Let
![p](/media/m/1/c/8/1c85c88d10b11745150467bf9935f7de.png)
be a prime number and
![f](/media/m/9/9/8/99891073047c7d6941fc8c6a39a75cf2.png)
an integer polynomial of degree
![d](/media/m/f/7/d/f7d3dcc684965febe6006946a72e0cd3.png)
such that
![f(0) = 0,f(1) = 1](/media/m/4/6/b/46b2ac6a2c3ba2330ec6b40f2b7d0de0.png)
and
![f(n)](/media/m/d/3/e/d3e47283bffbbf24c97f0c6474d5a82d.png)
is congruent to
![0](/media/m/7/b/8/7b8b0b058cf5852d38ded7a42d6292f5.png)
or
![1](/media/m/a/9/1/a913f49384c0227c8ea296a725bfc987.png)
modulo
![p](/media/m/1/c/8/1c85c88d10b11745150467bf9935f7de.png)
for every integer
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
. Prove that
![d\geq p - 1](/media/m/4/1/2/412cc47921fae502fb50defab1973776.png)
.
%V0
Let $p$ be a prime number and $f$ an integer polynomial of degree $d$ such that $f(0) = 0,f(1) = 1$ and $f(n)$ is congruent to $0$ or $1$ modulo $p$ for every integer $n$. Prove that $d\geq p - 1$.