MHB Proving operations of congruence modulo m

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If integers a, b, and m are greater than zero and a is congruent to b modulo m, then it follows that a^n is congruent to b^n modulo m for all positive integers n. The congruence is denoted as a ≡ b (mod m), and the proof can be approached using mathematical induction on n. The inductive step requires showing that if the statement holds for n, it also holds for n+1. Additionally, expanding (b + km)^n using the binomial theorem can help in deriving the conclusion. This discussion emphasizes the importance of understanding congruences and their properties in modular arithmetic.
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If a, b and m > 0 are integers such that a % b (mod m), then a^n % b^n (mod m) for all positive integers n. I don't know how to go about it, any help would be greatly appreciated.
 
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By '%', do you mean congruent? That's typically written
$$a \equiv b \;( \text{mod} \; m),\qquad \text{and}
\qquad a^{n} \equiv b^{n} \;( \text{mod} \; m).$$
Use induction on $n$ to prove this. What will you need to show the inductive step?
 
Welcome to MHB, crypt50! :)

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?
 
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