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herraotic
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Let a,b>1 be integers such that for all n>0 we have a^n-1|b^n-1. Then b is a natural power of a. I can't find a solution.
It doesn't look right to me. As far as I can tell you have a|b and nothing else. In other words if b/a is an integer, then so is (b/a)^{n}.Let a,b>1 be integers such that for all n>0 we have a^n-1|b^n-1. Then b is a natural power of a. I can't find a solution.
Since your hypothesis is for all n, then for n=2, you are assuming a|b, which means b/a is an integer. Since all positive integer powers of integers are integers, there seems to be nothing else there.Well mathman doesn't actually give a counterexample or prove a|b unless I'm missing something in, "In other words ...", so no progress so far.
Converse is easy to prove, and the result looks very plausible, but what is the origin of the question? I.e. is this likely to be a Sundy afternoon problem?
Well mathman doesn't actually give a counterexample or prove a|b unless I'm missing something in, "In other words ...", so no progress so far.
Converse is easy to prove, and the result looks very plausible, but what is the origin of the question? I.e. is this likely to be a Sundy afternoon problem?
Your first step fails: the hypothesis that b is not a power of a does not imply that b is divisible by a prime that doesn't divide a.Actually, I think my previous post is a complete solution to the problem. I wanted to acknowledge your contribution.
Petek
Your first step fails: the hypothesis that b is not a power of a does not imply that b is divisible by a prime that doesn't divide a.
e.g. consider b = 12 and a = 6.
I agree that Mathman and I were probably reading the problem incorrectly, because given thatI've just read Dodo's note and the previous replies now make more sense. But I think what he says is certainly correct. It is the case that
[itex](\exists k\in N)b=a^k\implies(\forall n\in N)a^n-1|b^n-1[/itex]
I think we're being asked to prove the converse. The question wouldn't make much sense read the other way.
The converse is likely to be a lot more awkward, which is why I wanted to get an idea of how much more awkward before I started thinking about it.