Timeline for Does the diophantine equation $\,\prod_{k=1}^n(p_k^{x_k}-1)=y^2\,$ have always at least a solution for $\,n\gt2\,$?
Current License: CC BY-SA 4.0
6 events
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| Dec 27, 2020 at 17:46 | comment | added | Will Jagy | @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. | |
| Dec 27, 2020 at 7:07 | comment | added | Tomita | 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 $(10, 4, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1), y=475169587200$ | |
| Dec 27, 2020 at 6:06 | comment | added | Mike Bennett | 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. | |
| Dec 26, 2020 at 23:31 | comment | added | Mike Bennett | For $n=9$, one has $(10,4,3,1,1,1,2,1,2)$ and $y=141419520$. | |
| Dec 26, 2020 at 22:02 | comment | added | Henri Cohen | It's immediate to find a solution for $n=8$, but I haven't found any for $n=9$. | |
| Dec 26, 2020 at 20:39 | history | asked | Augusto Santi | CC BY-SA 4.0 |