Quadratic Equations with Integer Roots

The standard form of the quadratic equation is $x^2+6ax-a^4=0$ .

Solve it using the quadratic formula: $x=\displaystyle\frac{-6a±\sqrt{36a^2+4a^4}}{2}$ .

Simplify by pulling $4a^2$ from under the square root: $x=\displaystyle\frac{-6a±2a\sqrt{9+a^2}}{2}$ .

Then, $x=-3a±a\sqrt{9+a^2}=a(±\sqrt{9+a^2}-3)$ , which has to be an integer. Since $a$ is an non-zero integer, this means that $\sqrt{9+a^2}$ is an integer.

In other words, there exists an integer $k$ such that $k^2=9+a^2$ .

This can be rearranged to $k^2-a^2=9$ . Factorize the left-hand side: $(k+a)(k-a)=9$ .

The only integer factorizations of $9$ is $9\cdot1=3\cdot3=-9\cdot-1=-3\cdot-3$ . The solutions we get for $a$ is $a=-4,0,4$ . Since we are interested in the product of all non-zero integers for $a$ , the answer is $-16$ .