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**asviola** Assume m is prime and a is some integer between 0 and m-1, prove there exists some positive integer r, such that a^t ≡ 1(mod m) if and only if t=nr, n=0,1,2,3,...

I started this proof by first noting that m divides (a^t - 1) from the definition of modulus, and therefore (a^t-1)=km where k exists in the set of positive ints. Then I added 1 to both sides, a^t=km+1 and took the log to get t*log(a)=log(km+1) --> t=rn so this is rn*log(a)=log(km+1),

therefore r=log(km+1)/(n*log(a)) --> r=log(km+1-a)/n but, I'm not sure where to go from here, or if I'm even headed in the right direction...