With k being a natural number such that p=2k−1 is a prime number (i.e 3, 7, 31) Define and list all positive integers which divide N;
N=2k−1p
So the positive integers that divide N must divide 2k−1 or p. Now p is prime, so the only thing that can divide it is itself and 1. The only things that can divide 2k−1 are multiples of two.
So divisors of 2k−1 are 2k−1,2k−2,2k−3,2k−4...2k−(k−2),2k−(k−1),2k−(k−0)
Thus this tells me the positive integers which divide N (I'm assuming there's an infinite number of possibilities because I'm not sure if there is or isn't) are;
Sorry for the long explanation and I hope you understand. Let me know if I've gone wrong somewhere. I can't do the bolded bit, how exactly do I prove that last statement? Any help/hints appreciated.
It's probably (slightly) easier to try to show that the sum of all the factors, including N, sum to 2N.
Hey thanks. Is that just a general equation I'm meant to know or something? And I'm not sure if I can do what you proposed as the question says without including N as a divisor.
I'm not exactly sure how I can use your useful piece of information. I know that
Thus this tells me the positive integers which divide N (I'm assuming there's an infinite number of possibilities because I'm not sure if there is or isn't) are;
Hey thanks. Is that just a general equation I'm meant to know or something? And I'm not sure if I can do what you proposed as the question says without including N as a divisor.
It'd be handy to know, yes I'm sure you can do it that way, as long as you subtract off N at the end.