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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 $$\sum_{x=0}^{p-1}\left(\dfrac{ax+b}{p}\right)=\sum_{x=0}^{p-1}\left(\dfrac{x}{p}\right)=0$$

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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.

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  • $\begingroup$ if $f$ is integral coefficients,is old result? can you post solution?Thanks $\endgroup$
    – math110
    Commented Apr 18, 2022 at 0:27
  • $\begingroup$ @msexkac If $f$ has integral coefficients, then it acts on residue classes mod $p$. That is, $f(x)\bmod{p}$ only depends on $x\bmod{p}$. This is trivial from the fact that reducing mod $p$ is a ring homomorphism from $\mathbb{Z}$ to $\mathbb{Z}/p\mathbb{Z}$. Now, as $(a,p)=1$, the map $x\mapsto ax+b$ permutes the residue classes mod $p$. So your two sums contain the same terms but in different order. So they are equal. Same story when $f$ has degree less than $p$, but in this case it is harder to show that $f$ acts on residue classes mod $p$. $\endgroup$
    – GH from MO
    Commented Apr 18, 2022 at 0:35

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