Let $f(x)$ be an integer-valued polynomial (when $x\in \mathbb{Z}$, then $f(x)\in \mathbb{Z}$), and $a,b$ be positive integers, and $p$ be a prime number with $(a,p)=1$.


Show that
$$\sum\_{x=0}^{p-1}\left(\dfrac{f(ax+b)}{p}\right)=\sum\_{x=0}^{p-1}\left(\dfrac{f(x)}{p}\right)$$


If $f(x)=x$, it is well known
$$\sum\_{x=0}^{p-1}\left(\dfrac{ax+b}{p}\right)=\sum\_{x=0}^{p-1}\left(\dfrac{x}{p}\right)=0$$



Sums of Legendre symbols with integer-valued polynomials
The statment is false. For a counterexample, take $f(x)=\binom{x}{3}$, $p=3$, $a=2$, $b=0$.


The statement is true when $f$ has degree less than $p$, or when $f$ has integral coefficients.


