Prove that there does not exist any function $f:\mathbb{R}\to\mathbb{R}$ such that $$f(f(x))=x^2-2016$$ for all real number $x$.
My attempt :
Substitute $x=f(x)$ in $f(f(x))=x^2-2016$,
we have $f(f(f(x)))=f(x)^2-2016$
so $f(x^2-2016) = f(x)^2-2016$ ---[1]
$f$ is onto $(-2016, \infty)$
Since $x^2 \geq 0$ so $x^2-2016 \geq -2016$
$f(f(x))\geq -2016$
substitute $x=-x$ in[1], $f(x)^2-2016=f(-x)^2-2016$
so $f(x)^2 =f(-x)^2$
Please suggest how to proceed.
By the answer linked in the comments, no such function exists. Using the notation at the linked MSE response, you would have $\Delta(f) = 0 - 4(1)(-2016) > 1$, which rules out existence of a function satisfying the desired criterion (cf. Theorem 6 in the following pdf).