Timeline for Are there nontrivial rational solutions of $x^{n-m}=(1+t^m)/(1+t^n)$?
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19 events
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| Feb 3, 2018 at 11:02 | vote | accept | mehdi baghalaghdam | ||
| Feb 3, 2018 at 11:02 | |||||
| S Feb 2, 2018 at 7:01 | history | suggested | CommunityBot | CC BY-SA 3.0 |
Shortened title. Added tag.
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| Feb 2, 2018 at 4:50 | review | Suggested edits | |||
| S Feb 2, 2018 at 7:01 | |||||
| Feb 1, 2018 at 23:07 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
Added LaTeX formula delimiters in title
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| Feb 1, 2018 at 22:51 | answer | added | Gerhard Paseman | timeline score: 1 | |
| Feb 1, 2018 at 17:16 | comment | added | Gerhard Paseman | Yes but it could be an error. Let t=r/s and get an expression r^2+ s^2 times s^2 in the numerator and r^4 + s^4 in the denominator, and worry about cancellation. Later I will see about recovering my thought process. Gerhard "Working On Different Recovery Presently" Paseman, 2018.02.01. | |
| Feb 1, 2018 at 16:35 | comment | added | Max Alekseyev | @GerhardPaseman: Indeed, the equation $X^4 + Y^4 = 2Z^2$ has the only integer solution $(X,Y,Z)=(u,u,u^2)$, which is coprime only for $u=\pm 1$. However, if I've not mistaken, the case $(n,m)=(4,2)$ reduces to a different equation: $X^2 + Y^2 = 2Z^4$. Do you have some other reduction in mind? | |
| Jan 31, 2018 at 22:06 | comment | added | Gerhard Paseman | You need to do a little more work to take the right hand side and convert it to a fraction of square integers. When you do this, the primary issue that arises is if (part of) the denominator is twice a square. This is routine algebraic manipulation. If you do not see this, my first guess is that this is the wrong forum for your comments. Gerhard "Likely There Aren't Interesting Solutions" Paseman, 2018.01.31. | |
| Jan 31, 2018 at 21:30 | comment | added | mehdi baghalaghdam | please explain more about the first case.of course I know that Fermat case has not solution... And what do you think about the second example? | |
| Jan 31, 2018 at 21:23 | comment | added | mehdi baghalaghdam | I do not understand how the problem converts to the above problem. By letting t=s/r we get x^2(s^4+r^4)=t^4+s^2r^2. then? | |
| Jan 31, 2018 at 21:13 | comment | added | Gerhard Paseman | By representing t as a fraction with coprime integers s and r. You should get that 2 is the only possible prime divisor that cancels. Gerhard "That And Some Algebra Too" Paseman, 2018.01.31. | |
| Jan 31, 2018 at 21:10 | comment | added | mehdi baghalaghdam | thanks for your comments.how do you convert my problem to Fermat problem? | |
| Jan 31, 2018 at 21:02 | comment | added | Gerhard Paseman | I think your first example (2,4) asks for coprime integers r and s with $r^4 + s^4$ being twice a square. With the exception of r=s=1 (and maybe one example of Ljunggren), I believe there are no examples. It was shown by Fermat that the sum of two biquadrates is never a square. I think you can find literature supporting my belief. Gerhard "Have Faith In The Impossibilities" Paseman, 2018.01.31 | |
| Jan 31, 2018 at 20:40 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
latex fix
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| Jan 31, 2018 at 20:37 | history | edited | mehdi baghalaghdam | CC BY-SA 3.0 |
added 1 character in body
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| S Jan 31, 2018 at 20:29 | history | suggested | jeq | CC BY-SA 3.0 |
Added MathJax
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| Jan 31, 2018 at 20:27 | review | Suggested edits | |||
| S Jan 31, 2018 at 20:29 | |||||
| Jan 31, 2018 at 20:21 | history | edited | mehdi baghalaghdam | CC BY-SA 3.0 |
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| Jan 31, 2018 at 20:06 | history | asked | mehdi baghalaghdam | CC BY-SA 3.0 |