Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the following theorem:

For any $a, b \in \mathbb{Z}^+$, there exist $m, n \in \mathbb{Z}$ such that $m > n$ and $a\ |\ b^m - b^n$.

What's the best way to prove it? I have an idea (and I know it's true because of that idea), but I don't know how rigorous it is to constitute a proof.

share|improve this question
    
Probably should rule out $a=0$, since then $a=0$, %b=2$ is counterexample. –  André Nicolas Jul 12 '13 at 14:48
    
Positive integers. Thanks again. –  Joe Z. Jul 12 '13 at 14:49

1 Answer 1

up vote 8 down vote accepted

Consider the remainders of $b^0$, $b^1$, $b^2$, ... when divided by $a$. Since this is an infinite sequence with finitely many possible values we must have $n$ and $m$ such that the remainder of the division of $b^n$ by $a$ is the same as the one of $b^m$. This is equivalent to $a|b^m-b^n$. In fact it is enough to consider $0\leq n,m\leq a$.

share|improve this answer
    
Haha, the method I was thinking of involved fractional representations in base $b$. Yours is much more elegant. –  Joe Z. Jul 12 '13 at 14:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.