Find constant when a quadratic equation has natural roots

Question

Given $a\in\mathbb{R}$, such that the two roots in $$f(x)=x^2+ax+(2a+319)$$ are both positive integers. Find $a$.

I've found that$$D=a^2-4\times(2a+319) = a^2-8a-1276$$

Since the two roots are integers, $$D=N^2, n\in\mathbb{N}$$ $$a^2-8a-1276=N^2$$

That's all I got. How can I move further?

Answer 1

Actually, $D=a^2-8a\color{red}-1276$

You have\begin{align}a^2-8a-1276=N^2&\iff(a-4)^2-1292=N^2\\&\iff1292=(a-4-N)(a-4+N).\end{align}So, for any way of writing $1292$ as $k\times l$ ($k,l\in\Bbb Z$), solve the system$$\left\{\begin{array}{l}a-4-N=k\\a-4+N=l.\end{array}\right.$$For instance, if $k=-34$, and $l=-38$, you will get $a=-32$ and the roots will be $15$ and $17$.

Answer 2

Let $f(x)=(x-r)(x-s)$. Then by Vieta, \begin{align*} r+s & = -a \\ rs & = 2a+319. \end{align*} Adding yields the Diophantine equation $$ rs+2r+2s=319, $$ which has the solution $(r,s)=(15,17),(17,15)$ in positive integers (there are only a few cases to check since $r,s\le 105$ and both $r,s$ are odd). So $a=-32$ is the only solution.