Prove $9^n \equiv 9 \pmod{24}\quad \forall n \in \Bbb N^*$

Question

I accidentally found that $9^n\equiv 9\pmod{24} \quad \forall n \in \Bbb N^*$

But don't know how to prove it in an exact way.

Please help

Answer 1

You can prove it using induction:

Initial case: It's trivial that $9\equiv 9 \text{ mod(24)}.$

Induction: We suppose $9^n\equiv 9 \text{ mod(24)}$ for some $n\in\mathbb{N}.$ Then, we'll see if it verifies for $9^{n+1}$: $$9^{n+1}=9^n\cdot 9,$$ and using modular arithmetic properties we see that $$(9^n\cdot 9) \text{ mod(24)}\equiv(9^n \text{ mod(24)}) \cdot(9\text{ mod(24)}),$$ so $$(9^n\cdot 9) \text{ mod(24)}\equiv 9\cdot 9=81 \equiv 9 \text{ mod(24)}.$$ So we have proved that $9^{n+1}\equiv \text{ mod(24)}$, and by induction the statement is proved.

Answer 2

$24\mid 72\mid 9^{\large n}\!-9\,$ is easy: factor out $\,\color{#0a0}{c=9}\,$ via mod distributive law

$$\begin{align} ca\:\!\bmod cm \,&=\, \color{#0a0}c\,(a\bmod m)\\[.2em] \Rightarrow\ \ 9^n-9\,\bmod\:\!\ 72 \,&=\, \color{#0a0}9\,(\color{#c00}9^{n-1}\!-1\bmod 8)\\[.2em] &=\, 9(\,\color{#c00}1^{n-1}\!-1\bmod 8)\ \ {\rm by}\ \ \color{#c00}{9\equiv 1}\!\!\!\!\pmod{\!8}\\[.2em] &=\, 9\,( 0) \end{align}\quad\ $$

Remark $ $ Numbers like $\,9\,$ above with $\,a^2=a\,$ are called idempotents. This implies $\,a^{\large n} = a\,$ for all $\,n\ge 2\,$ either as above or by a simple induction $\,a^{\large n+1}\! = a\,a^{\large n} = a\cdot a = a.\,$ More conceptually: note $a$ is a fixed point of $\,f(x) = ax\,$ i.e. $\,f(a) = a,\,$ and fixed points always stay fixed on iteration by a simple induction: if $\,\color{#c00}{f^n(a) = a}\,$ then $\,f^{\large n+1}(a) = f(\color{#c00}{f^{\large n}(a)}) = f(\color{#c00}a) = a$.

The $\!\bmod\!\!$ Distributive Law allows us to do this induction purely arithmetically, by cancelling out a factor to reduce it to the trivial induction $\,\color{#c00}1^n\equiv 1,\,$ exactly as in the special case above, i.e.

$$\begin{align} a^n-a\,\bmod\, a(a\!-\!1) \,&=\, \color{#0a0}a\,(\color{#c00}a^{n-1}\!-1\bmod a\!-\!1)\\[.2em] &=\, a(\,\color{#c00}1^{n-1}\!-1\bmod a\!-\!1)\ \ {\rm by}\ \ \color{#c00}{a\equiv 1}\!\!\!\!\pmod{\!a\!-\!1}\\[.2em] &=\, a\,( 0) \end{align}\!\!\!\!\!\!\!\! $$

therefore: $\ m\mid a(a\!-\!1)\mid a^{\large n}\!-a\ \Rightarrow\ a^n\equiv a\pmod{\!m}$

Or $\ \color{#c00}{a^{\large n}\equiv a}\pmod{\!(a\!-\!1)a},\,$ by it is true mod $\,a\!-\!1\!:$ $\,1^{\large n}\!\equiv 1\,$ & $\bmod a\!:\,0^{\large n}\!\equiv 0,\,$ thus it's also true $\!\bmod {\rm lcm}(a\!-\!1,a) = (a\!-\!1)a\,$ by CCRT, so the congruence persists $\!\bmod m\mid (a\!-\!1)a$.

Or $ $ by the Factor Theorem: $\, a\!-\!1\mid a^{\large n-1}\!-1\,\Rightarrow\, m\mid a(a\!-\!1)\mid a(a^{\large n-1}\!-1) = a^n-a$.

Idempotents $\!\bmod m\,$ play a fundament role in factorization of integers (and rings) since they correspond to splittings of $n$ into coprime factors, cf. last paragraph here.

The $\!\bmod\!$ Distributive Law is essentially an operational form of CRT = Chinese Remainder Theorem. Unlike CRT it is formulated operationally which often makes it more convenient to apply in concrete calculations like those above.

Answer 3

$$\begin{array}{rcl}9^n-9&=&9(9^{n-1}-1)\\&=&9(9-1)(9^{n-2}+9^{n-3}+\ldots+9+1)\\&=&72(9^{n-2}+9^{n-3}+\ldots+9+1)\end{array}$$

so $9^n-9$ is divisible not only by $24$ but by $72$ as well.