james digol wrote: Let $f$ be a polynomial function with integer coefficients and $p$ be a prime number. Suppose there are at least four distinct integers satisfying $f(x) = p$. Show that $f$ does not have integer zeros. If $f(x)$ has an integer root, WLOG consider $f(0)=0$ and so $f(x)=xg(x)$ $f(t)=p$ implies then $t|p$ and so $t\in\{-p,-1,1,p\}$ $f(-p)=p$ $\implies$ $g(-p)=-1$ and $f(p)=p$ $\implies$ $g(p)=1$ and so $2p|2$, impossible Hence the result