In $U_{24}$, find a smallest positive integer $n$ such that $[7]^n=[1]$

Question

In $U_{24}$, find a smallest positive integer $n$ such that $[7]^n=[1]$

We have $24\mid 7^n-1\implies 4\mid (7^{n-1}+7^{n-2}+\dots+1)$. From intuition, $n=2$. Is there any way to find such $n$?

Answer 1

$(6m\pm1)^2=36m^2\pm12m+1$

$=24m^2+24\cdot\dfrac{m(m\pm1)}2+1$

as the product of two consecutive integers is always divisible by $2$

Answer 2

Since $[7]^1\neq[1]$ and $[7]^2=[1]$, the answer is $2$ indeed.

Answer 3

Use the Chinese remainder theorem: $$\mathbf Z/24\mathbf Z\simeq\mathbf Z/3\mathbf Z\times\mathbf Z/8\mathbf Z$$ and observe that $7\equiv 1\bmod 3$, $7\equiv -1\bmod 8$, so it has order $1\bmod3$, order $2\bmod 8$. We may conclude that, modulo $24$, it has order $\operatorname{lcm}(1,2)=2$.

Answer 4

Even before intuition comes into play, note that order of the group is $\phi(24)=\phi(8)\times \phi(3)=4\times2=8$. (Theorem: Order of the group $U_n$ is given by the value of Euler function $\phi(n)$).

So order of any element has to be $2,4, or, 8$. So repeatedly square 2 until you get 1. So nothing formidable.