# Some help with a number theory problem

1. Feb 28, 2010

### herraotic

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.

2. Mar 1, 2010

### herraotic

34 views and no solutions???

Come on physicsforums, you're making me think this is a beta board and I should go look to more able forums.

3. Mar 1, 2010

### mathman

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.

4. Mar 27, 2010

### Martin Rattigan

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?

5. Mar 27, 2010

### mathman

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.

6. Mar 27, 2010

### ramsey2879

Didn't give a counterexample. Mathman said that all that was shown was a|b. In other words 2|6 is a counter example since 6 is not a power of 2. Any other pair of integers where the smaller divides the larger and the larger is not a power of the smaller is also a counter example or are we missing something here?

7. Mar 27, 2010

### dodo

I suspect that the premise is not being read correctly. I think it's not a^(n-1)|b^(n-1), but (a^n)-1|(b^n)-1. Had it been the former, it would have probably been expressed as a^n|b^n for n>=0. Besides, it's more interesting this way. :P

8. Mar 27, 2010

### Martin Rattigan

But $2^n-1|6^n-1$ is false for n=2, that would be 3|35! So I don't see how 2 and 6 are a counterexample. To get a counterexample you would need a pair of numbers $a,b$ such that $(\forall n\in N)a^n-1|b^n-1$ but where $\neg(\exists n\in N)b=a^n$.

Isn't that correct?

9. Mar 27, 2010

### Martin Rattigan

I'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

$(\exists k\in N)b=a^k\implies(\forall n\in N)a^n-1|b^n-1$

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.

Last edited: Mar 28, 2010
10. Mar 27, 2010

### Hurkyl

Staff Emeritus
Isn't the theorem straightforward once you know b=ka?

11. Mar 28, 2010

### Martin Rattigan

No.

Do we know b=ka?

12. Mar 28, 2010

### Hurkyl

Staff Emeritus
Try substituting it, then simplifying things.

mathman already asserted that b=ka. Skimming the details in my head, we might need to look at gcd(a,b) first, though; I'm not sure.

13. Mar 28, 2010

### Martin Rattigan

But I don't think mathman was reading the question as intended when he said that. (See last few posts.)

14. Mar 28, 2010

### Hurkyl

Staff Emeritus
I had convinced myself that was true yesterday (and I hadn't yet noticed that mathman also asserted it). Lemme see if I can remember how it worked -- I recall it being a straightforward "suppose p divides one of the numbers. Then it divides the other" argument, but I could have made a mistake in my head.

15. Mar 28, 2010

### Hurkyl

Staff Emeritus
Ah, yes. Suppose p divides b, but does not divide a. You can derive a contradiction by looking at things mod p....

Of course, that doesn't show b and a have the same prime factors yet. But maybe this is all I noticed when I thought about it before.

16. Mar 28, 2010

### Martin Rattigan

This entry referred to an edit to #9 which didn't appear to have happened. I now notice it did happen so I've deleted the previous irrelevant text in this entry.

Last edited: Mar 28, 2010
17. Mar 28, 2010

### Martin Rattigan

Hurkyl - If you derive a contradiction from $p\mid b,p\nmid a$ this shows neither that $b$ must be a multiple nor power of $a$. E.g. $b=18,a=12$.

The crux of the problem is that you are guaranteed a sequence such as:

$$3^0-1|(3^3)^0-1$$
$$3^1-1|(3^3)^1-1$$
$$3^2-1|(3^3)^2-1$$
$$3^3-1|(3^3)^3-1$$
$$\dots$$

You need to show that a sequence such as:

$$3^0-1|5^0-1$$
$$3^1-1|5^1-1$$
$$3^2-1|5^2-1$$
$$3^3-1|5^3-1$$
$$\dots$$

will fail at some point.

Last edited: Mar 28, 2010
18. Mar 28, 2010

### Hurkyl

Staff Emeritus
No, I meant p divides b but not a; that was the case in which I could get the contradiction.

19. Mar 28, 2010

### Martin Rattigan

Of course if $a,b$ are both prime as in the example I gave, it is obvious the sequence will fail when $n=b-1$.

20. Mar 28, 2010

### Martin Rattigan

In fact if $(a,b)=1$ the sequence will fail for $n=\phi(b)$.

Last edited: Mar 28, 2010