For all polynomials $q\in\mathbb Z[X]$ and all $n\in \mathbb N$ $$\exists k\in\mathbb N: q(p_n)\equiv q(-2k)\pmod{p_{n+1}}$$ Where $p_n$ is the $n$-th prime.
This conjecture is tested for random polynomials and seems to hold. Are there counterexamples? Is it a known conjecture? Is it trivial?
Because $p_n\equiv p_n-p_{n+1}\pmod {p_{n+1}}$, you can take $k=\frac{1}{2}(p_{n+1}-p_n)$, which will give you $p_n-p_{n+1}=-2k$. Thus, $q(p_n)\equiv q(p_n-p_{n+1})=q(-2k)\pmod {p_{n+1}}$.
This doesn't work for $k=1$, but then $q(2)\equiv q(-4)\pmod 3$ and you can take $k=2$, for example.
We have used the fact that $a\equiv b\pmod n$ implies $q(a)\equiv q(b)\pmod n$ for any $a,b,n\in\mathbb Z$, $n\gt 0$ and any $q\in\mathbb Z[x]$.