Timeline for If $2^x $and $3^x$ are integers, must $x$ be as well?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 10, 2010 at 14:08 | comment | added | Jacques Carette | Indeed - I'll have to think about it some more. I think my argument at least rules out some cases (so reducing the problem to positive m and q=0), but then get 'stuck'. I was originally thinking that the hard cases would come from $q\neq 0$ or $m\lt 0$! | |
| Mar 10, 2010 at 4:58 | comment | added | Qiaochu Yuan | You are still assuming that m is a power of 3. This is not necessary if a counterexample to the conjecture exists. | |
| Mar 10, 2010 at 3:54 | history | edited | Jacques Carette | CC BY-SA 2.5 |
fix circularity
|
| Mar 10, 2010 at 3:47 | comment | added | Jacques Carette | Right - fixing that. | |
| Mar 10, 2010 at 3:46 | comment | added | Qiaochu Yuan | Er... you assumed that m was a power of 3 to prove that m is a power of 3. | |
| Mar 10, 2010 at 3:37 | history | edited | Jacques Carette | CC BY-SA 2.5 |
expand answer into what seems like a solution
|
| Mar 9, 2010 at 6:41 | history | made wiki | Post Made Community Wiki by Kim Morrison | ||
| Mar 9, 2010 at 3:51 | history | answered | Jacques Carette | CC BY-SA 2.5 |