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?

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    $\begingroup$ Am I right guessing that $\;p_n\;$ is the $\;n\,-$ th prime? $\endgroup$
    – DonAntonio
    Feb 17, 2018 at 22:47
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    $\begingroup$ Please edit your question so that one can understand it without any guesswork. $\endgroup$ Feb 17, 2018 at 22:52
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    $\begingroup$ Maybe the quantifiers are throwing me, but isn't this trivial? After all $q(p_n)\equiv q(p_n+mp_{n+1})\pmod {p_{n+1}}$ for all $m$ so just take $m$ to be any negative number (setting it's parity as needed). $\endgroup$
    – lulu
    Feb 17, 2018 at 22:58
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    $\begingroup$ $p_n=p_{n+1}-2k\equiv-2k\,(\textrm{mod }p_{n+1})$, right? So then $q$ of the LHS $\equiv$ $q$ of the RHS, for any $q$. $\endgroup$
    – Ant
    Feb 17, 2018 at 23:00
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    $\begingroup$ @JohnChessant: yes of course. Thanks! $\endgroup$
    – Lehs
    Feb 17, 2018 at 23:12

1 Answer 1


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 $q\in\mathbb Z[x]$.


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