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- Thread starter crypt50
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- #2

- Jan 26, 2012

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$$a \equiv b \;( \text{mod} \; m),\qquad \text{and}

\qquad a^{n} \equiv b^{n} \;( \text{mod} \; m).$$

Use induction on $n$ to prove this. What will you need to show the inductive step?

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- #3

- Mar 5, 2012

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Assuming you meant what Ackbach suggested, here's an alternative way.

The expression $a \equiv b \pmod m$ means that there is a $k \in \mathbb Z$ such that $a=b+km$.

This implies that $a^n=(b+km)^n$.

Can you expand the right hand side with the binomial theorem?

If so, what can you conclude?