# Which of $[0]_3, [1]_3, [2]_3$ is $[5^k]_3$ equal to?

Let $$k\in \mathbb{N}$$. Which of $$[0]_3, [1]_3, [2]_3$$ is $$[5^k]_3$$ equal to? Prove your answer.

Below is my proof so far. I figured out what it equals when $$k$$ is even or odd, which is hopefully correct. And I know from this then a case for both even and odd must be made. I am just unsure of how you would translate how $$k$$ being even or odd affects the outcome into a proof? So far in class we have worked on multiplication tables for $$\mathbb{Z_n}$$ but not much incorporation into proofs.

Proof: Let $$k\in \mathbb{N}$$. Then $$[5^k]_3$$ is either $$[1]_3$$ or $$[2]_3$$, depending on $$k$$. Specifically: $$\begin{cases}[1]_3 & k \text{ odd} \\ [2]_3 & k \text{ even}\end{cases}$$. Case 1: Let $$k$$ be odd.

Case 2: Let $$k$$ be even.

(Note: $$[a]_n$$ is the congruence class of $$a$$ modulo $$n$$.)

• did you mean $[1]_3$ for $k$ even, and $[2]_3$ for $k$ odd? – J. W. Tanner Oct 17 at 6:23

See that $$[5]_3=[-1]_3$$ and therefore $$[5^k]_3=[(-1)^k]_3$$.

Now it's quite straightforward: we have $$(-1)^k=1$$ for $$k$$ even and $$(-1)^k=-1$$ for $$k$$ odd.

Hence the solution $$[5^k]_3=[1]_3$$ for $$k$$ even and $$[5^k]_3=[-1]_3=[2]_3$$ for $$k$$ odd.

Write $$5^k=(3+2)^k=\sum_{j=0}^k\binom{k}{j}3^j2^{k-j}=2^k+\sum_{j=1}^k \binom{k}{j}3^j2^{k-j}$$ to deduce that for $$k\geq 1$$ it is the case that $$5^k=2^k$$ modulo $$3$$.

Case 1: Let $$k$$ be odd. Then $$k=2n+1$$ for some $$n,$$

so $$[5^k]_3=[5^{2n+1}]_3=[(5^2)^n]_3[5]_3=[1^n]_3[5]_3=[2]_3.$$

Case 2: Let $$k$$ be even. Then $$k=2n$$ for some $$n,$$

so $$[5^k]_3=[5^{2n}]_3=[(5^2)^n]_3=[1^n]_3=[1]_3.$$