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 $f(x) = f(x-c)$
for all $x$, 
where $c$ is a constant,
and $f$ is a polynomial,
then
$f$ is constant.
Proof:
Let
$g(x) = f(x)-f(0)$.
Then $g(0) = 0$.
Also,
$g(-nc) = g(0) = 0$
for all positive integers $n$.
Since a polynomial of
degree $d$ has at most $d$
distinct zeros,
and $g$ has an infinite
number of zeros,
$g$ must be zero.
Therefore
$f(x) = f(0)$
for all $x$.
(If there is anything
original here,
I apologize.) 
