Infinite exponentiation $n^{n^{n^{...^n}}} \equiv m \pmod q$ , find m?

Question

let $(n,q) \in \mathbb N^{*^2}$

I was wondering if it was possible to find a function $f_q$ such that :

$f_q(n)=m$ where $m$ is such that $n^{n^{...^n}} \equiv m \mod q$

or at least an easy way to find m, where $n^{n^{...^n}}$ denotes "infinite" exponentiation, assuming of course it has a "limit" if you see what I mean (that's to say, if you exponentiate long enough, $n^{n^{n^{...^n}}} \equiv m \mod q$, m becomes constant. Is it always the case ? I don't know, but it has to be at least periodic since m belongs somewhere between $0$ and $q-1$... so it has to repeat itself after some exponentiations) To make things easier, let's only consider n for which $n^{n^{n^{...^n}}}$ tends to $m \mod q$ constant

I have made some trials and it seems like it often has a "limit" ($m$ constant)

For example, $f_{10}(7)=3$ since :

$7^7 \equiv 49*49*49*7 \equiv 9*9*9*7 \equiv 81*63 \equiv 3$ mod 20

$7^{3+20k} \equiv 343*(343*343*343*7)^{2k} \equiv 3*(3*3*3*7)^{2k} \equiv 3*81^k \equiv$ 3 mod 20

So $7^{7^{...^7}} \equiv 3$ mod 20, and of course $\equiv 3$ mod 10

I have done this for other numbers, and could provide them with proof but it's a very exhausting process when n becomes big and I couldn't find any obvious pattern...

Sorry if the explanation is a bit messy, the maths I am trying to do isn't very conventional... and probably not very useful either, but I'm still curious... If additional explanation is needed, I'll just edit and add some information

Thanks a lot for anyone ready to help me tackle this not-so-easy arithmetic problem... To be honest I don't think there is a general solution, but I'd be very grateful if someone could find f for even a small subset of $N$.

One tricky problem here is that exponentiation is not really a modular operation in the exponent place; we generally have $a^b\bmod q \ne a^{b\bmod q}\bmod q$. So the naive argument that you always reach a period doesn't seem to work. — hmakholm left over Monica, Aug 15 '15 at 15:42
How are you calculating this? Is $2^{2^{2^{2}}}=65536$ or $256$? — Zach466920, Aug 15 '15 at 15:45
Perhaps there's a way to find a restriction on n and q so that it works for sure if we have exponentiated long enough ? I'll try to think about the two comments posted so far, thanks. — Evariste, Aug 15 '15 at 15:45
Zach466920 - You are inquiring about n=2, but q has to be defined, for example let's say q=10. We have $2^{2^2}=16=6+10k$, k odd. Since $2^{6+10k} \equiv 64(1024)^k \equiv 4^{k+1}$ mod 20, k odd, we have $2^{6+10k} \equiv 16^p$ p even. 1616=256, so we have a constant loop and $2^{6+10k} \equiv 16$ mod 20. With a similar reasoning to the one I showed in my post, we get $f_{10}(2)=6$ (I hope I didn't make any mistakes...) — Evariste, Aug 15 '15 at 15:55
For natural numbers $a,n$, there is a number $k$, such that $a\uparrow \uparrow m \equiv a\uparrow \uparrow k$ (mod $n$). for all $m\ge k$. So, the sequence of residues becomes eventually stationary. You can reduce the base modulo $n$, the first exponent modulo $\phi(n)$, the second exponent modulo $\phi(\phi(n))$ and so on. But the calculation is not easy in general, especially for high modules. — Peter, Aug 15 '15 at 15:57
As soon as the nested $\phi$-function has the value $1$, you can remove the corresponding exponent and all higher ones. — Peter, Aug 15 '15 at 16:03
@Evariste I think Zach466920 is wondering whether you're grouping your operations as $2^{(2^{(2^2)})}$ or as $((2^2)^2)^2$. — Milo Brandt, Aug 15 '15 at 16:03
The second version would be far easier to handle, but I do not hink the OP means that. — Peter, Aug 15 '15 at 16:07
Milo Brandt - Nevermind then, I meant $2^{2^{2^2}}=2^{16}=65536$ and not $2^{222}=2^8=256$ — Evariste, Aug 15 '15 at 16:07
@Peter: You should probably post that as an answer. (Unfortunately, implementing it seems to require computation of totients, which is thought to be hard, but it may be the best we can get here). — hmakholm left over Monica, Aug 15 '15 at 16:09
@Evariste Note, that $a\uparrow \uparrow n$ is a common notation for $\large a^{a^{a^...}}$ with $n$ $a$'s in the power tower. — Peter, Aug 15 '15 at 16:10
@Henning I would do that, but I think it is not enough for an answer. But if you do think, it is, I can do it. — Peter, Aug 15 '15 at 16:11
For $n = 2$ see What's the formula for the fixed point of a power tower of 2s modulo n?. For $n = 3$, see Do the last digits of exponential towers really converge to a fixed sequence?. — J.-E. Pin, Aug 15 '15 at 16:15
See also How can I effectively compute tetration mod a? on mathoverflow. — J.-E. Pin, Aug 15 '15 at 16:19

score 5 · Accepted Answer · answered Aug 15 '15 at 16:14

Yes, this is always possible, though it's not necessarily trivial to prove. Let's start with the easier case: assume $n$ and $q$ are coprime. Then, the following identity defines $f_q(n)$ uniquely (mod $q$): $$f_q(n)\equiv n^{f_{\phi(q)}(n)}\pmod q$$ where $\phi(q)$ is Euler's totient function, equaling the number of integers between $0$ and $q$ which are coprime to $q$. For instance, to obtain $f_{10}(7)=3$ using this, we note that $\phi(10)=4$ and $\phi(4)=2$ and $\phi(2)=1$ and then using the recursion (noting that $f_1$ is define mod $1$, so can be chosen arbitrarily): $$f_1(7)\equiv 0$$ $$f_2(7)\equiv 7^0\equiv 1$$ $$f_4(7)\equiv 7^1 \equiv 3$$ $$f_{10}(7)\equiv 7^3 \equiv 3$$ Since that function is decreasing, applying the second identity finitely many times eventually reduces to $q=1$. Notice, as rough bound on how many exponents we need, we can say $$f_{q}(n)=\underbrace{n^{n^{n^{\ldots ^{n}}}}}_{q\text{ times}}$$ but this is much harder to computer than just using the recurrence relation.

The crucial fact used here is that, for any $q$ and large enough $k$, the following is true: $$a^k\equiv a^{k+\phi(q)}\pmod q$$ meaning exponentiation is eventually periodic, with period $\phi(q)$, so knowing the infinite power tower's value mod $\phi(q)$ suffices to determine it mod $q$. The nicest proof of this fact I know is that the integers coprime to $q$ taken mod $q$ form a group under multiplication of order $\phi(q)$ and $\{a^0,a^1,a^2,\ldots,\}$ is a subgroup thereof so its order $k$ divides $\phi(q)$ due to Lagrange's theorem. This implies $a^k=a^0$ and hence $a^{\phi(q)}=a^0$, which implies the desired result. One may notice that a similar thing works when $n$ is not coprime to $q$ - exponentiation $n^x$ is still eventually periodic mod $q$. One may prove, as a loose bound, that if $x>\log_2(q)$ then $n^{x}\equiv n^{x+\phi(q)}\pmod{q}$, which allows us to compute $f$ in a similar manner as long as the representative we choose of $f_{\phi(q)}(n)$ mod $\phi(q)$ is big enough.

Infinite exponentiation $n^{n^{n^{...^n}}} \equiv m \pmod q$ , find m?

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