Divisibility relations

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

#Algebra #NumberTheory #AlgebraicIdentities #Recursion #Divisibility

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 5 years, 10 months ago

This should do.

Little Ostrich - 5 years, 10 months ago

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 - 5 years, 10 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.

Little Ostrich - 5 years, 10 months ago

Hey I did not understand how did you find $b=a^{k}$

Department 8 - 5 years, 10 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.

Potsawee Manakul - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$