# Prove that n is square or doubled square

• Oct 1st 2012, 06:54 AM
Lukaszm
Prove that n is square or doubled square
Hey,
N is positive integer number. Prove: if summation of all positive divisors of n is uneven, then n is square or doubled square of integer number.
So far I have proved that in this case n is a square of integer number. How?

First of all, every natural number can be shown as:
Attachment 24992
and we should know the equation for summation of all positive divisors:
Attachment 24990
which is equal to
Attachment 24991
As far as I know $\alpha = \lambda$
Example: Divisors of number 15: 1,3,5,15 Sum=24
From my equation: 15=3*5 Sum=(1+3)(1+5)=4*6=24
I know that one example doesn't prove that this equation is true, nevermind.

So if the sum of all postive divisors is uneven then every bracket should be uneven (if one was even then the product would be even).
Attachment 24989
Then we should have "pairs" inside all brackets, as shown on the picture, so that in every bracket sum+1 is uneven.
From that we can deduce that $\alpha$should be even (there are only "pairs of elements"+1 in brackets)- hope you understand what I mean.
If $\alpha$ is even then it can be written as 2t (t is natural number) and then our initial equation would look like this:
Attachment 24993 (click to enlarge)
That means n is square of natural number. But in my task there is also "doubled square". I found out that n would be doubled square if n is 2 to the uneven power.
E.g.
Attachment 24994
But how can I prove that?
Thanks for your replies and sorry for mistakes. If something is unclear- simply ask (Happy)
Regards
Lukasz
• Oct 1st 2012, 12:54 PM
johnsomeone
Re: Prove that n is square or doubled square
Your conclusion that $\alpha$ had to be even assumed that the prime was an odd number. No such restriction applies when p=2. The factor will be odd no matter the $\alpha$ when p=2.
That's where the "or double a square" part will show up.