Timeline for Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Jul 7, 2014 at 22:03 | history | bounty ended | CommunityBot | ||
| Jul 6, 2014 at 20:19 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
deleted some text
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| Jul 6, 2014 at 11:02 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
added more texxt
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| Jul 5, 2014 at 0:39 | comment | added | Geoff Robinson | Once you have agreed a convention that makes $f$ a bijection, it makes sense to iterate $f$ as often as you like. If $2$ generates the multiplicative group (mod $p$), then $x \to 2^{x}$ is a well defined bijection from $\{1,2, \ldots, p-1 \}$ to itself. | |
| Jul 4, 2014 at 20:40 | comment | added | H A Helfgott | Interpret the question as loosely or as strictly as you wish, as long as it helps to make it possible to say something meaningful. Want to define $f(x)=2^x$ for $x\in {0,1,\dotsc,p-2}$, $f(x) = 0$ for $x=p-1$, and $f(x) = f(\overline{x})$ elsewhere, where $\overline{x}$ is the element of $\{0,1,\dotsc,p-1\}$ congruent to $x$ modulo $p$? Be my guest. | |
| Jul 4, 2014 at 18:16 | comment | added | Rodrigo | My point is that, as defined, the cycles of length $3$ of $f$, for instance, don't say anything about the fixed points of $2^{2^{2^x}}$. In order to iterate $f$, we need to reduce the exponent modulo $p-1$, not $p$, and so on. | |
| Jul 4, 2014 at 18:08 | comment | added | Geoff Robinson | You have to adopt the convention that you read $x$ as being between $1$ and $p-1$ when you calculate $2^{x},$ then it becomes well defined. The OP already stated that in comments below his question. that is why you get $f(x)f(-x) =2$, because you have to interpret $f(-x)$ as $2^{p-x}$ for $0 < x < p.$ | |
| Jul 4, 2014 at 18:01 | comment | added | Rodrigo | I think the question (if interpreted exactly as stated) can't be solved by studying the map $f:x \rightarrow 2^x\, mod\, p$, since $x$ mod $p$ doesn't define $2^x$ mod $p$. Trying such approach would require us to study integers modulo $p\phi(p)\phi(\phi(p))$...etc. However this approach would probably give us some trouble in working with integers specifically in the interval $(0,p)$. | |
| Jul 4, 2014 at 7:45 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Tidying up
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| Jul 4, 2014 at 7:39 | history | undeleted | Geoff Robinson | ||
| Jul 4, 2014 at 7:37 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
updated answer
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| Jul 3, 2014 at 15:39 | history | deleted | Geoff Robinson | via Vote | |
| Jul 3, 2014 at 15:38 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
added 118 characters in body
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| Jul 3, 2014 at 15:37 | history | undeleted | Geoff Robinson | ||
| Jul 3, 2014 at 14:58 | history | deleted | Geoff Robinson | via Vote | |
| Jul 3, 2014 at 14:47 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
typos
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| Jul 3, 2014 at 12:55 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
typos and minor textual changes
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| Jul 3, 2014 at 12:42 | history | answered | Geoff Robinson | CC BY-SA 3.0 |