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Divisibility relations

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

Note by Jessica Wang
2 years, 4 months ago

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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

Potsawee Manakul - 2 years, 4 months ago

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This should do.

Little Ostrich - 2 years, 4 months ago

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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.

Siddhartha Srivastava - 2 years, 4 months ago

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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.

Little Ostrich - 2 years, 4 months ago

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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.

Potsawee Manakul - 2 years, 4 months ago

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

Lakshya Sinha - 2 years, 4 months ago

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