Timeline for Integer Solutions of $x+y^n = y + x^m$ for $n < m$
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| May 6, 2023 at 21:19 | comment | added | Nimish | This paper in the Journal of Mathematics from 1993 seems to have a proof: sciencedirect.com/science/article/pii/S0022314X83710413 | |
| Feb 20, 2014 at 6:04 | vote | accept | Willie Wu | ||
| Feb 20, 2014 at 5:27 | answer | added | Mike Bennett | timeline score: 11 | |
| Feb 20, 2014 at 4:33 | comment | added | Willie Wu | Pipe $P(v,R)$ is principal if for each $r\in R$, either $r$ does not divide $v$ or $(v-r)/r$ is not a product of elements in $R$. For more info, please refer to sites.google.com/site/basicpipetheory | |
| Feb 20, 2014 at 4:31 | comment | added | Willie Wu | For pipe $P(v,R)$, the number of elements in $R$ is called order of the pipe $P(v,R)$. | |
| Feb 20, 2014 at 4:29 | comment | added | Willie Wu | Let $R$ be a finite set of integers $\geq$ 2 and $v$ be an integer larger than any integer in $R$. The doublet $(v, R)$ is called a pipe if $v-r$ is a product of at least two elements in $R$ (multiplicities are taken into account) for each $r\in R$. We denote this doublet $(v, R)$ as $P(v, R)$ if it is a pipe. | |
| Feb 20, 2014 at 4:26 | comment | added | Gerry Myerson | What's a pipe? What's a principal pipe? What's a principal pipe of order 2? | |
| Feb 20, 2014 at 3:23 | comment | added | Robert Israel | These are all the solutions with $2 \le n < m \le 200$ and $2 \ge x,y \le 20000$. | |
| Feb 20, 2014 at 1:17 | history | edited | Willie Wu | CC BY-SA 3.0 |
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| Feb 20, 2014 at 0:27 | history | edited | Willie Wu | CC BY-SA 3.0 |
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| Feb 19, 2014 at 23:08 | history | edited | Willie Wu | CC BY-SA 3.0 |
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| Feb 19, 2014 at 23:02 | history | edited | Willie Wu | CC BY-SA 3.0 |
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| Feb 19, 2014 at 22:19 | comment | added | Willie Wu | Thanks. So far I have: $(n,m) = 1$, $y = \lceil x^{m/n} \rceil$ and $x = \lfloor y^{n/m} \rfloor$ | |
| Feb 19, 2014 at 21:52 | comment | added | Neil Strickland | Your equation is equivalent to $x(x^{m-1}-1)=y(y^{n-1}-1)$. If $p$ is a prime that divides $x-1$, then the $p$-adic valuation of $x(x^{m-1}-1)$ is $v_p(x-1)+v_p(m-1)$. If $p$ does not divide $x-1$ or $x$, then there are similar but more complicated statements that also involve the order of $x$ modulo $p$. You could probably get quite far with this kind of analysis. | |
| Feb 19, 2014 at 20:16 | comment | added | Willie Wu | Most likely that $(n-1)\mid(m-1)$ is true. | |
| Feb 19, 2014 at 19:59 | comment | added | The Masked Avenger | Such solutions are likely to involve consecutive powers. You might try looking at results of Pillai. Ribenboim's book on Catalan's equation has more, but I do not think it has this specific form. | |
| Feb 19, 2014 at 19:57 | comment | added | Willie Wu | Let $v = x + y^n = y + x^m$, then I found only 8 solutions for $v < 2^{63}$. | |
| Feb 19, 2014 at 19:51 | history | edited | Willie Wu | CC BY-SA 3.0 |
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| Feb 19, 2014 at 19:50 | comment | added | Willie Wu | I should add the condition: $n < m$ here. Let | |
| Feb 19, 2014 at 19:48 | history | edited | Claudio Gorodski |
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| Feb 19, 2014 at 19:45 | comment | added | Claudio Gorodski | What about $n=1$ or $m=1$? Why do you believe there are no other solutions if $n>1$ and $m>1$? Just because you cannot find them? | |
| Feb 19, 2014 at 19:32 | review | First posts | |||
| Feb 19, 2014 at 19:45 | |||||
| Feb 19, 2014 at 19:14 | history | asked | Willie Wu | CC BY-SA 3.0 |