Timeline for Special arithmetic progressions involving perfect squares
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 12, 2012 at 19:56 | comment | added | Noam D. Elkies | You're welcome. You saw that by now M.Stoll has completely solved the 4-term problem, even over the rationals. For the 3-term problem, you write that you "checked manually" up to $10^9$; how did you do that? I used the factorization of $t^2-1$ in $bc+1=t^2$ to search up to $t = 10^8$ in a few hours, still finding only the Pell solutions up to $(a,b,c) = (7865521, 58709048, 109552575)$; so $t=10^9$ (if that's what you meant) is feasible too, though it would take a while (or a number of CPU's running in parallel) to finish − and it's certainly not "manual" in the usual sense of "by hand"... | |
| Feb 12, 2012 at 5:10 | comment | added | Cosmin Pohoata | Thanks for the reply! This is essentially what I did to claim that I "checked it manually": I verified the correspondence between the solutions of the problem in question and $x^{2}-3y^{2}=1$ up to $10^{9}$ and then just assummed this to be true. I was hoping however to find some slick way of doing the four number case without refering to the three case :). | |
| Feb 12, 2012 at 5:07 | vote | accept | Cosmin Pohoata | ||
| Feb 12, 2012 at 5:11 | |||||
| Feb 11, 2012 at 21:43 | comment | added | Noam D. Elkies | I was reminded of this 1,3,8,120 puzzle too, because Martin Gardner wrote about it in one of his columns decades ago and reported that the nonexistence of a fifth positive integer was proved only with difficulty (including several 1000+ digit computations, back when that was impressive). But there are infinitely many such integer quadruplets. A few Google searches turned up vixra.org/pdf/0907.0024v1.pdf with citations going as far as Euler and Diophantus! But naturally none of these are four-term arithmetic progressions... | |
| Feb 11, 2012 at 20:37 | comment | added | Gerhard Paseman | Isn't there a similar problem involving 1,3,8,120, that says something like there is no fifth member such that the product of any two is one less than a square? It might be also that there are finitely many quadruplets, with the nontrivial ones being not arithmetic progressions. Gerhard "Ask Me About Obscure Puzzles" Paseman, 2012.02.11 | |
| Feb 11, 2012 at 19:39 | history | answered | Noam D. Elkies | CC BY-SA 3.0 |