Prove of $m \equiv_4 n \rightarrow 123^m \equiv_{10} 33^n$

Question

I want to prove that, if $m \equiv_4 n$ for all $m,n \in \mathbb{Z}$, then $123^m \equiv_{10} 33^n$

I have no idea how to prove something like that

Answer 1

Start with $$\varphi(10)=4 \tag{1}$$ where $\varphi(n)$ is Euler's totient function. Then, because $$123 \equiv 33 \pmod{10} \Rightarrow 123^m \equiv 33^m \pmod{10} \tag{2}$$ But, it is given that $m \equiv n \pmod{4} \iff m=q\cdot4+n$, for some $q$. Because $\gcd(33,10)=1$, from Euler's theorem $$33^{\varphi(10)}\equiv 1\pmod{10} \overset{(1)}{\iff} 33^4 \equiv 1\pmod{10}$$ or $$33^{q\cdot4} \equiv 1^{q}\equiv 1\pmod{10} \Rightarrow 33^{q\cdot4+n} \equiv 33^{n}\pmod{10}$$ or $$33^m \equiv 33^n \pmod{10} \tag{3}$$ now, putting together $(2)$ and $(3)$ $$123^m \equiv 33^n \pmod{10}$$

Answer 2

Hint:

The last digit of $3^1$ is $3$. The last digit of $3^2$ is $9$. The last digit of $3^3$ is $7$. The last digit of $3^4$ is $1$. The last digit of $3^5$ is $3$ again.

Answer 3

$\bmod 10\!:\ \color{#c00}{3^{\large 4}\equiv 1}\,\Rightarrow\,3^{\large m}\! = 3^{\large n+4k}\!\equiv 3^{\large n}(\color{#c00}{3^{\large 4}})^{\large k}\!\equiv 3^{\large n}\ $ by Congruence Product / Power Rules