Pinko wrote: We are given a polynomial $f(x)=x^6-11x^4+36x^2-36$. Prove that for an arbitrary prime number $p$, $f(x)\equiv 0\pmod{p}$ has a solution. $f(x)=(x^2-2)(x^2-3)(x^2-6)$ and so : If $\left(\frac 2p\right)=+1$, $\exists x$ such that $x^2-2\equiv 0\pmod p$ If $\left(\frac 3p\right)=+1$, $\exists x$ such that $x^2-3\equiv 0\pmod p$ If $\left(\frac 2p\right)=\left(\frac 3p\right)=-1$, then $\left(\frac 6p\right)=\left(\frac 2p\right)\left(\frac 3p\right)=+1$ And so $\exists x$ such that $x^2-6\equiv 0\pmod p$ Q.E.D.