View Single Post
Old 2018-01-14, 21:22   #6
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts
Default

Quote:
Originally Posted by lukerichards View Post
I think this is sufficient?
NB Spoiler below.













If  f(x, n) = \sum_{k=1}^{(x)} (2^{n})^{k-1}

Then, for example:
\ f(3, n) = (2^{n})^2 + (2^{n}) + 1

 f(x, n) is an integer for all integer values of x and n.
And:
 ({2^n} - 1)f(3, n) = (2^{n})^3 - 1

Generalised:

({2^n} - 1)f(x, n) = (2^{n})^x - 1
This is an integer for all integer values of n and x.

Now, if n|m, then m = x * n, where m, n and x are all integers.

So:

2^{m} = 2^{xn} = (2^n)^x

Therefore:

2^{m} - 1 = 2^{xn} - 1 = (2^n)^x - 1

As above:

 (2^n)^x - 1 = ({2^n} - 1)f(x, n)

It follows:

 \frac{(2^n)^x - 1}{({2^n} - 1)}\ = f(x, n)

So:

 \frac{2^m - 1}{({2^n} - 1)}\ = f(x, n)

And f(x) is an integer, so M_n divides M_m when n divides m.
There are
SPOILER
tags

But I know I can beat that in simplicity.

Note the reccurence:

[TEX]M_n=2M_{n-1}+1=2M_{n-1}+M_1[/TEX]

This leads to [TEX]2^y(M_n)+M_y=M_{n+y}[/TEX] which when reccursively applied shows the statement.

Last fiddled with by science_man_88 on 2018-01-14 at 21:50
science_man_88 is offline   Reply With Quote