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Is there a set of triple-primes satisfying the following equation?

$p_1p_2+p_2p_3+p_3p_1+p_1+p_2+p_3=2^β,p_1p_2p_3=2^α-1,α>β.$ I have checked the first 11 numbers that no one satisfy the above condition. It's easy to check that when all the prime numbers $p_i$ are in the form of $4k+3,$ then we have $p_i=2^{s_{i}}k_i+2^{s_{i}}−1$

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    $\begingroup$ In other words, $(p_1+1)(p_2+1)(p_3+1) = 2^\alpha + 2^\beta$. $\endgroup$
    – Max Alekseyev
    Commented Mar 27, 2016 at 23:41
  • $\begingroup$ Crossposted at MSE. $\endgroup$
    – Dietrich Burde
    Commented Mar 28, 2016 at 15:27

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Yes, there is precisely one such triple of distinct primes with $\alpha \leq 500$, namely $(3, 11, 31)$.

It would be a surprising coincidence if there are further such triples for larger $\alpha$.

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    $\begingroup$ I've checked all $\alpha \leq 1200$ -- there are no new triples. $\endgroup$
    – Max Alekseyev
    Commented Mar 27, 2016 at 23:32
  • $\begingroup$ if we let $\alpha>\beta+1,gcd(2^{\alpha}-1,2^{\beta}+1)=1$.that means there almost no such primes satisfying the equation,at least for $\alpha \leq 1200$ is true. $\endgroup$
    – qian feng
    Commented Mar 28, 2016 at 2:11

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