Timeline for Algorithm to count number of positive integer solutions of $x^2(8x-3)=y^2z$?
Current License: CC BY-SA 3.0
10 events
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Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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Aug 3, 2013 at 23:46 | comment | added | Kirill | @DavidSpeyer No, it has the form $p(x)=y^2z$, not homogeneous. | |
Aug 3, 2013 at 20:53 | comment | added | David E Speyer | Is that meant to be $8x-3z$ (so that we get a homogenous cubic)? | |
Aug 3, 2013 at 17:42 | answer | added | user64494 | timeline score: 1 | |
Aug 3, 2013 at 12:35 | answer | added | Daniel Loughran | timeline score: 2 | |
Aug 3, 2013 at 12:06 | comment | added | Gerry Myerson | @Noam, yes, I was responding to the second question, about fixed $x$. | |
Aug 3, 2013 at 10:47 | answer | added | John R Ramsden | timeline score: 1 | |
Aug 3, 2013 at 4:36 | comment | added | Noam D. Elkies | @Gerry Myerson: That's the solutions for a given $x$, not for all $x,y,z$ with $x+y+z=n$. You can estimate the number of positive integer solutions of $x=yz$ with $x\leq n$ without factoring each $x$. | |
Aug 2, 2013 at 22:54 | comment | added | Gerry Myerson | Considering how closely the number of solutions is tied to the factorization of $x^2(8x-3)$, I'd be surprised if there were any way to count solutions that wouldn't be more or less equivalent to factoring $x^2(8x-3)$. Note, for example, that determining whether a number is squarefree is about as hard as factoring. | |
Aug 2, 2013 at 22:35 | history | asked | Kirill | CC BY-SA 3.0 |