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For each positive integer $n$ let $P_n=\prod_{k=1}^n p_k$, where $p_k$ is the $k$th prime.

Question. Is my following conjecture true?

Conjecture. For any integer $n>1$, there are $k,m\in\{1,\ldots,n-1\}$ such that $$P_n\equiv P_k\pmod n\ \ \text{and}\ \ P_n\equiv -P_m\pmod n.$$

For example, $P_{32}\equiv P_{23}\pmod{32}$ and $P_{32}\equiv -P_8\pmod{32}$.

I have verified the conjecture for all $n=2,3,\ldots,70000$. When $n$ is squarefree, the conjecture holds trivially. I'm unable to prove the conjecture fully.

For the motivation of the conjecture, one may look at Conjecture 1.5 and Remark 1.7 in my paper available from http://dx.doi.org/10.1016/j.jnt.2013.02.003.

Your comments are welcome!

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  • $\begingroup$ As you noted, $P_n$ is always divisible by $\operatorname{rad}(n)$, so dividing everything by $\operatorname{rad}(n)$ gives a question about products of primes modulo $d = n/\!\operatorname{rad}(n)$. If $p_i | n$, then $i \log(i) \approx p_i \leq n$, so $i \leq n/W(n)$, so for $n \geq 70000$ this forces $i \leq n/8$. For such $p_i$ you need $i \leq k, m$, so this still leaves $k,m \in \{n/8,\ldots,n\}$. $\endgroup$
    – R. van Dobben de Bruyn
    May 30, 2020 at 23:41
  • $\begingroup$ That said, according to this paper, not much is known about the distributions of consecutive primes modulo an integer. So you probably really want to exploit that your intervals are rather large compared to the modulus $d = n/\!\operatorname{rad}(n)$. $\endgroup$
    – R. van Dobben de Bruyn
    May 30, 2020 at 23:42

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