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?
 A: 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 $f = 0$.
A: 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 $f\equiv 0$.
A: You simply get that 
$$\{f(x)=ax^2+bx+c\mid f(1)=0, f(x)=f(x-1)\}=\{0\}$$
which is actually correct.
But in other way, you can say that $$\{f(x)\in \mathbb R_2[x]\mid \deg f=2,\  f(1)=0,\  f(x)=f(x-1)\}=\emptyset$$
A: You're not doing anything wrong. Think about it this way: $f(x) = f(x-1)$ means that the graph of $f(x)$ stays the same when you shift it to the right by 1 unit. Is it possible for any parabola or line of nonzero slope to have this property?
A: 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.) 
