for $n=2,3,9$. The corresponding values for $a$ are $-1/4$ and $1/3$.
A: If the roots are $p$ and $q$, then
$$ p+q = \frac{4}{a} $$
$$ pq = \frac{9}{a} $$
Dividing gives
$$ \frac{1}{p} + \frac{1}{q} = \frac{4}{9} $$
If you can find all integer solutions to this equation, you can plug in to get the possible values for $a$.
It's not to hard to do an exhaustive search over the small values of $(p,q)$ until you reach the point where it's obvious nothing larger could be a solution.

Alternatively, you can boil down the roots to a single integer, e.g
$$ (p+q)^2 = \frac{16 + 18 a}{a^2} $$
so the problem boils down to finding $a$ such that $a^2 \mid (16 + 18 a)$.
(here, divisibility means that the r.h.s. is an integer multiple of the l.h.s.)
A: I propose a graphical understanding of this question:
Consider it as looking for the intersection points of these two curves 
$$\cases{y=4x-9\\y=ax^2}$$


*

*The first one is a fixed straight line on which we define points with integer abscissas $P_k(k,4k-9) \ \ (k \neq 0).$

*the second one is a variable parabola with an "opening" coefficient $a$ we have to constrain in order that the parabola passes through the different points $P_k$. And this is possible for every values of $k$ by equating the ordinates (for all $k \neq 0$)
$$a k^2 = 4 k -9 \ \ \ \Longleftrightarrow \ \ \ a=\dfrac{4k-9}{k^2}$$
