Hi, I'm new here. Just a quick question

Is there a theorem that states that if gcd(a,m)=1 and $\displaystyle a^x \equiv a^y (mod m)$, then $\displaystyle a^{x-y} \equiv 1(mod m)$?

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- Oct 22nd 2009, 12:57 PMbrogersquick question
Hi, I'm new here. Just a quick question

Is there a theorem that states that if gcd(a,m)=1 and $\displaystyle a^x \equiv a^y (mod m)$, then $\displaystyle a^{x-y} \equiv 1(mod m)$? - Oct 22nd 2009, 01:34 PMproscientia
I don’t know if there is a name for this theorem, but the result can be quite easily established by considering the multiplicative group of the units of the ring $\displaystyle \mathbb Z_m.$

- Oct 22nd 2009, 01:43 PMbrogers
thanks, I just needed to know if that was a real theorem