Who can post the solution of this problem?

HUY VAN wrote: Who can post the solution of this problem? This is set of all number stisfying

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We claim that, all such primes $p$ are Mersenne primes, namely, $p=2^k-1$ for a suitable $k$ (which itself is a prime). For convenience, recall that, the set $M_p$ of all Mersenne primes, less than $10000$ are $\{3,7,31,127,8191\}$. First, let $S=M_p\setminus \{p^*\}$. Note that, $q+1\mid \prod_{k}(p_k+1)=2^u$ for some $u$. In particular, $q+1$ is a Mersenne prime. If $p^*\notin S$, it holds that $S\cap M_p = \varnothing$, which yields that, it is impossible for such a $q$ to exist. Hence, $p^*\in S$, and therefore, $q=p^*$. Now, conversely, we shall prove that, all $p^*\in M_p$ satisfies the conditions. We begin with $p^*=3$. Note that, since $k\geq 2$, we immediately have $(p_1+1)(p_2+1)\cdots (p_k+1)$ is divisible by $3$, thus it suffices to take $q=3$. If $p^*=7$, take a subset $S$, not containing $7$. If $3\notin S$, we can simply take $q=3$. If $3\in S$, then note that $(p_1+1)(p_2+1)=4(p_2+1)$, and is divisible by $8$. Hence, we can take $q=7$. Now, suppose $p^*=31$, and take a subset. Again, if $3\notin S$, take $q=3$. If $3\in S$, and $7\notin S$, then take $q=7$. If $3,7\in S$, then $(p_1+1)\cdots (p_k+1)$ is divisible by $(3+1)(7+1)=32$. Thus, we may take $q=31$. Now, let $p^*=127=2^7-1$. Again, if $3\notin S$, take $q=3$, if $3\in S$, and $7\notin S$, take $q=7$. Now, suppose $3,7\in S$. If $31\notin S$, take $q=31$. Otherwise, if $31\in S$, then $(3+1)(7+1)(31+1)=2^{10}$ divides $(p_1+1)\cdots (p_k+1)$. Thus, we can take $q=127$. Finally, if $q=8191=2^{13}-1$, then do the same analysis, if $3,7,31,127\in S$, then $2^{17}\mid (p_1+1)\cdots (p_k+1)$, hence we can take $q=8191$.