Timeline for Elementary constraints for the solutions of $z^{n-2}y(y+z)=x^n$?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| May 25, 2019 at 11:17 | comment | added | joro | @Wojowu Your reduction from FLT is slightly wrong: $X=a c b^{n-2}$ | |
| May 25, 2019 at 5:39 | comment | added | joro | @Wojowu Thanks. Is the converse true: all points on C_n are on the Fermat curve. By the argument about equating exponents of p in the LHS and RHS Y,Z are n-th powers. This leads to Z_1^(n*(n-2)) Y_1^n (Y_1^n+Z1^n)=X^n. All factors except (Y_1^n+Z1^n) are n-th powers so it must be n-th power too. | |
| May 24, 2019 at 18:23 | comment | added | Wojowu | It makes sense, and you can't prove there are no solutions without FLT - if $a^n+b^n=c^n$ with $a,b,c$ coprime, then $X=b^{n-2}c,Y=a^n,Z=b^n$ is a solution to your equation. | |
| May 24, 2019 at 15:28 | comment | added | joro | @Wojowu I edited, does it make sense now? | |
| May 24, 2019 at 15:27 | history | edited | joro | CC BY-SA 4.0 |
want elementary proof
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| May 24, 2019 at 15:20 | comment | added | joro | @Wojowu Thanks, will edit! The other answer assumes FLT and I am interested in elementary proof, not using heavy stuff like FLT and modularity. | |
| May 24, 2019 at 14:40 | comment | added | Wojowu | What partial results? There are no solutions, what else is there to say? | |
| May 24, 2019 at 14:39 | comment | added | joro | @Wojowu Indeed. I am mainly interested in extending the partial results, if this gets closed will move it in the other question. | |
| May 24, 2019 at 14:35 | comment | added | Wojowu | If $(X,Y,Z)$ was a nontrivial solution, then $x=X/Z,y=Y/Z$ would satisfy $x^n=y(y+1)$. How does the answer to the other question not answer this one? | |
| May 24, 2019 at 14:20 | history | edited | joro |
edited tags
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| May 24, 2019 at 13:14 | history | asked | joro | CC BY-SA 4.0 |