Prove the following theorem.

Thm: n is an even perfect # iff n = 2^(m-1)*(2m - 1) where m >= 2 and 2^m - 1 is prime.

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- March 29th 2007, 01:47 PMflash101Proof Perfect Numbers
Prove the following theorem.

Thm: n is an even perfect # iff n = 2^(m-1)*(2m - 1) where m >= 2 and 2^m - 1 is prime. - March 29th 2007, 02:02 PMAfterShock
Ahh, this will be long. Here goes...

Proof (<==)

n = 2^(m - 1)*(2^m - 1) where m >= 2, 2^m - 1 is prime. Show n is an even perfect number.

Note that since 2^m - 1 is prime implies m is prime.

Let sigma(n) = sum of divisors.

sigma(n) = 2n

m >= 2 implies m - 1 >= 1 implies 2|n implies n is even.

From 2|n we know 2^{exponent}*(2^m - 1)

sigma(n) = sigma(2^(m - 1)*(2^m - 1)) = (*fill in here after noting the following*-------- that something: sigma(2^(m - 1))*sigma(2^m - 1)

gcd(2^(m - 1), 2^m - 1) = 1, and therefore we can separate sigma over each piece since sigma is multiplicative.

We know 2^m - 1 is odd; (there are no odd divisors of 2^(m - 1)

From the red we know (2^m - 1)/2 - 1 * {note sigma(p) = 1 + p} and therefore (2^m -1 + 1) .. since 2^m - 1 is prime, remember.

= 2^m*(2^m - 1)

= 2*2^(m - 1)*(2^m - 1) = 2n (do you see how I got this?)

See next post for (==> part of proof) - March 29th 2007, 02:11 PMAfterShock
This part is a little harder...at least I think.

Proof (==>) n is an even perfect number, so sigma(n) = 2n

Let n = 2^(s)*t where s >= 1 and t is odd.. note we can do this for any number.. do you see why?

sigma(n) = sigma(2^s*t) = sigma(2^s)*sigma(t) = [2^(s + 1) - 1]/[2 - 1]*sigma(t)

= (2^(s + 1) - 1)*sigma(t)

Note: red above that gcd(2^s, t) = 1

Moving on...

2n = (2^(s + 1) - 1)*sigma(t)

2*2^s*t

2^(s + 1)*t = (2^(s + 1) - 1)*sigma(t)

Note (2^(s + 1) - 1) is odd

So, since (2^(s + 1) - 1) is odd, 2^(s + 1)|sigma(t)

So there exists a positive integer q such that sigma(t) = 2^(s + 1)*q

2^(s + 1)*t = (2^(s + 1) - 1)*(2^(s + 1)*q)

Cancel the 2^(s + 1)'s..

t = (2^(s + 1) - 1)*q

Recall that we started with n = 2^s*t

n = 2^s*(2^(s + 1) - 1)*q

Show q = 1 and that (2^(s + 1) - 1) is prime

Note that q|t and q < t

Add q to both sides:

t + q = (2^(s + 1) - 1)*q + q

t + q = 2^(s + 1)*q = sigma(t) !!

If q does not equal 1, q is a divisor of t with 1 < q < t which implies that sigma(t) >= 1 + q + t

Contradiction!!!

Therefore, q = 1 and sigma(t) = t + q = t + 1

Therefore, t is prime and t = 2^(s + 1) - 1

Hence, n = s^2*(2^(s + 1) - 1) where 2^(s + 1) - 1 is prime and 2 >= 1.

**Q.E.D.**