$p\cdot q\cdot r=a^3-1+43\cdot (b^2-1)$
p, q, r are primes.
a, b integers>0.
Is this equation a Mordell equation? Has this equation infinitely many solutions?
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2$\begingroup$ Any motivation for this question? To a non-expert it looks completely arbitrary. $\endgroup$– abxCommented Mar 26, 2020 at 19:35
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2$\begingroup$ Is there any reason you picked 43? $\endgroup$– Robert IsraelCommented Mar 26, 2020 at 21:12
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1$\begingroup$ Please don't ask random "unmotivated" questions. Note that this site was created for professional mathematicians and PhD students to help their research efforts. Of course anyone can ask and answer questions, but this is the level expected here. $\endgroup$– GH from MOCommented Mar 26, 2020 at 22:39
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1$\begingroup$ @Robert, perhaps OP is in the $43$ cabal that has been asking questions involving $43$ here and on m.se for a while now. E.g., math.stackexchange.com/questions/3574600/… and math.stackexchange.com/questions/3566264/… and math.stackexchange.com/questions/3564437/… and more. $\endgroup$– Gerry MyersonCommented Mar 26, 2020 at 22:57
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1$\begingroup$ @GerryMyerson I don't know about a cabal but from the quirks of idiom in certain recent questions, e.g. mathoverflow.net/questions/355415/… , my guess is that a single user is creating unregistered accounts, using them to post these rather unmotivated questions, and then just creating new accounts the next time a question occurs to them $\endgroup$– Yemon ChoiCommented Mar 28, 2020 at 1:17
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1 Answer
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Consider e.g. $b=1$ and $a = 2k+1$, so the equation becomes $pqr =2 k (4k^2 + 6k + 3)$. The generalized Bunyakovsky conjecture implies there are infinitely many primes $k$ such that $4 k^2 + 6 k + 3$ is prime, so that we have a solution with $p=2, q=k, r=4k^2 + 6 k + 3, a=2k+1, b=1$. To date all nontrivial cases of that conjecture, including this one, remain open.
answered Mar 26, 2020 at 21:37