Your conclusion that 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 when p=2.
That's where the "or double a square" part will show up.
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:
and we should know the equation for summation of all positive divisors:
which is equal to
As far as I know
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).
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 should be even (there are only "pairs of elements"+1 in brackets)- hope you understand what I mean.
If is even then it can be written as 2t (t is natural number) and then our initial equation would look like this:
(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.
But how can I prove that?
Thanks for your replies and sorry for mistakes. If something is unclear- simply ask
Regards
Lukasz
Your conclusion that 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 when p=2.
That's where the "or double a square" part will show up.