PNP01 – Euclid: If 2p – 1 is
prime, then 2p – 1(2p – 1) is a perfect number
01
|
assume 2p – 1 is prime
|
assumption
|
02
|
σ(2p – 1(2p
– 1))
|
find the sum of all the divisors of 2p – 1(2p –
1)
|
03
|
= σ(2p – 1)σ(2p
– 1)
|
σ(ab) = σ(a)σ(b) if a and b are relatively prime
|
04
|
= (2p – 1)σ(2p
– 1)
|
σ(2n) = 2n + 1 – 1
|
05
|
= (2p – 1)(2p
– 1 + 1)
|
σ(prime) = prime + 1, and 2p – 1 is prime
|
06
|
= (2p – 1)(2p)
|
simplify
|
07
|
= (2p)(2p
– 1)
|
commutative
|
08
|
= 2(2p – 1(2p
– 1))
|
laws of exponents
|
09
|
2p – 1(2p
– 1) is perfect
|
if n = 2σ(n), then n is perfect
|
PNP02 – Euler: If a perfect number is even,
it is in the form of 2p – 1(2p – 1), where 2p
– 1 is prime
01
|
assume
n is an even
perfect number |
assumption
|
02
|
n = 2kx, where 2k and
x are relatively prime
|
properties of an even number
|
03
|
σ(n) = 2n
|
definition of a perfect number
|
04
|
σ(2kx) = 2(2kx)
|
substitute n = 2kx
|
05
|
σ(2kx) = 2k+1x
|
laws of exponents
|
06
|
σ(2k)σ(x) = 2k+1x
|
σ(ab) = σ(a)σ(b) if a and b are relatively prime
|
07
|
(2k+1 – 1)σ(x) = 2k+1x
|
σ(2n) = 2n + 1 – 1
|
08
|
x = y(2k+1 – 1)
|
2k+1 – 1 is odd, and cannot divide into 2k+1,
so it must divide into x
|
09
|
σ(x) = 2k+1x/(2k+1 –
1)
|
divide by (2k+1 – 1)
|
10
|
σ(x) = 2k+1y
|
substitute x = y(2k+1 – 1)
|
11
|
x + y + d = 2k+1y
|
x and y and others are divisors of x, so σ(x) = x + y + d
|
12
|
y(2k+1 – 1) + y + d = 2k+1y
|
substitute x = y(2k+1 – 1)
|
13
|
2k+1y – y + y + d = 2k+1y
|
distribute
|
14
|
2k+1y + d = 2k+1y
|
simplify
|
15
|
d = 0
|
subtract 2k+1y
|
16
|
x = 2k+1 – 1 must be prime
|
x has no other divisors d
|
17
|
x = 2p – 1 must be prime
|
substitute k = p – 1
|
PNP03 – If a perfect number is even (and
therefore in the form of 2p – 1(2p – 1)), then it can be
represented in binary as p ones followed by p – 1 zeros
01
|
assume
n is an even perfect number
|
assumption
|
02
|
n = 2p – 1(2p – 1)
|
Euler’s theorem (PNP02)
|
03
|
multiplying a number q with a number that
is a 1 followed by r zeros results in a number that starts with q and is
followed by r zeros
|
rules of multiplication
|
04
|
2p – 1 in binary is a 1 followed
by p – 1 zeros
|
rules of binary
|
05
|
2p – 1 in binary is p ones
|
rules of binary
|
06
|
n in binary is p ones followed by p – 1
zeros
|
substitution
|
PNP04 – If a perfect number is even (and
therefore in the form of 2p – 1(2p – 1)), then it is the
sum of all the integers from 1 to 2p – 1
01
|
assume
n is an even perfect number
|
assumption
|
02
|
n = 2p – 1(2p – 1)
|
Euler’s theorem
|
03
|
n = ½(2p)(2p – 1)
|
divide 2p – 1 by 2
|
04
|
n = ½(2p – 1)(2p)
|
commutative
|
05
|
n = ½k(k + 1)
|
substitute k = 2p – 1
|
06
|
n is
the sum of all integers from 1 to k
|
Σ1kk
= ½k(k + 1)
|
07
|
n is the sum of all integers from 1 to 2p
– 1
|
substitute k = 2p – 1
|
PNP05 – If a perfect number is even and not
6 (and therefore in the form of 2p – 1(2p – 1)), then it is
the sum of the first 2(p−1)/2 odd cubes
01
|
assume
n is an even perfect number and not 6
|
assumption
|
02
|
n = 2p – 1(2p – 1)
|
Euler’s theorem
|
03
|
n = (2p – 1)2p – 1
|
commutative
|
04
|
let x2 = 2p – 1 and y2
= 2p – 1
|
define
variables
|
05
|
n =
x2y2
|
substitute x2 = 2p – 1 and y2 = 2p
– 1
|
06
|
y = 2(p – 1)/2
|
square
root y2 = 2p – 1 (p also must be an odd number
for this, which is why the perfect number cannot be 6, the only instance
where p is even).
|
07
|
x2 – 2y2 = (2p
– 1) – 2(2p – 1)
|
substitute x2 = 2p – 1 and y2 = 22p
– 1
|
08
|
x2 – 2y2 = 2p
– 1 – 2p
|
laws of exponents
|
09
|
x2 – 2y2 = -1
|
simplify
|
10
|
x2y2 is the sum of
the first y cubes
|
If x2 – 2y2 = -1, then x2y2
is the sum of the first y cubes
|
11
|
n is the sum of the first y cubes
|
substitute n = x2y2
|
12
|
n is the sum of the first 2(p−1)/2
cubes
|
substitute y = 2(p – 1)/2
|
PNP06 – If a perfect number is even and not
6 (and therefore in the form of 2p – 1(2p – 1)), then it is
in the form of 1 + 9(T(2^p-2)/3), where T(2^p-2)/3 is the
sum of all the integers from 1 to 2^p – 2 / 3
01
|
assume
n is an even perfect number and not 6
|
assumption
|
02
|
n = 2p – 1(2p – 1)
|
Euler’s theorem
|
03
|
n = 22p – 1 – 2p – 1
|
distribute
|
04
|
n = 1 + 22p – 1 – 2p – 1
– 1
|
add 1 and subtract 1
|
05
|
n = 1 + ½(22p – 2p –
2)
|
factor out a ½
|
06
|
n = 1 + ½(2p – 2)(2p
+ 1)
|
factor 22p – 2p – 2
|
07
|
n = 1 + ½∙9∙1/3∙1/3(2p
– 2)(2p + 1)
|
multiply
by 9, 1/3, and 1/3
|
08
|
n
= 1 + 9∙½∙1/3(2p
– 2)1/3(2p + 1)
|
commutative
|
09
|
n
= 1 + 9∙½∙1/3(2p
– 2)1/3(2p – 2 + 3)
|
substitute
1 = -2 + 3
|
10
|
n
= 1 + 9∙½((2^p
– 2)/3)((2^p – 2)/3 + 1)
|
distribute
1/3
|
11
|
n
= 1 + 9∙½k(k
+ 1)
|
substitute k = (2^p – 2)/3, (p also must be an
odd number for this, which is why the perfect number cannot be 6, the only
instance where p is even)
|
12
|
n
= 1 + 9Tk
|
Tk = Σ1kk = ½k(k + 1) = the sum of all
integers from 1 to k
|
13
|
n
= 1 + 9T(2^p-2)/3
|
substitute k = (2^p – 2)/3, T(2^p-2)/3 is
the sum of all the integers from 1 to
2^p – 2 / 3 |
PNP07 – If a perfect number is even and not
6, then it has a digital root of 1
01
|
assume
n is an even perfect number and not 6
|
assumption
|
02
|
n = 1 + 9T(2^p-2)/3
|
proved above (PNP06)
|
03
|
n mod 9 = 1 mod 9 + 9T(2^p-2)/3
mod 9
|
mod 9 the equation n = 1 + 9T(2^p-2)/3
|
04
|
n mod 9 = 1
|
laws of mod
|
05
|
the digital root of n is 1
|
PNP08 – The sum of reciprocals of all the
divisors of every perfect number is equal to 2.
01
|
1/d1
+ 1/d2 + … + 1/dn-1 + 1/dn
|
sum of reciprocals of all the divisors of a perfect number n
|
02
|
=
n/n(1/d1 + 1/d2
+ … + 1/dn-1 + 1/dn)
|
multiply and divide by n
|
03
|
=
1/n(n/d1 + n/d2
+ … + n/dn-1 + n/dn)
|
distribute n
|
04
|
=
1/n(dn + dn – 1 + … d2
+ d1)
|
a number divided by its divisor is another divisor
|
05
|
=
1/n(d1 + d2 + … dn – 1
+ dn)
|
commutative
|
06
|
=
1/nσ(n)
|
σ(n)
is the sum of all the divisors of n
|
07
|
=
1/n∙2n
|
by
definition of a perfect number, σ(n) = 2n
|
08
|
=
2
|
simplify
|
PNP09 – Any number raised to a perfect number
exponent is equal to the product of all its perfect roots.
01
|
xn
|
any number raised to a perfect number exponent n
|
02
|
=
x2n
– n
|
substitute n = 2n – n
|
03
|
=
xσ(n)
– n
|
by
definition of a perfect number, σ(n) = 2n
|
04
|
=
x(d1 + d2 + … dn-2 + dn–1 + dn) – n
|
σ(n)
is the sum of all the divisors of n
|
05
|
=
xd1 + d2 + … dn-2 + dn–1
|
The greatest divisor of n is n itself, so dn = n,
therefore dn – n = 0
|
06
|
=
xdn-1 + dn-2 + … + d2 + d1
|
commutative
|
07
|
=
xn/d2 + n/d3 + … + n/dn-1 + n/dn
|
a number divided by its divisor is another divisor
|
08
|
=
d2√xn∙d3√xn∙…∙dn-1√xn∙dn√xn
|
laws
of exponents
|
PNP10 – If a perfect number is even, then
it ends in 6 or 8
01
|
assume
n is an even perfect number
|
assumption
|
02
|
n = 2p – 1(2p – 1)
|
Euler’s theorem
|
03
|
p is odd or p = 2
|
p is prime by Euler’s theorem
|
04
|
if p = 2:
|
assumption
|
05
|
n = 22 – 1(22 – 1)
= 6
|
substitute p = 2
|
06
|
n ends in 6
|
definition
|
07
|
if p is odd:
|
assumption
|
08
|
p = 2q + 1
|
definition of odd
|
09
|
n = 22q(22q + 1 –
1)
|
substitute p = 2q + 1
|
10
|
n = 4q(22q + 1 – 1)
|
laws of exponents
|
11
|
n = 4q(2∙22q
– 1)
|
laws of exponents
|
12
|
n = 4q(2∙4q
– 1)
|
laws of exponents
|
13
|
4q ends in 4 or 6
|
last digit cycle of 4
|
14
|
if 4q ends in 4:
|
assumption
|
15
|
n mod 10 = 4(2∙4 – 1) mod 10 = 28 mod 10 =
8
|
substitute 4q mod 10 = 4
|
16
|
n ends in 8
|
definition of mod 10
|
17
|
if 4q ends in 6:
|
assumption
|
18
|
n mod 10 = 6(2∙6 – 1) mod 10 = 66 mod 10 =
6
|
substitute 4q mod 10 = 6
|
19
|
n ends in 6
|
definition of mod 10
|
20
|
n ends in 6 or 8
|
conclusion of all options
|
PNP11 – If a perfect number is even and has
more than 1 digit, then it ends in 16, 28, 36, 56, 76, or 96
01
|
assume
n is an even perfect number
|
assumption
|
02
|
n = 2p – 1(2p – 1)
|
Euler’s theorem
|
03
|
p > 2
|
n has more than one digit
|
04
|
p is odd
|
p > 2 and p is prime by Euler’s theorem
|
05
|
p = 2q + 1
|
definition of odd
|
06
|
n = 22q(22q + 1 – 1)
|
substitute p = 2q + 1
|
07
|
n = 4q(22q + 1 – 1)
|
laws of exponents
|
08
|
n = 4q(2∙22q – 1)
|
laws of exponents
|
09
|
n = 4q(2∙4q – 1)
|
laws of exponents
|
10
|
4q ends in 4, 16, 64, 56, 24,
96, 84, 36, 44, or 76
|
last 2 digit cycle of 4
|
11
|
if 4q ends in 4:
|
assumption
|
12
|
n mod 100 = 4(2∙4 – 1) mod 100 = 28 mod 100
= 28
|
substitute 4q mod 100 = 4
|
13
|
n ends in 28
|
definition of mod 100
|
14
|
if 4q ends in 16:
|
assumption
|
15
|
n mod 100 = 16(2∙16 – 1) mod 100 = 496 mod
100 = 96
|
substitute 4q mod 100 = 16
|
16
|
n ends in 96
|
definition of mod 100
|
17
|
if 4q ends in 64:
|
assumption
|
18
|
n mod 100 = 64(2∙64 – 1) mod 100 = 1728 mod
100 = 28
|
substitute 4q mod 100 = 64
|
19
|
n ends in 28
|
definition of mod 100
|
20
|
if 4q ends in 56:
|
assumption
|
21
|
n mod 100 = 56(2∙56 – 1) mod 100 = 616 mod
100 = 16
|
substitute 4q mod 100 = 56
|
22
|
n ends in 16
|
definition of mod 100
|
23
|
if 4q ends in 24:
|
assumption
|
24
|
n mod 100 = 24(2∙24 – 1) mod 100 = 1128 mod
100 = 28
|
substitute 4q mod 100 = 24
|
25
|
n ends in 28
|
definition of mod 100
|
26
|
if 4q ends in 96:
|
assumption
|
27
|
n mod 100 = 96(2∙96 – 1) mod 100 = 8736 mod
100 = 36
|
substitute 4q mod 100 = 96
|
28
|
n ends in 36
|
definition of mod 100
|
29
|
if 4q ends in 84:
|
assumption
|
30
|
n mod 100 = 84(2∙84 – 1) mod 100 = 5628 mod
100 = 28
|
substitute 4q mod 100 = 84
|
31
|
n ends in 28
|
definition of mod 100
|
32
|
if 4q ends in 36:
|
assumption
|
33
|
n mod 100 = 36(2∙36 – 1) mod 100 = 2556 mod
100 = 56
|
substitute 4q mod 100 = 36
|
34
|
n ends in 56
|
definition of mod 100
|
35
|
if 4q ends in 44:
|
assumption
|
36
|
n mod 100 = 44(2∙44 – 1) mod 100 = 3828 mod
100 = 28
|
substitute 4q mod 100 = 44
|
37
|
n ends in 28
|
definition of mod 100
|
38
|
if 4q ends in 76:
|
assumption
|
39
|
n mod 100 = 76(2∙76 – 1) mod 100 = 3876 mod
100 = 76
|
substitute 4q mod 100 = 76
|
40
|
n ends in 76
|
definition of mod 100
|
41
|
n ends in 16, 28, 36, 56, 76, or 96
|
conclusion of all options
|
No comments:
Post a Comment