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Mathematics 18 Online
OpenStudy (dan815):

why is n a prime if 2^n-1 a prime and why is (2^n-1)*(2^(n-1)) a perfect number if it is

OpenStudy (dan815):

i know that 2^n-1 = sum of k from 0 to n-1 of 2^k

OpenStudy (dan815):

and thats about it xD

OpenStudy (kainui):

Yeah that's basically how it works, you can look at that geometric series as being the binary representation. So for example \[\Large 2^5-1=11111_2\] See there are 5 ones there. Let's do a non prime \[\Large 2^6-1 = 111111_2 = (10101*11)_2=(1001*111)_2\] So you can see how we are really splitting it up into multiplication of its factors, 3*2 or 2*3, but you can't do that if it's prime. Although what you have 2^n-1 isn't always prime if n is prime, it's just more likely.

OpenStudy (kainui):

Yeah the second one, cause he said: \[\Large 2^n-1 = \sum_{k=0}^{n-1}2^k\]

OpenStudy (dan815):

if 2^n-1 is prime then is n always a prime?

OpenStudy (dan815):

even though if n is a prime 2^n-1 doesnt have to be

OpenStudy (kainui):

Oh wait let me reread what you originally had.

OpenStudy (kainui):

Yeah ok I see I thought you meant n is prime means 2^n-1 is always prime, but you said the opposite 2^n-1 is primes means n is prime. This is definitely true, cause think about what I said about the binary. I can give a better explanation if you don't understand what I mean.

OpenStudy (dan815):

im just thinking like suppose there is a factor in binary is that the same as having factors in decimal?

OpenStudy (kainui):

Suppose that 2^n-1 was prime, but n was composite, then that would mean in base 2 we could factor it out into 111...111 and 10001000 type of a number which means 2^n-1 is composite, a contradiction.

OpenStudy (dan815):

ok i guess thats a silly qn nvm

OpenStudy (dan815):

just seems easier to find factors in binary though

ganeshie8 (ganeshie8):

factors and primality doesn't depend on number system

OpenStudy (dan815):

how do u know that a prime number of 1s cant be factored in binary

OpenStudy (kainui):

No, you're thinking of it backwards now like I was originally.

OpenStudy (kainui):

We already know that the number 111111...111 is prime. Since this is 2^n-1 this means n has to be prime as well, since composite n always produces composite numbers.

OpenStudy (dan815):

okk i see! right after knowing its prime!

OpenStudy (dan815):

buutt still ... wait

OpenStudy (dan815):

composite n always produces composite numbers lemme see that

OpenStudy (dan815):

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