Find the largest $m$ such that $n^{12}\equiv 1\mod m$ for all $n$ relatively prime to $m$? (using only Euler's totient theorem)

Question

I was able to show that if $(n,72)=1$ then $n^{12}\equiv 1\mod {72}$, and now the second part of this question seeks to find the largest $m$ such that $n^{12}\equiv 1\mod m$ for all $n$ relatively prime to $m$. Based on the first part we must have $m\geq 72$. But I have no idea on how to proceed; any hint would be vastly appreciated. Maybe using Euler's totient function?

@J.W.Tanner but how can I find the largest one (or prove that 144 is the largest one if that's the case)? — Stupid Questions Inc, Mar 05 '20 at 20:02
$m=65520$ seems to work (it's the product of $16, 9, 5, 7, $ and $13$, which work) — J. W. Tanner, Mar 05 '20 at 20:13

J. W. Tanner · Answer 1 · 2020-03-05T21:23:22.130

2

We are looking for the largest $m$ whose Carmichael function is $12$.

The Carmichael function of $m$ is the least common multiple

of the Carmichael function of the prime powers that are factors of $m$.

The Carmichael function of $p^k$ is $\phi(p^k)$ for $p$ odd or $p^k\in\{2,4\},$

and it is $\dfrac{\phi(2^k)}2$ for $p=2$ and $k\ge3$, where $\phi$ is Euler's totient function.

The Carmichael function divides $12$ for $m=16$, $m=9$, $m=5$, $m=7$, and $m=13$,

and not for any higher prime powers.

Therefore, the largest $m$ is $16\times9\times5\times7\times13=65520$.

edited Mar 05 '20 at 21:23

answered Mar 05 '20 at 20:51

J. W. Tanner

60,406

Is it possible to offer an answer using only the Euler totient function, his theorem, etc? – Stupid Questions Inc Mar 05 '20 at 20:58
I think you need the Carmichael function (at least in concept), as the other (incorrect) answer illustrates: $\phi(72)=24,$ but $\lambda(72)=6$, which divides $12$ – J. W. Tanner Mar 05 '20 at 21:07
Typo. In the final product you forgot the $5$. – Mar 05 '20 at 21:22
Thanks very much, @ChrisCuster; I edited accordingly – J. W. Tanner Mar 05 '20 at 21:23

score 0 · Answer 2 · answered Mar 05 '20 at 20:40

0

just an idea

For each $ n $ relatively prime to $ m $, Euler's theorem allows us to write

$$n^{\phi(m)} \equiv 1 \mod m$$

So, you can look for the largest $ m $ such that $$\phi(m)=12$$

answered Mar 05 '20 at 20:40

hamam_Abdallah

62,951

good idea, but i don't think that's sufficient, for instance $\phi(72)=24$ and yet $n^{12}\equiv 1\mod{72}$ is also satisfied, – Stupid Questions Inc Mar 05 '20 at 20:49

Find the largest $m$ such that $n^{12}\equiv 1\mod m$ for all $n$ relatively prime to $m$? (using only Euler's totient theorem)

2 Answers2