Timeline for Prove that the equation $2^a - 2^b - 1=3^c$ has no integral solution with $a,b\geq 3$
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Sep 5, 2021 at 3:04 | vote | accept | Nya | ||
| Sep 5, 2021 at 1:03 | review | Close votes | |||
| Sep 11, 2021 at 6:51 | |||||
| Sep 5, 2021 at 0:21 | history | edited | GH from MO | CC BY-SA 4.0 |
edited title
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| Sep 5, 2021 at 0:21 | answer | added | GH from MO | timeline score: 3 | |
| Sep 5, 2021 at 0:16 | history | edited | GH from MO | CC BY-SA 4.0 |
edited title
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| Sep 5, 2021 at 0:00 | comment | added | markvs | Is it suppose to be $a,b>2$? | |
| Sep 4, 2021 at 23:34 | comment | added | Nya | Oh, 3^3 = 2^5 - 2^2 - 1 perfectly matches, I got my question answered, thank you very much | |
| Sep 4, 2021 at 23:30 | history | edited | Nya | CC BY-SA 4.0 |
deleted 1 character in body
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| Sep 4, 2021 at 23:29 | comment | added | Nya | Thank you, I forgot to clarify that the single zero in binary format is a must | |
| Sep 4, 2021 at 23:29 | comment | added | Andreas Blass | A bit less trivially, $3^3=2^5-2^2-1$. | |
| Sep 4, 2021 at 23:28 | history | edited | Nya | CC BY-SA 4.0 |
added 10 characters in body
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| Sep 4, 2021 at 23:28 | comment | added | Andreas Blass | $3^1=2^3-2^2-1$ | |
| S Sep 4, 2021 at 23:24 | review | First questions | |||
| Sep 5, 2021 at 7:42 | |||||
| S Sep 4, 2021 at 23:24 | history | asked | Nya | CC BY-SA 4.0 |