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

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.

• Probably no, but looks hopeless. There should be many primes for which $2p+1$ is prime and $4p-1$ is too large. Dec 27, 2020 at 12:39
• 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. Dec 27, 2020 at 17:51
• 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 $k$. Mar 3, 2021 at 10:54 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 $$1$$.