Prove that :
$f(1)+f(-1)-2(f(0)+1)\equiv 0\pmod{2f(-1)}$ Problem : 
$$f(t)=t^{3}+\alpha t^{2}+\beta t+\gamma , \alpha ,\beta  \operatorname{and}  \gamma \in\mathbb{Z}$$ 
with root $x_{1},x_{2},x_{1}x_{2}$
Question is :
Prove that :

$$f(1)+f(-1)-2(f(0)+1)\equiv 0\pmod{2f(-1)}$$

My try: 
We have : 
$$\begin{cases}x_{1}+x_{2}+x_{1}x_{2}=-\alpha \\x_{1}x_{2}+x_{1}^{2}x_{2}+x_{1}x_{2}^{2}=\beta \\x_{1}^{2}x_{2}^{2}=-\gamma\end{cases}$$
Also : 
$$f(1)+f(-1)-2(f(0)+1)=2(\alpha -1)$$
$$2f(-1)=2(\alpha -\beta +\gamma -1)$$ 
So : 
$$f(1)+f(-1)-2(f(0)+1)\equiv \beta -\gamma\pmod{2f(-1)}$$
But we have from system equation
$$\beta -\gamma =x_{1}x_{2}(1+x_{1}+x_{2}+x_{1}x_{2})=x_{1}x_{2}(1-\alpha )$$
the problem done just how we prove $x_{1}x_{2}\in\mathbb{Z}$ ??

COMMENT.-I am afraid your question is not true. Choose, for instance, $(x_1,x_2)=(3,5)$ so $$f(t)=t^3-23t^2+135t+225$$ and $f(1)+f(-1)-2(f(0)+1)=-48$ and $2f(-1)=-768$. However if you interchange the modulus you have $48$ divides $768$ and maybe you could have a correct question.
