n IS A positive integer, a perfect number is equal to the sum of its proper positive divisors, that is, the sum of its positive divisors excluding the number itself. if p and 2^p-1 are primes, then a perfect number is ?

Question

rock_mit182 · Answer

the expression is : $$2^{p} -1$$

tkhunny · Answer

I'm not certain we have a question, here.  If p is prime, does it have any "Proper Divisors"?

rock_mit182 · Answer

yeah but how am i going to find the expression for a perfect number given

p and$$2^{p}-1$$

rock_mit182 · Answer

i guess since they're prime numbers the only divisor they colud have is one, but how i make the proof and find the expression for the perfect number ?

tkhunny · Answer

Like I said, I'm not sure we have a question.

tkhunny · Answer

Oh, we get to count "1" in the proper divisors.  We just don't get to count n.

rock_mit182 · Answer

for example if i multiply $$p * 2^{p} +1$$ that would be a perfect number ?

rock_mit182 · Answer

if so how am i going to proof that

tkhunny · Answer

Note: It wants "A" perfect number, not "THE" perfect number.

You had better not multiply anything like that.  Remember your Order of Operations.  Try $p\cdot \left(2^{p} - 1\right)$

It's proper divisors are what?

tkhunny · Answer

Answer my question.  What are the proper divisors of that product?

rock_mit182 · Answer

i mean i saw the possible answer in wikipedia: "Euclid proved that$$2^{p-1}(2^{p}-1)$$

rock_mit182 · Answer

but i don't understand how to get there
from my question

tkhunny · Answer

Why are you not answering my question?

$p\cdot (2^{p} - 1)$ has proper divisors $1,\;p,\;and\;2^{p}-1$

Do you believe?

rock_mit182 · Answer

yes i know, but since has to to be a perfect number, couldn't include p and 2^p-1

rock_mit182 · Answer

@ranga @mathmale @whpalmer4

rock_mit182 · Answer

@wio

tkhunny · Answer

You're not seeing it.  Outside the box.  This course will require nonstandard thinking from your brain.  You have to see outside where you have been, before.

$p\cdot\left(2^{p} - 1\right)$ is a lovely number.  Let's just talk about this number for a moment.  Let's call this number "Steve".

Steve has Proper Divisors of $1,\;p,\;and\;2^{p}-1$

If Steve is equal to the sum of Steve's proper divisors, then Steve is a Perfect Number.  Can we track down a specific value for Steve?

$Steve\;=\;p\cdot\left(2^{p}-1\right)\;=\;1 + p + \left(2^{p}-1\right)$

Can we do anything with that?  Steve is lonely.  He's having an identity crisis.

whpalmer4 · Answer

It's as simple as 1, 2, 3...