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?

Choosing $ a = 1$, $b = -3$, $c = 5$ gives two positive integer roots to $f(x) = 2$ ($x = 1$ and $x = 2$) and $[f^2(1)+f^2(-1)]/2 = 23$. This statement appears to be false.
If $f^2(x)$ means $f(x)^2$, then the statement fails for $a =1$, $b = -6$, $c = 11$, where the two roots are 2 and 4 and $[f(1)^2+f(-1)^2]/2 = 157$.
