×

# Divisibility relations

Prove: If $$a^{n}-1|b^{n}-1$$, then there exists $$k$$ such that $$b=a^{k}$$.

Note by Jessica Wang
1 year, 5 months ago

## Comments

Sort by:

Top Newest

Suppose that $$a,b,n$$ are integers

$\frac{b^{n}-1}{a^{n}-1}=m$

let $$b=a^{k}$$

$\frac{a^{nk}-1}{a^{n}-1}=m$

if $$n|k$$,

$$a^{nk}-1 = (a^n-1)(a^k+a^{k-n}+a^{k-2n}+...a^n+1)$$

Hence, $\frac{b^{n}-1}{a^{n}-1}=\frac{(a^n-1)(a^k+a^{k-n}+a^{k-2n}+...a^n+1)}{a^{n}-1}=a^k+a^{k-n}+a^{k-2n}+...a^n+1=m$

Thus, when $$k$$ is any number which can be divided by $$n$$, $$b=a^k$$

*not sure if this is correct, waiting for others to complete · 1 year, 5 months ago

Log in to reply

This should do. · 1 year, 5 months ago

Log in to reply

Your proof is not correct.

Let A: $$a^n -1 | b^n -1$$.

and B : $$b = a^k$$

You have to prove that A implies B. What you've proven is that B implies A. Both are not the same thing. You can see this with the help of an example.

Suppose we want to prove " every integer x is even". (Obviously false).

Take A: x is an integer

and B: x is an even number.

We see that B implies A, but A does not imply B. · 1 year, 5 months ago

Log in to reply

He has done nothing wrong. He was very careful with the wording, knew the limits of his approach and even hinted, his proof wouldn't be complete. The assumption $$b = a^k$$ is perfectly fine. It was not postulated, $$k$$ would be a integer. Proving $$k$$ being an integer is the actual "difficult" part. There just missed the part for showing $$n \ !| \ k$$ contradicts A. However I added that part, you may bother reading my post with the included link. · 1 year, 5 months ago

Log in to reply

I'm sorry - my solution above is wrong as we cannot assume that $$b=a^k$$, but we instead have to prove that the statement $$a^n-1|b^n-1$$ implies $$b=a^k$$. I've asked help from my friend and his suggestion is 'every prime that can divide $$b$$ must also divide $$a$$'. However, I still have no idea how to prove. · 1 year, 5 months ago

Log in to reply

Hey I did not understand how did you find $$b=a^{k}$$ · 1 year, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...