Timeline for Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 8, 2014 at 7:14 | comment | added | H A Helfgott | That's nice - yes, the question we are discussing is really about the cycles of $f_g(u)$, in the paper's notation. It looks like my (self-)answer improves on Theorem 6. Unfortunately, the paper contains no bounds on what it calls $N_g(4)$ (which is what we are trying to bound non-trivially here). | |
| Jul 8, 2014 at 7:12 | comment | added | Lev Glebsky | Here are an estimates (From above and, I think, very imprecise)of number of solutions of $q^{x^{x^x}}=x \mod p$ based on similar constructions. There are some annoying details due to $x+k (\mod p)= x+k (\mod p-1)$ or $x+k(\mod p-1) +1$. So, $2^{x+k}= 2^x2^k$ or $22^x2^k\mod p$. I don't think that there is an easy generalization for points of period 4. I think that the proof goes through due to the group $\langle a,b,w\;|\;b^{-1}ab=b^2,w^{-1}aw=b, w^3=1\rangle$ is finite. (We may define "almost action" of this group, with $xa=x+1$, $xb=2x$, $xw=2^x\mod p$) | |
| Jul 8, 2014 at 6:45 | comment | added | Lev Glebsky | Here an estimates of | |
| Jul 5, 2014 at 17:53 | comment | added | Sean Eberhard | The term $2^y + k$ in the displayed equation preceding (**) should be $2^y 2^{l_i} + k$, but I don't think that affects at all the validity of this nice proof. One thing I was worried about was the possibility of $2$ having small order modulo $p-1$, so that if you take $\gg p$ values $x$ and pass to $y= 2^{2^x}$ you don't necessarily have $\gg p$ values $y$. But in fact if $2$ doesn't have order $\gg p$ modulo $p-1$ then you're immediately done because $2^{2^{2^x}}$ would have small image. Similarly the assumption that $2$ is primitive mod $p$ is superfluous. | |
| Jul 4, 2014 at 20:43 | comment | added | H A Helfgott | Note that the argument above is in some sense "local", in that relies entirely on comparisons between the values of functions at $x$ and at $x+k$, where $k$ is bounded by a constant. Something tells me that this will not be so straightforward for the quadruple power. | |
| Jun 28, 2014 at 21:30 | history | answered | H A Helfgott | CC BY-SA 3.0 |