Timeline for Diophantine problem
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Feb 16, 2011 at 23:23 | comment | added | Gerhard Paseman | Instead of keeping up with the changes, I'll just suggest the idea above: use n+i for x_1, n+j for x_2, n+k for x_3, or something similar, and see where that takes you. If you develop it and want more suggestions, you can respond here. Again, good luck. Gerhard "Going To Play Elsewhere Now" Paseman, 2016.02.16 | |
| Feb 16, 2011 at 23:18 | comment | added | Gerhard Paseman | Oops: In the comment above, I mean s = (i+j+k) now, not s=(i+j+k+1). Perhaps you should do it yourself. Gerhard "Ask Me About System Design" Paseman, 2011.02.16 | |
| Feb 16, 2011 at 23:13 | comment | added | Gerhard Paseman | In the new version (using s and t), we have s > 0, 2sn +t =a, and ijk -sn^2 = b. With n given, one can get s and t in terms of a and b nicely. (I'll let you do that.) The same general advice holds. Good Luck. Gerhard "Ask Me About System Design" Paseman, 2011.02.16 | |
| Feb 16, 2011 at 21:24 | history | edited | Gerhard Paseman | CC BY-SA 2.5 |
fixed strike
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| Feb 16, 2011 at 21:14 | history | answered | Gerhard Paseman | CC BY-SA 2.5 |