## Theorem euro [was: Theorem eight]

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• PS: This _very_ old theorem has an impeccable European lineage. Euclid proved (Book IX, Proposition 36) that ... In symbols: N(p) = 2^(p-1)*M(p) is perfect if
Message 1 of 1 , Jan 5, 2002
PS: This _very_ old theorem has an impeccable European lineage.

Euclid proved (Book IX, Proposition 36) that

> If as many numbers as we please beginning from a unit are set
> out continuously in double proportion until the sum of all
> becomes prime, and if the sum multiplied into the last makes
> some number, then the product is perfect.

In symbols: N(p) = 2^(p-1)*M(p) is perfect if
M(p) = 1 + 2 + 4 + 8 + ... + 2^(p-1) = 2^p-1 is prime.

Proof: For any p, the sum of the divisors of N(p) is
sigma_1(N(p)) = sigma_1(2^(p-1))*sigma_1(M(p)) with
sigma_1(2^(p-1)) = M(p). If M(p) is prime, then
sigma_1(M(p)) = 1 + M(p) = 2^p and hence
sigma_1(N(p)) = 2*N(p), which is the definition of perfection.

Euler proved, about 2000 years later, that every
_even_ perfect number is of the form 2^(p-1)*M(p)
where M(p) is prime.

Mind you, we might have to wait another 2000 years for
some non-european (on Alpha Centuri perhaps?) to provide
a proof that there is no _odd_ perfect number...

David
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