Prove that if m and n are positive integers and if ab ≡ 1 (mod m) and

a^n ≡ 1 (mod m), then b^n ≡ 1 (mod m).

Any help would be greatly appreciated! I am lost with this one but I probably am just missing a very simple step. Thanks!

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- Jan 20th 2010, 11:20 AMFatherMikeCongruence Proof
Prove that if m and n are positive integers and if ab ≡ 1 (mod m) and

a^n ≡ 1 (mod m), then b^n ≡ 1 (mod m).

Any help would be greatly appreciated! I am lost with this one but I probably am just missing a very simple step. Thanks! - Jan 20th 2010, 12:30 PMDinkydoe
Given is $\displaystyle ab\equiv 1$ mod m. This implies that $\displaystyle (ab)^n\equiv 1$ mod m.

Thus if $\displaystyle a^n\equiv 1$ mod n what does this say about $\displaystyle (ab) ^n\equiv b^n $ mod m? - Jan 20th 2010, 12:30 PMDrexel28