Let $f$ be a real continuous function such that $f(a+x) - f(x)$ is polynomial for all real $a$. Show that $f$ is polynomial.
I tried without success showing that $f$ is differentiable over $\mathbb{R}$, I also tried to use the fact that $\sum_{i=0}^n i^k $ is a polynomial in $n$ to write $f(ka)$ as a polynomial for fixed $a$ but it didn't bring much.
Lastly, I noticed that their exists an integer $N$ and a non trivial interval $]a,b[$ such that all of the polynomials $f_c$ are of degree $n$ for $c\in ]a,b[$ but I don't know how to use that.
Any hints are welcome ! Thanks
If $f_a(x)$ is a polynomial in $x$ of degree $n \ge 1$, and $b \ne 0$, then $f_a(x+b) - f_a(x) = f(x+a+b) - f(x+a) - f(x+b) + f(x)$ is a polynomial in $x$ of degree $n-1$. But note that this is also $f_b(x+a) - f_b(x)$. This implies that all the polynomials $f_a(x)$ for $a \ne 0$ have the same degree. Similarly, the leading coefficient of $f_a(x)$ must be $ac$ where $c$ is the same for all $a \ne 0$.
Now let $g(x) = f(x) - c x^{n+1}/(n+1)$. Then $g_a(x) = f_a(x) - c \frac{(x+a)^{n+1} - x^{n+1}}{n+1}$. $g$ has the same property as $f$, except $g_a(x)$ has lower degree than $f_a(x)$. In a finite number of steps, we get to a continuous function $h$ where $h_a(x)$ is constant for each $a$. That constant is $h_a(0) = h(a) - h(0)$. From this we can prove that $h$ is a polynomial of degree $\le 1$, and working backwards we get that $f$ is a polynomial.