# $2^{3^{4^{…^{n}}}} \equiv 1$ (mod $n+1$)

I remember when I started learning modular arithmetics I found a tetration equation stated as follows

$2^{3^{4^{...^{n}}}} \equiv 1$ (mod $n+1$)

I am wondering how could this be proved, I tried this but I got lost:

$2^{3^{4^{5}}}$ (mod $6$) $\equiv 2^{3^{4^{5}} mod 5}$(mod $6$)

How can I prove it?

• You need to give a condition for $n$. When $n=3$, $2^3$ is not congruent to $1$ mod $3+1$ – TheSimpliFire Aug 30 '18 at 18:08
• I hadn't been given any condition :0 – alienflow Aug 30 '18 at 19:51
• Have you tried the power rule? Where this boils down to 2 ^ 3 * 4 *..n. Then you have 2 * 2 * 2... *2 which could probably be boiled down to 2(something n + 1) or the equivalent [needs more though] for which you can get 1(modn + 1). I don't have a lot of time right now but that would be how i go about it. – Andrew Scott Evans Mar 11 at 22:47
• @AndrewScottEvans unparenthesized powers don't work that way. – Roddy MacPhee 2 days ago
• $2^{3^4}\equiv 2\ne1\pmod5$ – Piquito yesterday

$$a^{b^{c^{d^e}}}\equiv f \bmod 65536$$
We first mod the base a, by 65536 (see polynomial remainder theorem). We, then take$$b^{c^{d^e}}\equiv g \bmod (\phi(65536)=32768)$$ by taking b mod 32768. and then we take $${c^{d^e}}\equiv h \bmod (\phi(32768)=16384)$$ etc.
In your case, any time n is odd, n+1 will be even and then because $$y\equiv b \bmod m \implies y=mx+b$$ and the distributive law, we can factor out a 2 when we rearrange via subtraction on one side. but 1 doesn't have a 2 as a divisor. We therefore get, that only even n could ever make this congruence true.