Functional equation polynomial problem: $x^2f(x-1)=(x-1)^2f(x)$

Question

Let $f(x)$ be a polynomial such that $$x^2f(x-1)=(x-1)^2f(x)$$ which one of the polynomials $(a)$ $x(x-1)$, $(b)$ $2016x$ or $(c)$ $2016x^2$ can $f(x)$ be?

How should one approach this? I cannot see any options to use other than trying to find some roots, but even that seems very hard here.

Answer 1

$\frac {f(x-1)}{(x-1)^2}= \frac{f(x)}{x^2}$ $ \\ \Rightarrow \frac{f(x)}{x^2}= C \\ or \ f(x) = C{x^2}$

Answer 2

Evaluating $$x^2f(x-1)=(x-1)^2f(x)$$ at $x=0$ implies $f(0)=0$. Now by differentiating both sides of the equation we find $$2xf(x-1)+x^2f'(x-1)=2(x-1)f(x)+(x-1)^2f'(x).$$ Again, plug $x=0$ to deduce $f'(0)=0$. Hence, the only possible candidate is $f(x)=2016x^2$.