Does the equation in positive integers $\,(n,\,y)$ $$\prod_{k=1}^n(p_k^2-1)=y^2$$ only have the solution $(3,\,24)\,$?

I asked a more general question here.

The computational complexity of the problem can be slightly reduced considering that $$(2^2-1)(3^2-1)(5^2-1)=24^2$$ $$24|p^2-1\;\;\;\;\;p\gt3$$ Therefore, one can consider the equivalent problem $$\frac{\prod_{k=1}^{2n+3}(p_k^2-1)}{24^{2(n+1)}}=z^2$$ and the known solution $\,(n,\,z)=(0,\,1)$.

Up to $\,n=5\cdot10^4-2\,$ I have not found any other solution.

Many thanks.

  • 2
    $\begingroup$ Probably no, but looks hopeless. There should be many primes for which $2p+1$ is prime and $4p-1$ is too large. $\endgroup$ Dec 27, 2020 at 12:39
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    $\begingroup$ To expand on @FedorPetrov 's comment. the density of Germain primes en.wikipedia.org/wiki/Safe_and_Sophie_Germain_primes strongly suggest that for any such product which is sufficiently large, one will have at least prime p where $p^2 -1$ is divisible by a very big prime tha has not shown up anywhere else in the product. $\endgroup$
    – JoshuaZ
    Dec 27, 2020 at 17:51
  • $\begingroup$ This may have something to do with Mihailescu theorem (former Catalan conjecture), if we manage to prove $p_{k}^{2}-1$ has to be a perfect power as well as a divisor of $y$ for some $k$. $\endgroup$ Mar 3, 2021 at 10:54

1 Answer 1


Not an answer - only posting here since no pictures can be placed in comments

enter image description here

That the $n$th row of the above picture has black pixel at the $m$th position means that the $m$th prime has odd multiplicity in $\prod_{k\leqslant n}(\mathrm{Prime}_k^2-1)$. So a counterexample would mean having an entirely white horizontal line.

The widening black streak to the right actually comes from a string of $1$s; that is, the larger and larger amount of highest prime divisors all have multiplicity $1$. Should be not entirely impossible to show this. Actually, it obviously suffices to show that the highest prime divisor has multiplicity $1$.

  • $\begingroup$ I'm not sure this graph is helpful: I think a random number of a given size will display a similar behavior (i.e. listing the parity of the times each prime divides it, one would get a row similar to the ones in this graph for the corresponding size). In particular it will have several "ones" on the right. $\endgroup$ Dec 28, 2020 at 11:08
  • $\begingroup$ @JoelMoreira Well the fact is the number of ones to the right grows for this particular sequence; you may interpret this as saying that this sequence consists of "typical" numbers for their size; if this can be made rigorous it would suffice to prove what one needs. $\endgroup$ Dec 28, 2020 at 16:11

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