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

Many thanks!

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

Many thanks!

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TopNewestSuppose 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 – Potsawee Manakul · 1 year, 2 months ago

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This should do. – Little Ostrich · 1 year, 2 months ago

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@Little Ostrich Thanks old pal. This was a problem in the IMO Chinese camp once upon a time... Seems that I imagined it to be too hard. – Jessica Wang · 1 year, 2 months ago

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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. – Siddhartha Srivastava · 1 year, 2 months ago

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– Little Ostrich · 1 year, 2 months ago

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.Log in to reply

– Potsawee Manakul · 1 year, 2 months ago

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.Log in to reply

– Jessica Wang · 1 year, 2 months ago

O.oLog in to reply

– Lakshya Sinha · 1 year, 2 months ago

Hey I did not understand how did you find \(b=a^{k}\)Log in to reply

@Lakshya Sinha – Jessica Wang · 1 year, 2 months ago

He first assumed the statement is true, then found the value of m, which is applicable. Therefore the statement is true. (Am I right? Need some confirmation as well...)Log in to reply

– Jessica Wang · 1 year, 2 months ago

Thank you! Yes I think your solution is correct, as I currently cannot find any loopholes in it.Log in to reply