Quote:
Originally Posted by lukerichards
I think this is sufficient?
NB Spoiler below.
If
Then, for example:
is an integer for all integer values of x and n.
And:
Generalised:
This is an integer for all integer values of n and x.
Now, if nm, then m = x * n, where m, n and x are all integers.
So:
Therefore:
As above:
It follows:
So:
And f(x) is an integer, so divides when n divides m.

There are
SPOILER
tags
But I know I can beat that in simplicity.
Note the reccurence:
[TEX]M_n=2M_{n1}+1=2M_{n1}+M_1[/TEX]
This leads to [TEX]2^y(M_n)+M_y=M_{n+y}[/TEX] which when reccursively applied shows the statement.