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P.G.Walsh proved in this paper that the diophantine equation $\,(2^{x_1}-1)(3^{x_2}-1)=y^2\,$ has no solution in positive integers $\,x_1$, $\,x_2\,$ and $\,y$.

If we generalize the previous equation to the following (in $\,n+1$ variables: $\,x_1,\,x_2,\,...,\,x_n,\,y$) $$\prod_{k=1}^n(p_k^{x_k}-1)=y^2$$ ($p_k$ being the prime of index $\,k$) positive integer solutions appear. For example: $$n=3\;\;\rightarrow\;\;(2,2,2)\;\;\;\;y=24$$ $$n=4\;\;\rightarrow\;\;(2,1,1,1)\;\;\;\;y=12$$ $$n=5\;\;\rightarrow\;\;(2,4,1,1,1)\;\;\;\;y=240$$ $$n=6\;\;\rightarrow\;\;(1,4,1,1,1,1)\;\;\;\;y=480$$ $$n=7\;\;\rightarrow\;\;(1,4,1,1,1,1,1)\;\;\;\;y=1920$$ Does the diophantine equation $$\prod_{k=1}^n(p_k^{x_k}-1)=y^2$$ have always (at least) a solution for $\,n\gt2\,$?

Many thanks.

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    $\begingroup$ It's immediate to find a solution for $n=8$, but I haven't found any for $n=9$. $\endgroup$
    – Henri Cohen
    Commented Dec 26, 2020 at 22:02
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    $\begingroup$ For $n=9$, one has $(10,4,3,1,1,1,2,1,2)$ and $y=141419520$. $\endgroup$
    – Mike Bennett
    Commented Dec 26, 2020 at 23:31
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    $\begingroup$ For $n \leq 14$, it is not too hard to find solutions. For $n=15$ (and likely $n \geq 15$), it seems to be rather trickier (one ends up playing a game of primitive divisor "whack-a-mole"). I suspect that no solutions exist, but proving this would be, I would suppose, beyond current technology. $\endgroup$
    – Mike Bennett
    Commented Dec 27, 2020 at 6:06
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    $\begingroup$ For $n=10$, solution is $(10, 4, 3, 2, 2, 2, 1, 1, 1, 1), y=1319915520,$ For $n=11$, solution is $(10, 10, 3, 3, 3, 3, 1, 1, 1, 1, 1), y=6310351111680,$ For $n=12$, solution is $(10, 10, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1), y=37862106670080,$ For $n=13$, solution is $(10, 4, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1), y=13000639905792000,$ For $n=14$, solution is $(10, 4, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1), y=475169587200$ $\endgroup$
    – Tomita
    Commented Dec 27, 2020 at 7:07
  • $\begingroup$ @MikeBennett one assumes that primes can be chosen, instead of consecutive beginning at $2,$ to arrange square for all the $x_i = 1,$ separate problem all $x_i = 2,$ and so on. Actually, I've done something similar, make a list of the first many primes and all prime factors of each $p-1$ for one problem or $p^2 - 1;$ then linear algebra over the field of two elements. Ordinary row reduction works very well. $\endgroup$
    – Will Jagy
    Commented Dec 27, 2020 at 17:46

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