amparvardi wrote: Let $p(x)$ be a cubic polynomial with integer coefficients with leading coefficient $1$ and with one of its roots equal to the product of the other two. Show that $2p(-1)$ is a multiple of $p(1)+p(-1)-2(1+p(0)).$ Let the roots be $m$, $n$ and $mn$ then $p(x)=(x-m)(x-n)(x-mn)=x^3-(m+n+mn)x^2+mn(1+m+n)x-(mn)^2$ So $m+n+mn$, $mn(1+m+n)$ and $(mn)^2$ are integers. Then $mn(1+m)(1+n)=mn(1+m+n)+(mn)^2$ is an integer. Also $(1+m)(1+n)=1+(m+n+mn)$ is an integer. So $mn=\frac{mn(1+m)(1+n)}{(1+m)(1+n)}$ is a rational number. But $p(x)$ is a monic and integer polynomial so if it has a rational root then that root is an integer. Hence $mn\in\mathbb{Z}$ so $(1+mn)$ is an integer. Now $2p(-1)=-2(1+m)(1+n)(1+mn)$ (which is an integer) is a multiple of $-2(1+m)(1+n)$ (which is an integer) because $1+mn$ is an integer. $p(1)+p(-1)-2(1+p(0))=(1-m)(1-n)(1-mn)-(1+m)(1+n)(1+mn)-2$ $+2m^2n^2$ So $p(1)+p(-1)-2(1+p(0))=-2(1+m+n+mn)=-2(1+m)(1+n)$ Hence $2p(-1)$ is a multiple of $p(1)+p(-1)-2(1+p(0)),$ as required.

This proof is somawhat messier than the one of @above Let $p(x)=x^3+a_3x^2+a_2x+a_1$ and let the roots be $r_1, r_2, r_1r_2$ \par Then by Vieta's formulas we have that: \begin{align} a_1 &= -(r_1r_2)^2\\ a_2 &= r_1r_2+{r_1}^2r_2+{r_2}^2r_1\\ a_3 &= -(r_1+r_2+r_1r_2) \end{align}Now we have that \begin{align*} p(1)+p(-1) &= 2a_3+2a_1\\ -2(1+p(0)) &= -2-2a_1 \end{align*}Adding up we get $2(a_3-1)$ then it is enough to show that $a_3-1\mid p(-1)$ \par Since $p(-1)=-1+a_3-a_2+a_1$ we can substitute the roots and factor it to obtain \[-r_1r_2(1+r_1r_2)-r_1(1+r_1r_2)-r_2(1+r_1r_2)-r_1r_2-1=(1+r_1r_2)(-r_1r_2-r_1-r_2-1)\]Which clearly is a multiple of $a_3-1=(-r_1r_2-r_1-r_2-1)$ if and only if $(1+r_1r_2)$ is a integer. \par So now I will prove that $(1+r_1r_2)\in \mathbb{Z}$ \par First notice that $a_2-a_1=r_1r_2(1+r_1+r_2+r_1r_2)$ is an integer and since $a_3=-(r_1+r_2+r_1r_2)$ we have that $r_1r_2 \in \mathbb{Q}$ but $a_1=-(r_1r_2)^2$ thus $r_1r_2$ is an integer as is $r_1r_2+1$. This concludes the proof $\blacksquare$