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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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?
Originally Posted by FatherMike 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! We have that $\displaystyle 1\equiv a^n\equiv \left(ab\right)^n=a^nb^n\equiv 1\cdot b^n=b^n\text{ mod }m$
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