Timeline for Product of arbitrary Mersenne numbers
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 6, 2019 at 16:34 | comment | added | hookah | By the way, you don't need the case $m$ even. $2^n$ must also be a square, forcing $n$ to be even, yielding that $m=n+1$ is odd, and thus, $2^m-3\equiv -1\pmod{3}$, contradiction. | |
| May 4, 2019 at 14:51 | comment | added | Gareth McCaughan | D'oh! Of course they are. You're quite right. Sorry. | |
| May 4, 2019 at 13:21 | comment | added | Seva | @GarethMcCaughan: No, I am not missing anything. The numbers $2^m-2^{m-n}-1$ and $2^n$ are coprime; if the product of two co-prime numbers is a square, then so is each of them. | |
| May 4, 2019 at 12:55 | comment | added | Gareth McCaughan | Not quite necessarily odd -- we could on the face of it have n=2. But in every other case n is odd. | |
| May 4, 2019 at 12:48 | comment | added | Gareth McCaughan | Aren't you missing a factor of $2$? $n$ is going to be an odd number, so it's $2(2^m+2^{m-n}-1)$ that's a square. | |
| May 4, 2019 at 10:28 | history | edited | Seva | CC BY-SA 4.0 |
deleted 89 characters in body
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| May 4, 2019 at 10:22 | history | answered | Seva | CC BY-SA 4.0 |