Let $m,n$ be positive integers. Prove that $(2^m-1)^2|(2^n-1)$ if and only if $m(2^m-1)|n$.

Source:Russia 1997

## m and n

- Phlembac Adib Hasan
**Posts:**1016**Joined:**Tue Nov 22, 2011 7:49 pm**Location:**127.0.0.1-
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### Re: m and n

**Hint:**

- zadid xcalibured
**Posts:**217**Joined:**Thu Oct 27, 2011 11:04 am**Location:**mymensingh

### Re: m and n

$2^m-1|2^n-1 \longleftrightarrow m|n$.Let $n=km$. As $2^m-1|\frac{2^n-1}{2^m-1}$

and $2^{m(k-1)}+2^{m(k-2)}+.............+1 \equiv k \equiv 0 (mod 2^m-1)$

\[\Longleftrightarrow 2^m-1|k\]

\[\Longleftrightarrow m(2^m-1)|n\]

and $2^{m(k-1)}+2^{m(k-2)}+.............+1 \equiv k \equiv 0 (mod 2^m-1)$

\[\Longleftrightarrow 2^m-1|k\]

\[\Longleftrightarrow m(2^m-1)|n\]

### Re: m and n

https://www.awesomemath.org/assets/PDFs ... _Lemma.pdf

A general version, and applicable in many criteria. Particularly, this problem becomes one liner

A general version, and applicable in many criteria. Particularly, this problem becomes one liner

One one thing is neutral in the universe, that is $0$.

- Fm Jakaria
**Posts:**79**Joined:**Thu Feb 28, 2013 11:49 pm

### Re: m and n

Ithink problem 3 has a crack in the solution. How can we assume a and d relatively prime?Masum wrote:https://www.awesomemath.org/assets/PDFs ... _Lemma.pdf

A general version, and applicable in many criteria. Particularly, this problem becomes one liner

You cannot say if I fail to recite-

the umpteenth digit of PI,

Whether I'll live - or

whether I may, drown in tub and die.

the umpteenth digit of PI,

Whether I'll live - or

whether I may, drown in tub and die.