A perfect number \(P\) is equal to the sum of its divisors (where the divisors include \(1\), but not \(P\) itself).

**Euclid**: If \(2^n-1\) is prime then \(2^{n-1}(2^n-1)\) is perfect

**Euler**:*all*even perfect numbers are of the form \(2^{p-1}(2^p-1)\), where \(2^p-1\) is a Mersenne prime (and so \(p\) is prime).**6**= 1 +**2**+**3**= 2 (2^{2}-1)**28**= 1 + 2 +**4**+**7**+ 14 = 2^{2}(2^{3}-1)**496**= 1 + 2 + 4 + 8 +**16**+**31**+ 62 + 124 + 248 = 2^{4}(2^{5}-1)**8128**= 1 + 2 + 4 + ... +**64**+**127**+ ... + 4064 = 2^{6}(2^{7}-1)**33,550,336**= 1 + ... +**4096**+**8191**+ ... + 16775168 = 2^{12}(2^{13}-1)**8,589,869,056**= 1 + ... +**65536**+**131071**+ ... + 4294934528 = 2^{16}(2^{17}-1)**137,438,691,328**= 1 + ... +**262144**+**524287**+ ... + 68719345664 = 2^{18}(2^{19}-1)- ...

- Every even perfect number ends in a '6' or an '8'.
- All even perfect numbers are triangular numbers.
- 6 = 1 + 2 + 3 = 1 + 2 + (2
^{2}-1) - 28 = 1 + 2 + 3 + ... + 7 = 1 + 2 + 3 + ... + (2
^{3}-1) - 496 = 1 + 2 + 3 + ... + 31 = 1 + 2 + 3 + ... + (2
^{5}-1) - 8128 = 1 + 2 + 3 + ... + 127 = 1 + 2 + 3 + ... + (2
^{7}-1) - 33,550,336 = 1 + 2 + 3 + ... + 8191 = 1 + 2 + 3 + ... +
(2
^{13}-1) - 8,589,869,056 = 1 + 2 + 3 + ... + 131071 = 1 + 2 + 3 +
... + (2
^{17}-1) - 137,438,691,328 = 1 + 2 + 3 + ... + 524287 = 1 + 2 + 3
+ ... + (2
^{19}-1) - ...
- general result: \( n: \mathbb{N} \vdash 2^{n-1} (2^n-1) = 1 + 2 + 3+ \ldots + (2^n-1)\)

- 6 = 1 + 2 + 3 = 1 + 2 + (2
- Every even perfect number, other than 6, is the sum of consecutive
odd cubes.
- (6, with
=2, does not fit the pattern)*p* - 28 = 1
^{3}+ 3^{3}= 1^{3}+ (2^{(3+1)/2}-1)^{3} - 496 = 1
^{3}+ 3^{3}+ 5^{3}+ 7^{3}= 1^{3}+ 3^{3}+ 5^{3}+ (2^{(5+1)/2}-1)^{3} - 8128 = 1
^{3}+ 3^{3}+ ... + 15^{3}= 1^{3}+ 3^{3}+ ... + (2^{(7+1)/2}-1)^{3} - 33,550,336 = 1
^{3}+ 3^{3}+ ... + 127^{3}= 1^{3}+ 3^{3}+ ... + (2^{(13+1)/2}-1)^{3} - 8,589,869,056 = 1
^{3}+ 3^{3}+ ... + 511^{3}= 1^{3}+ 3^{3}+ ... + (2^{(17+1)/2}-1)^{3} - 137,438,691,328 = 1
^{3}+ 3^{3}+ ... + 1023^{3}= 1^{3}+ 3^{3}+ ... + (2^{(19+1)/2}-1)^{3} - ...
- conjecture: \( n: \mbox{Odd}\, ?{\vdash}\, 2^{n-1} (2^n-1) = 1^3 + 3^3 + \ldots + (2^{(n+1)/2}-1)^3 \)

- (6, with

- No
*odd*perfect numbers are known, but if one does exist, a lot is known about it:- it is a perfect square multiplied by an odd power of a single prime
- it has at least 8 distinct prime factors
- it has at least 75 prime factors (not necessarily distinct)
- its largest prime factor greater that 10
^{7} - its second largest prime factor is greater that 10
^{4} - its third largest prime factor is greater that 10
^{2} - it is divisible by a prime
component greater that 10
^{20}

^{300}.

[*My thanks to Douglas Iannucci and Joshua Zelinsky for some of this information, some of which is from the work of Kevin Hare.*]

- Conway and Guy.
*The Book of Numbers*. chapter 5 - Devlin.
*All the Math that's Fit to Print*. chapters 113, 138 - Hardy & Wright.
*Theory of Numbers*. section 16.8