$f(x)=x^3+ax^2+bx+p=0$ has no integral solutions.

Question

let $p$ be a prime number, does the polynomial:$f(x)=x^3+ax^2+bx+p=0$ have any integral solution if $p>a>2$ and $ x>2 $?

I concluded that there was none on the basis that $p>a$ and a prime. Is my reasoning flawed?

Answer 1

Since $b$ is a free variable you're asking if $x^3+ax^2+p$ can be a multiple of $x$. Sure, $x = p$.

It's probably easy to name a case where there isn't an integral solution as well. So your conditions do not imply whether there is an integral solution, either way.