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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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.
Ahh, this will be long. Here goes...Quote:
Originally Posted by flash101
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)
This part is a little harder...at least I think.Quote:
Originally Posted by flash101
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.