sonnens@mprvaxa.UUCP (10/27/83)
The following proof is by Dickson (1911), from the book
"First Course in Theory of Numbers" by Harry N. Wright.
Note. An arithmetic function f is said to be multiplicative if, when a and b
are relatively prime, f(a*b) = f(a) * f(b). It is fairly easy to prove that
the sum-of-divisors function sigma (called s here), is multiplicative.
Let m be an even perfect number, so that s(m) = 2m, and write m in the form
n-1
2 * q, where q is odd and n > 1.
n-1 n-1 n n
Then s(m) = s(2 * q) = s(2 ) * s(q) = (2 - 1) * s(q) = 2m = 2 * q.
n n
Let d = s(q) - q, so that s(q) = q + d. Then (2 - 1) * (q + d) = 2 * q
can be solved to give: q = d(2^n - 1).
This shows d to be a proper divisor of q, so that if d > 1,
s(q) >= q + d + 1 > q + d = s(q), -- a contradiction.
Therefore d = 1, and s(q) = q + d = q + 1, which is only true when q is a prime.
We conclude that q = d(2^n - 1) = 2^n - 1, so the perfect number is of the form:
n-1 n
2 * (2 - 1).rjnoe@ihlts.UUCP (10/31/83)
Seeing q = d * (2^n - 1) is easy enough, but how do you then get the s(q) >= q + d + 1 part of the inequality? How does it follow? -- Roger Noe ...ihnp4!ihlts!rjnoe
sonnens@mprvaxa (11/03/83)
Since n > 1, d < q. If d > 1, then 1, d, and q are distinct divisors of q.