Timeline for Are there integer solutions of $m^4+m^2n^2+n^4=k^2$?
Current License: CC BY-SA 4.0
3 events
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| Nov 22 at 11:26 | comment | added | Peter Mueller | @Wolfgang Mordell does not give a precise reference, but alludes that this might go back to Fermat or Euler. I had tried the more obvious factorizations $(m^2+mn+n^2)(m^2-mn+n^2)=k^2$ and $(m^2+n^2+k)(m^2+n^2-k)=m^2n^2$, but for neither could I find an infinite descent. | |
| Nov 22 at 9:48 | comment | added | Wolfgang | Excellent! The "crucial trick" is an idea of genius... | |
| Nov 21 at 13:16 | history | answered | Peter Mueller | CC BY-SA 4.0 |