Finf $f(x)$ which is a second degree polynomial, such that $f(1)=0$ and $f(x) = f(x-1)$ I must find a function $f(x) = ax^2+bx+c$ such that:
$$f(1) = a+b+c=0\\f(x)=f(x-1)\implies ax^2+bx+c = a(x-1)^2+b(x-1)+c\implies\\ax^2+bx+c = ax^2+(-2a+b)x+a-b+c\implies\\a = a, b = -2a+b, c = a-b+c$$
but this results fo $a=b=c=0$. What am I doing wrong?

If $f$ is a polynomial and $f(1)=0$ together with $f(x)=f(x-1)$, then $$0=f(1)=f(2)=f(3)=f(4)=\ldots $$
and $f$ has an infinite number of real roots. That implies $f\equiv 0$.
