Timeline for Are there nontrivial rational solutions of $x^{n-m}=(1+t^m)/(1+t^n)$?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 3, 2018 at 21:17 | comment | added | Max Alekseyev | @mehdibaghalaghdam: Anyway, reduction proposed by Gerhard Paseman works better here. I've shown mine just out of curiosity. | |
| Feb 3, 2018 at 21:08 | comment | added | Max Alekseyev | @mehdibaghalaghdam: $(u^2,u^2,u)$ is not the only solution to $X^2+Y^2=2Z^4$. E.g., $((u^2+1)(u^2+2u-1),(u^2+1)(u^2-2u-1),u^2+1)$ is another one. | |
| Feb 3, 2018 at 20:37 | comment | added | mehdi baghalaghdam | Yes. This is right. thank you very much for your comments. we know that the soution of the equation x^2+y^2=2z^4 is (u^2,u^2,u). Then we get (2r^2p^2-s^2q^2=2s^2p^2-s^2q^2 , i.e., r=s or t=1 which leads to the trivial solution t=r/s=1. Then the case 2,4 has only trivial solution. | |
| Feb 3, 2018 at 19:12 | comment | added | Max Alekseyev | @mehdibaghalaghdam: Indeed, I lost a couple of coefficients. The correct equation is $$(2r^2p^2 - s^2q^2)^2 + (2s^2p^2-s^2q^2)^2 = 2(sq)^4.$$ | |
| Feb 3, 2018 at 11:02 | vote | accept | mehdi baghalaghdam | ||
| Feb 3, 2018 at 11:02 | |||||
| Feb 3, 2018 at 9:55 | comment | added | mehdi baghalaghdam | Dear Alekseyev ; very thanks, note that by letting x=p/q, t=r/s in the equation, we get the relation $p^2s^4+p^2r^4-q^2s^4-q^2s^2r^2=0$ which is not equivalent with $p^2s^4+p^2r^4-2r^2s^2q^2-2s^4q^2=0$.(this is obtained after some simplification in the above relation.) !! Am I saying right? | |
| Feb 2, 2018 at 18:33 | comment | added | mehdi baghalaghdam | thanks, but t=0,1,-1 are trivial solution. | |
| Feb 2, 2018 at 16:48 | comment | added | LSpice | Sorry, I don't know what I was thinking. Clearly $t = 1$ does indeed give a solution. | |
| Feb 2, 2018 at 16:07 | comment | added | mehdi baghalaghdam | thank you very much for your valuable comments. I think that there exist integers n,m such that the equation has nontrivial solutions. What do you think about this? | |
| Feb 1, 2018 at 23:52 | comment | added | Gerhard Paseman | Actually "gives" instead of "is". But I will leave it as is. Gerhard "Solutions Can Also Be Indirect" Paseman, 2018.02.01. | |
| Feb 1, 2018 at 23:32 | comment | added | LSpice | $x = 1$ instead of $t = 1$, I guess. | |
| Feb 1, 2018 at 22:51 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |