Product of roots of $ax^2 + (a+3)x + a-3 = 0$ when these are positive integers There is only one real value of $'a'$ for which the quadratic equation $$ax^2 + (a+3)x + a-3 = 0$$ has two positive integral solutions.The product of these two solutions is : 
Since the solutions are positive, therefore the product of roots and sum of root will be positive. This will give us two inequalities in $a$. Substitute the values of $a$ in the quadratic but I'm not getting my answer correct.
The two inequalities will be $$\frac{a-3}{a} > 0$$ and the other one will be $$\frac {a+3}{a} < 0$$
From the first one we get $a>3$ and from the second one we get $a< -3$.
I don't know how to proceed after this.
Kindly help.
 A: We need $$\frac{a+3}{a}\in\Bbb{Z}\ ,  \frac{a-3}{a}\in\Bbb{Z}$$ or $$1+\frac{3}{a}\in\Bbb{Z}\ , \ 1-\frac{3}{a}\in\Bbb{Z}$$ thus $\displaystyle \frac{3}{a}\in\Bbb{Z}$, means that $\displaystyle a=\frac{3}{m}$ where $m\in\Bbb{Z}$.
Now we can write the equation as $$\frac{3}{m}x^2+\left(\frac{3}{m}+3\right)x+\frac{3}{m}-3=0\implies x^2+(m+1)x+1-m=0$$
Using quadratic formula we have $$x_{1,2}=\frac{-m-1\pm\sqrt{m^2+6m-3}}{2}=\frac{-m-1\pm\sqrt{(m+3)^2-12}}{2}$$We need the expression under the square root to be an integer. There are only two squares with difference $12$ (those are $4,16$), hence we want $(m+3)^2=16\implies m=1\text{ or }m=-7$, thus $$a=3 \text{ or } a=-\frac{3}{7}$$ and as $a<0$ we have 

$$a=-\frac{3}{7}$$

A: Thanks to @rtmd, I fixed the solution. This only holds when $a=-\frac{3}{7}$. Here is the proof.
$\frac{a-3}{a}, \frac{a+3}{a} \in \mathbb{Z}$ implies $\frac{6}{a} \in \mathbb{Z}$. We can write $a=\frac{6}{n}$ for some integer $n$.
\begin{align*}
\frac{a-3}{a}= 1-\frac{n}{2} \in \mathbb{Z}\\
\frac{a+3}{a}= 1+\frac{n}{2} \in \mathbb{Z}
\end{align*}
Therefore $n$ is even. Assume $p,q$ are the positive integer solutions of the given quadratic equation.
\begin{align*}
pq= 1-\frac{n}{2} \in \mathbb{Z}\\
p+q= -1-\frac{n}{2} \in \mathbb{Z}
\end{align*}
This gives us 
\begin{align*}
pq-p-q&=2\\
(p-1)(q-1)&=3
\end{align*}
Therefore WLOG $p=4,q=2$. This only holds when $n=-14$ and $a=-\frac{3}{7}$
