Prove that $A = \dfrac{f^2(1)+f^2(-1)}{2}$ is a composite number 
Given $f(x) = a^{2016}x^2+bx+a^{2016}c-1$ where $a,b,c \in \mathbb{Z}$, suppose that the equation $f(x) = -2$ has two positive integer solutions. Prove that $A = \dfrac{f(1)^2+f(-1)^2}{2}$ is a composite number.

Let $g(x) = f(x)-2 = a^{2016}x^2+bx+a^{2016}c-3$, and let $-r,-s$ be the two positive integer roots of $g(x)$. Then by Vieta's Formula, $r+s = \dfrac{b}{a^{2016}}$ and $rs = \dfrac{a^{2016}c-3}{a^{2016}}$. Thus $a = \pm 1$ and so $f(x) = x^2+bx+c-1$ and so $f(1) = b+c$ and $f(-1) = c-b$. Then $$A = \dfrac{[(b+c)^2+b(b+c)+c-1]+[(c-b)^2+b(c-b)+c-1]}{2} = b^2+bc+c^2+c-1.$$ How do we prove this is composite?

The claim is not true: taking $a=1, \ b=2, \ c=3$, the equation $f(x) - 2 = 0$ is $x^2 + 2x = 0$, having both roots integers ($0$ and $-2$), but $A = \dfrac {5^2 + 1^2} 2 = 13$, clearly not a composite number.
