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• An odd perfect number, N, would have to be of the form N = p1^a1 p2^a2 . . . p_k^a_k where each of the p s are odd primes. The sum of the divisors of N,
Message 1 of 1 , Sep 27, 2007
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An odd perfect number, N, would have to be of the form

N = p1^a1 p2^a2 . . . p_k^a_k

where each of the p's are odd primes.

The sum of the divisors of N, ( including N) is the product

S = (1 + p1 + p1^2 + . . . + p1^a1) ( 1 + p2 + p2^2 + . . . + p2^a2) .
. . (1 + p_k + p_k ^2 + . . . + p_k ^a_k )

Thus an odd perfect number must satisfy the requirement

2 p1^a1 p2^a2 . . . p_k^a_k = (1 + p1 + p1^2 + . . . + p1^a1) ( 1 +
p2 + p2^2 + . . . + p2^a2) . . . (1 + p_k + p_k ^2 + . . . + p_k ^a_k )

where each p1, p2, . . . p_k is odd and greater than 2.

Thus (1 + p1 + p1^2 + . . . + p1^a1) ( 1 + p2 + p2^2 + . . . + p2^a2)
. . . (1 + p_k + p_k ^2 + . . . + p_k ^a_k ) / [ p1^a1 p2^a2 . . .
p_k^a_k ] = 2

Is it possible for (1 + p1 + p1^2 + . . . + p1^a1) ( 1 + p2 + p2^2 + .
. . + p2^a2) . . . (1 + p_k + p_k ^2 + . . . + p_k ^a_k ) / [
p1^a1 p2^a2 . . . p_k^a_k ]

to equal 2 when all the p1, p2, ... p_k are odd and greater than 2?

Since 2 = 2 mod 4, we see that the numerator must not = 0 mod 4, and so
exactly one of the a1, a2, ...a_k must be odd, and all the rest must be
even.

This shows that an odd perfect number must be an odd power of a prime
times a square.

I'm sure that many more elementary observations have been made about
requirements of odd perfect numbers.

I anticipate that the proof that odd perfect numbers do not exist would
follow closely the proof that even perfect numbers are a prime times a
power of 2.

I have not yet done a literature research on it. Does anyone here know
off the top of their head other tidbits about odd perfect numbers?

Kermit < kermit@... >
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