integer values of the parameter $a$ so that the equation $x^2 + ax + 2017^{2017}=0$ has $2$ integer polynomial roots?

Question

I have a problem that I can't solve:

How many integer values can the parameter $a$ assume so that the equation $x^2 + ax + 2017^{2017}=0$ has $2$ integer polynomial roots?

Answer 1

From $x^2+ax+2017^{2017}=0$, we get $a=-x-\frac{2017^{2017}}{x}$. Since $2017$ is a prime, $x$ can be from $1$ to $2017^{2017}$. However do note that the value of $a$ is the same for $x=k$ and $x=\frac{2017^{2017}}{k}$, where $k$ is a power of $2017$. Hence, the of nnumber of possible values of $a$ range from $x$ starting from $2017^0$ to $2017^{1008}$, where by the result of $a$ will be the same afterwards. Hence, there are $1009$ possible values of $a$.