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Over 2300 years ago Euclid
proved that If 2^{k}1 is a prime
number (it would be a Mersenne
prime), then 2^{k1}(2^{k}1) is a perfect
number. A few hundred years ago Euler
proved the converse (that every even perfect number has this form).
It is still unknown if there are any odd perfect numbers (but if there are,
they are large and have many prime factors).
 Theorem:
 If 2^{k}1 is a prime number, then 2^{k1}(2^{k}1)
is a perfect number and every even perfect number has this form.
Proof: Suppose first that p = 2^{k}1
is a prime number, and set n = 2^{k1}(2^{k}1).
To show n is perfect we need only show sigma(n)
= 2n. Since sigma is multiplicative
and sigma(p) = p+1 = 2^{k}, we know
sigma(n) = sigma(2^{k1})^{.}sigma(p)
= (2^{k}1)2^{k} = 2n.
This shows that n is a perfect number.
On the other hand, suppose n is any even perfect number
and write n as 2^{k1}m where m is an odd
integer and k>2. Again sigma is multiplicative so
sigma(2^{k1}m) = sigma(2^{k1})^{.}sigma(m)
= (2^{k}1)^{.}sigma(m).
Since n is perfect we also know that
sigma(n) = 2n = 2^{k}m.
Together these two criteria give
2^{k}m = (2^{k}1)^{.}sigma(m),
so 2^{k}1 divides 2^{k}m hence 2^{k}1
divides m, say m = (2^{k}1)M. Now substitute
this back into the equation above and divide by 2^{k}1 to get
2^{k}M = sigma(m). Since m and M are both divisors
of m we know that
2^{k}M = sigma(m) > m + M
= 2^{k}M,
so sigma(m) = m + M. This means that m is prime
and its only two divisors are itself (m) and one (M). Thus
m = 2^{k}1 is a prime and we have prove that the number
n has the prescribed form. 