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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 $n$.)

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

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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 $3$.

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

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