<s>[INST] 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?
 [/INST] Since $f(1) = 0$ and $f(x) = f(x-1)$, we have $f(2) = f(1) = 0$. Hence if $\deg f \leq 2$,
$$
f(x) = a(x-2)(x-1),
$$
and so
$$
a(x-2)(x-1) = a(x-3)(x-2) \\
a(x-1) = a(x-3)
$$
for all $x$. Hence, $a=0$, so $f = 0$.
</s>