Euclid notice a distinct pattern between the first four perfect numbers:
6=21(1+2)=2·3
28=22(1+2+22)=4·7
496=4(1+2+22+23+24 )=16·31 and
8128=26(1+2+22+23+24+25+26 )=64·127
Upon further inspection, notice that 2
3 and 2
5 are
missing in the sequence. This is because 2
3(1+2+2
2+2
3
)=8·15=120 and 2
5(1+2+2
2+2
3
+2
4+2
5)=32·63=2016, and in both cases, 8·15=120 and
32·63=2016, each case the second number is a composite number, i.e. 15=3·5 and
63=7·9, whereas when the product yields a perfect number, the second multiplier is a
prime number, i.e. 3, 7, 31, and 27. (Voight) From Euclid's Elements, he states "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" (Voight). Another words, a Perfect
number is a natural number where all of its proper divisors, or all of its divisors
excluding itself, add up to the number itself, where the number is in the form of
(2
n-1)(2
n-1). So, if the number itself is included, the divisors
of a perfect number would add up to two times that given number.
One other important element to perfect numbers, as discussed in the history
section, is their tie to Mersenne primes. A Mersenne prime is in the form 2
n-1, where n is a prime integer that makes 2
n-1 prime as well. The
formula (2
n-1)(2
n-1) is a perfect number if, and only if
2
n-1 is a Mersenne Prime. So, if n in (2
n-1)(2
n-
1) is prime AND if (2
n -1) is prime, then the equation (2
n-1)(2
n-1) yields a perfect number. Using all this information, a
proof of Euclid perfect number theorem can be done easily.
Before this is done, let's add up all the factors of, say, 496.
496=1+2+4+8+16+31+62+124+248
Now, let's add in 496 so we'll get:
2·496=1+2+4+8+16+31+62+124+248+496
Because we did this, 31 can be factored out of the second half of the equation from
above to yield:
⟹1+2+4+8+16+31(1+2+4+8+16)
Now, we can combine like terms, meaning, because we have (1)(1+2+4+8+16)+(31)
(1+2+4+8+16), we can take the 31 and the 1 and add them together with everything
else behind it.
⟹(1+31)(1+2+4+8+16)
⟹(32)(31)=992
⟹2(496)
The same thing can be done for every perfect number, and it can be proven in a
similar manner in a general sense. We start with the formula N =(2n -1)
(2n-1), where N is a Perfect Number iff n 𝜖 prime numbers and (2n -1)
is a Mersenne prime.
N=(2n-1)(2n-1)
The Factors of 2n-1 are:
2n-1=(1+21+22+23+…+2n-2+2n-1)
The factors of N
= 21(2n-1)+22(2n-1)+23(2n-1)+…+2n-2
(2n-1)+2n-1(2n-1)
When we combine the two, these yield:
⟹(1+21+22+23+…+2n-
2+2n-1)+21 (2n-1)+22
(2n-1)+23(2n-1)+…+2n-2
(2n-1)+2n-1(2n-1)=2·[(2n-
1)(2n-1)]
Simplifying slightly:
⟹(1+21+22+23+…+2n-
2+2n-1)+(2n 1)(1+21+22+23+…+2n-
2+2n-1)
Because the first half of the equation above is one times by all of the first parentheses,
we can combine like terms and add 1 and (2n-1), resulting in:
⤬ ⟹2n (1+21+22+23+…+2n-2+2n-1)
Now, Let T=1+21+22+23+…+2n-2+2n-1
So, 2T=21+22+23+…+2n-
1+2n
2T-T=2n-1=T
Using this we can substitute T, which is just 2n-1, into step ⤬. The result is as follows:
2n (2n-1)
Now, a 2 can be factored out to get 2 times our original equation:
2n(2n-1) ⟹ 2[(2n-1 )(2n-1)]
And thus, have proved that (2
n-1 )(2
n-1) is a perfect number. □
(Haran)
For a video of the proof, click
here.
Click
here for more about significance and applications of perfect numbers.