Timeline for Identity?: $\frac{\varphi(2^n-1)}{n}=\frac{2^{\varphi(2^n-1)}-1 \bmod (2^n-1)^2}{2^n-1}$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| 2 days ago | comment | added | joro | I asked related identities for (2^n+1)/3: mathoverflow.net/questions/483951/… | |
| Nov 27 at 16:45 | vote | accept | joro | ||
| Nov 26 at 17:47 | comment | added | joro | @FredHucht Thanks, I also noticed, that the main result of the answer works for random $a$ too. | |
| Nov 26 at 16:41 | comment | added | joro | Thanks, I think you are right. | |
| Nov 26 at 15:35 | comment | added | Fred Hucht | @joro Remember that Euler's totient fulfils $0<\varphi(m)<m$ for $m=2^n-1>1$. Your identity holds for arbitrary functions $\phi(m)=a n$ with this property. | |
| Nov 26 at 12:44 | comment | added | Petr Kucheryavy | @joro You want the identity $am = (2^{an}-1 \pmod{m^2})$. I assume that by equality over integers you mean that the right hand side is a number less than $m^2$. But $am < m^2$, that is why congruence $\mod m^2$ is enough. | |
| Nov 26 at 11:43 | comment | added | joro | @ChrisWuthrich I still fail to understand why congruence $\mod{m^2}$ in which there is exponent $a=(\varphi(2^n-1))/n$ implies equality over the integers. | |
| Nov 26 at 10:30 | comment | added | Chris Wuthrich | But $\varphi(2^n-1)/n<2^n-1$, so this is complete. It is a prime example of why formulating things as in (1) is harder to work with than congruences and inequalities. | |
| Nov 26 at 9:47 | comment | added | joro | Thanks, this is of interest. I edited clarifying that I am working over the integers and congruence is not a full answer to the question. | |
| Nov 25 at 19:42 | history | answered | Petr Kucheryavy | CC BY-SA 4.0 |