More Euler & Perfect Numbers (3/28)

1. More Euler & Perfect Numbers (3/28) • Again:Euler’s Theorem on Perfect Numbers. Every even perfect number is of the form 2p – 1 (2p – 1) where p and 2p – 1 are prime. • Proof. (This proof makes use of the properties of the sigma function.) • Suppose n is an even perfect number. Write n = 2km where (note!) k > 0 and m is odd (so GCD(2k, m) = 1). • Hence (n) = (2k) (m) = (2k+1– 1) (m), but also n is perfect, so we also have (n) = 2n = 2k+1m.

2. Proof continued • So we have two formulas for (n) which we can equate:2k+1m = (2k+1 – 1) (m) . • Crucial moment: Since 2k+1 – 1 is odd, we have that2k+1 divides (m), so (m) = 2k+1c for some c. • This then gives us a formula for m in terms of k and c:2k+1m = (2k+1 – 1) 2k+1c, and canceling 2k+1, we getm = (2k+1 – 1)c . • Note that if c = 1, we’re almost done. Note also (crucial moment again!) that c  m since k > 0 (!!). Just suppose c > 1, then 1, c, and m are distinct divisors of m.

3. Here comes the finish line! • Okay, so then 2k+1c = (m)  1 + c + m = 1 + c +(2k+1 – 1)c = 1 + 2k+1c . • I don’t think so! Hence our assumption that c > 1 must have false, so c = 1, and so m = 2k+1– 1 and n = 2k (2k+1 – 1). Almost done. • Finally, we claim m = 2k+1 – 1 must be prime. Since(m) = 2k+1 = 1 + (2k+1– 1) = 1 + m, the only divisors of m are 1 and itself! • Finally finally, since m = 2k+1 – 1 is prime, it must be that k + 1 is prime (why??), so set p = k+ 1, and we have n = 2p-1(2p– 1) as required. • QED!!!! (Whew!)