Timeline for Are there any solutions to $2^n-3^m=1$?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Sep 24, 2019 at 13:31 | history | edited | Ivan Izmestiev | CC BY-SA 4.0 |
Corrected spelling of aside and made other minor typographical improvements.
|
| S Sep 24, 2019 at 13:31 | history | suggested | J. W. Tanner | CC BY-SA 4.0 |
Corrected spelling of aside and made other minor typographical improvements
|
| Sep 24, 2019 at 12:55 | review | Suggested edits | |||
| S Sep 24, 2019 at 13:31 | |||||
| May 20, 2019 at 1:38 | comment | added | ErikR | @Junkie Updated link for Ivars Peterson's proof: archive.is/iRXz | |
| Jul 1, 2011 at 22:34 | answer | added | DavidLHarden | timeline score: 7 | |
| Jul 1, 2011 at 18:56 | comment | added | Gottfried Helms | Hmm, the question is already answered, and an answer is also accepted, so this is just an addendum. I'd simply reorder the equation into $2^n-1=3^m$ and use Euler's phi-function for the primefactors of the lhs and the powers of 3: to have a power k of 3 as factor of $2^n-1$ n must have the form $x*\varphi(3^k)=x*2*3^{k-1}$ where x is coprime to 2 and 3. After that, the lhs has additional (prime-)factors due to the $\varphi$-function for nontrivial n>1 except if n=6; here we can use the szigmondy-theorem or a simple comparision of the growthrate of the lhs and rhs, if k>2. | |
| Jul 1, 2011 at 13:22 | comment | added | Kevin | Thanks guys, that was much more interesting than I expected! | |
| Jul 1, 2011 at 13:21 | vote | accept | Kevin | ||
| S Jul 1, 2011 at 13:21 | vote | accept | Kevin | ||
| Jul 1, 2011 at 13:21 | |||||
| Jul 1, 2011 at 13:21 | vote | accept | Kevin | ||
| S Jul 1, 2011 at 13:21 | |||||
| Jul 1, 2011 at 12:34 | answer | added | Junkie | timeline score: 49 | |
| Jul 1, 2011 at 12:25 | comment | added | Junkie | Didn't Gersonides do this in 1343? en.wikipedia.org/wiki/Gersonides "One year later, at the request of the bishop of Meaux, he wrote The Harmony of Numbers in which he considers a problem of Philippe de Vitry involving so-called harmonic numbers, which have the form $2^m\cdot 3^n$. The problem was to characterize all pairs of harmonic numbers differing by 1. Gersonides proved that there are only four such pairs: (1,2), (2,3), (3,4) and (8,9)." Ivars Peterson gives an easy proof at maa.org/mathland/mathtrek_1_25_99.html | |
| Jul 1, 2011 at 12:19 | answer | added | Valerio Talamanca | timeline score: 13 | |
| Jul 1, 2011 at 12:12 | answer | added | Felipe Voloch | timeline score: 16 | |
| Jul 1, 2011 at 12:09 | comment | added | Todd Trimble | en.wikipedia.org/wiki/Catalan%27s_conjecture | |
| Jul 1, 2011 at 12:02 | history | asked | Kevin | CC BY-SA 3.0 |