RFSC 04-01
Revised
A PROOF OF THE ODD PERFECT NUMBER CONJECTURE
Simon Davis
Research Foundation of Southern California
8837 Villa La Jolla Drive #13595
La Jolla, CA 92039
Abstract. It is sufficient to prove that there is an excess of prime factors in the product
of repunits with odd prime bases defined by the sum of divisors of the integer N = (4k +
Q`
1)4m+1 i=1 qi2αi to establish that there do not exist any odd integers with equality between
σ(N ) and 2N . The existence of distinct prime factors in the repunits in σ(N ) follows from
2α +1
the properties of products of linear factors in the decomposition of qi2αi +1 −1 and qj j −1,
when qi , qj , 2αi + 1 and 2αj + 1 are odd primes. The occurrence of new prime divisors
4m+2
q
2αi +1
−1
in each quotient (4k+1)4k −1 , i qi −1 , i = 1, ...` also implies that the square root of the
product of 2(4k + 1) and the sequence of repunits will not be rational unless the primes
are matched. Although twelve solutions to the rationality condition for the existence of
)
odd perfect numbers are obtained, it is verified that these integers all satisfy σ(N
6= 2
N
because the repunits in the product representing σ(N ) introduce new prime divisors. It is
then proven that the prime factors of σ(N ) and 2N differ for all sets of primes {4k + 1, qi }
and exponents {4m + 1, 2αi }, and therefore, there exist no odd perfect numbers.
MSC Class: 11D61, 11K65
1. Introduction
The uniqueness of the prime decomposition of an integer allows for the comparison
between its magnitude and the sum of the divisors. Since the sum of the divisors of an
2α +1
Q`
qi i −1
(4k+1)4m+2 −1 Q`
2αi
by
the
product
odd integer N = (4k + 1)4m+1
q
i
i=1
i=1
4k
qi −1 ,
it is sufficient to determine the properties of the prime factors of the repunits to establish
that there exist no odd perfect numbers. The irrationality of
h
i 12
p
Q`
2(4k + 1) σ((4k + 1)4m+1 ) · i=1 σ(qi2αi ) would imply that σ(N ) cannot equal 2N .
It has been proven for a large class of primes {4k + 1; qi } and exponents {4m + 1; 2αi } that
the rationality condition is not satisfied. The irrationality of the square root of the product
of 2(4k + 1) and the sequence of repunits is not valid for all sets of primes and exponents,
however, and it is verified in §4 that the rationality condition holds for twelve odd integers.
The factorizations of these integers have the property that the repunits have prime divisors
which form interlocking rings, whereas, in general, the sequence of prime factors does not
close. The presence of a sequence of primes of increasing magnitude prevents any finite
)
odd integer from being a perfect number. To prove that σ(N
6= 2, it is necessary also to
N
obtain a lower bound for the number of prime divisors in σ(N ). Since the number of factors
q
2αi +1
−1
of i qi −1 is minimized when 2αi + 1 is prime, the exponent is presumed to be prime
throughout the discussion. It is demonstrated in Theorem 1 that any pair of repunits, with
prime bases and exponents satisfying a given inequality, do not have identical sets of prime
divisors. Then, either σ(N ) has an excess of prime divisors or constraints must be imposed
on {4k + 1; qi } and {4m + 1; 2αi } which have no integer solution. The non-existence of odd
perfect numbers also follows from the sequence of prime factors of increasing magnitude in
the factorization of σ(N ), when one of three specified relations is satisfied, and constraints
on the basis and exponents otherwise.
2. The Existence of Different Prime Divisors in the Repunit Factors of the
Sum of Divisors
The existence of distinct prime divisors in the quotients
1, ..., ` shall be used to prove the inequality
σ(N )
N
(4k+1)4m+2 −1
4k
and
2αi +1
qi
−1
qi −1 ,
i=
6= 2.
There are three known positive integer solutions to the exponential Diophantine equan
2
3
3
m
−1
−1
−1
−1
−1
= yy−1
, m, n > 1, 23 −1 = 66−1
, 31 = 25 −1 = 55−1
and 8191 = 213 −1 = 90
tion xx−1
90−1
[9][13][14] and no equalities of this type have been established yet for odd prime bases and
exponents. It can be assumed initially that
2α +1
qj j − 1
qi2αi +1 − 1
6=
qi − 1
qj − 1
1
(2.1)
If the inequality (2.1) holds, then either the sets of primitive divisors of
2α +1
qj j −1
qj −1
that
2αi +1
qi
−1
qi −1
and
are not identical or the exponents of the prime power divisors are different, given
2α +1
qi i −1
qi −1
2αj +1
qj
6=
−1
qj −1 .
qi2αi +1 − 1
2α +1
qj j
−1
Let qi > qj . Since
2αj +1
=
'
qi2αi +1 − 1 qi
2α +1
qi j
−1
qi2αi +1 − 1
2αj +1
qi
−1
2α +1
qj j
·µ
−1
qi
qj
−1
¶2αj +1 Ã
1+
Ã
+O
this equals
qi2αi +1
2α +1
qi j
−1
"µ
qi
qj
2αj +1
qi2αi +1 − qj
−1
!
2α +1 2αj +1
qj
qi j
!¸
1
2α +1 4α +2
qi j qj j
#
¶2αj +1
+²
−1
µ ¶2αj +1 2αj +1
2α +1
qi
− qj j − 1
qi
²'
2α +1 2α +1
qj
qi j qj j
2α +1
When qj < qi <
²·
2 2αj +1
qj i ,
qi2αi +1 − 1
2αj +1
qi
−1
²·
2αi +1
qi
−1
qi
−1
2αj +1
qi2αi +1 − 1 (qi
2αj +1
The prime divisors of (qi
2α +1
qj j
2αj +1
qj
2αj +1
− qj
4αj +2
qj
−1
2αj +1
− qj
(2.3)
< 1. The exact value of the remainder term is
2αj +1
=
(2.2)
2αj +1
− 1)(qj
2αj +1
− 1)(qj
2αj +1
(qj
− 1) − 1
− 1)
(2.4)
− 1) − 1 are different from the factors
4α +2 2α +1
of
− 1. Unless they are all cancelled by qj j (qi j − 1) an additional primitive
2α +1
q i −1
2α +1
2α +1
2α +1
divisor occurs in the factorization of i qi −1 . Since (qi j −qj j −1)(qj j −1)−1 ≡
2α +1
4α +2
−qi j 6≡ 0 (mod qj ), none of the divisors of the numerator are cancelled by qj j . It
2α +1 2α +1
2α +1
follows that these factors can be cancelled entirely only if qj j (qj j − 1) + 1|qi j − 1.
2α +1
However, the primitive divisors of qi j − 1 are congruent to 1 modulo 2αj + 1 and their
2αj +1
qi
−1
qj −1
2αj +1
qj
−1
and qj −1 is 2αj +1 when the
product is a(2αj +1)+1. The only other divisor of
exponent is prime. However, 2αj +1 cannot be chosen to be a prime divisor of both repunits
2α +1
since there will be a distinct prime divisor amongst the factors of qj j (q 2αj +1 − 1) + 1
2αj +1
and qi
− 1.
There are therefore two cases to consider:
2
2αj +1
(i) (qj − 1)qj
(2αj + 1)(b(2αj + 1) + 1) + 1; (qi − 1)(q(2αj + 1) + 1) with 2αj + 1 6 |qi − 1
2αj +1
The sets of prime divisors will be identical if a = (qj − 1)qj
(b(2αj + 1) + 1) which
2αj +1 2αj +1
2α +1
would imply that (qi − 1)((2αj + 1)qj
(qj
− 1) + 1) = qi j − 1. However,
2α +1 2α +1
2α +1 2α +1
(2αj + 1)qj j (qj j − 1) + 1 and qj j (qj j − 1) + 1 do not have common divisors
as 2αj + 1 < qj2α+1 .
2αj +1
(ii) (qj − 1)qj
(b(2αj + 1) + 1) + 1 with 2αj + 1 6 |qj − 1; (2αj + 1)(qi − 1)(a(2αj + 1) + 1)
2αj +1
These two integers will have the same prime factors only if 2αj +1|qj
qj = 2αj + 1. The equality (qj −
2α +1
1)qj j (b(2αj
+ 1) + 1) = c(2αj + 1) − 1, which is
2αj +1
necessary for a(2αj + 1) + 1 to be a divisor of (qj − 1)qj
be satisfied, because (2αj + 1) 6 |c(2αj + 1) − 1.
2αj +1
Consequently, qj
2αj +1
(q 2αj +1 −1)+1 6 |qi
2α +1
2 2αj +1
i
prime divisor if qj < qi < qj
, and therefore
(b(2αj + 1) + 1) + 1 cannot
−1 and the repunit
2αi +1
qi
−1
qj −1
has a distinct
.
h
h
Let qi − 1 = ph1 i1 ...phs is and qj − 1 = p1 j1 ...ps js . Then
(ph1 i1 ...phs is + 1)2αi +1 − 1
= 2αi + 1 + αi (2αi + 1)(ph1 i1 ...phs is ) + ... + (ph1 i1 ...phs is )2αi
his
h1s
p1 ...ps
h
h
(p1 j1 ...ps js + 1)2αj +1 − 1
h
ph1 1s ...ps js
h
h
= 2αj + 1 + αj (2αj + 1)(p1 j1 ...phs is ) + ... + (p1 j1 ...phs js )2αj
qi2αi +1
(2.5)
− 1. Consider the congruence c0 + c1 x + ... +
Let P be a primitive divisor of the
c2αi x2αi ≡ 0 (mod P ), ct > 0,
t = 0, 1, 2, ..., 2αj . There is a unique solution x0 to the congruence relation f (x) ≡ c0 +
c1 x + ... + cn xn ≡ 0 (mod P ) within [x0 − ², x0 + ²] [12], where
²f 0 (x0 ) +
³
Setting ct =
2αi +1
t+1
´
,
µ
¶
2αi + 1
f (qi − 1) =
+2
(qi − 1) + ... + 2αi (qi − 1)2αi −1
3
³
´
i −1
. Consequently, if
the constraint (2.6) implies a bound on ² of 1 − 2αi2+1 q2α
i
µ
¶
2
qi − 1
|qi − qj | < 1 −
2αi + 1
2αi
0
2αi + 1
2
¶
²2 00
f (x0 ) + ... < P
2!
(2.6)
µ
3
(2.7)
(2.8)
the prime P is not a common factor of both repunits. The number of repunits with
primitive divisor P will be bounded by the number of integer solutions to the congruence
relation.
An example of congruence with more than one solution less than P is
(2αi + 1)2αi (2αi − 1) 2 (2αi + 1)2αi (2αi − 1)(2αi − 2) 3
(2αi + 1) + αi (2αi + 1)x +
x +
x
3!
4!
(2αi + 1)2αi (2αi − 1)(2αi − 2)(2αi − 3) 4
+
x = 0 (mod 11)
5!
(2.9)
which is solved by x = 2, 4 when 2αi + 1 = 5. This is consistent with the inequality (2.8),
5
5
since ² = 0.074897796 when x0 = 2. Consequently, 3 2−1 = 11 · 11 and 5 4−1 = 11 · 71 both
5
have the divisor 11 and the distinct prime divisor arises in the larger repunit. Indeed, 3 2−1
q
2αk +1
−1
would not have a different prime factor from a repunit k qk −1 that has 11 as a divisor.
A set of primes {3, qk , ...} and exponents {5, 2αk , ...} does not represent an exception to
5
the Theorem 1 if 3 2−1 is chosen to be the first repunit in the product, introducing the
prime divisor 11. Each subsequent repunit then would have a distinct prime factor from
the previous repunit in the sequence.
2αj +1
For P to be a divisor of qj
h
h
h
−1, p1 j1 p2 j2 ...ps js also would have to satisfy the congruh
h
h
h
h
h
ence relation when αi = αj . If ph1 i1 ph2 i2 ...phs is = p1 j1 p2 j2 ...ps js + nP , then p1 j1 p2 j2 ...ps js is
another solution to the congruence. Therefore gcd(qi − 1, qj − 1)
min(hi1 ,hj1 )
min(his ,hjs )
= p1
...ps
|nP , and, as P cannot equal any of the prime divisors pik ,
min(hi1 ,hj1 )
min(his ,hjs )
gcd(qi −1, qj −1) = p1
...ps
|n. Then, qi −1 = qj −1+k gcd(qi −1, qj −1)P
qi −qj
for some integer k , and P | gcd(qi −1,qj −1) .
However, if P is a prime divisor of
2αi +1
qi
−1
qi −1
2αj +1
and
qj
−1
qj −1 ,
and αi = αj ,
P |(qi − qj )(qi2αi + (2αi + 1)qi2αi −1 qj + ... + (2αi + 1)qi qj2αi −1 + qj2αi )
(2.10)
Since gcd(qi − qj , qi2αi + (2αi + 1)qi2αi −1 qj + ... + (2αi + 1)qi qj2αi −1 + qj2αi ) = 1, either
P |qi − qj or
P |qi2αi + (2αi + 1)qi2αi −1 qj + ... + (2αi + 1)qi qj2αi −1 + qj2αi . When P |qi − qj , it may satisfy
the divisibility condition for the congruence relation. However, there also would be a
primitive divisor which is a factor of qi2αi + qi2αi −1 qj + ... + qj2αi −1 qi + qj2αi , and not qi − qj .
h
h
h
Then ph1 i1 ph2 i2 ...phs is < p1 j1 p2 j2 ...ps js + P . If there is only one solution to the congruence
2αi + 1 + αi (2αi + 1)x + ... + x2αi ≡ 0 (mod P ) less than P , a contradiction is obtained.
When one of the congruences is not satisfied for at least one primitive divisor P , this prime
is not a common factor of both repunits.
It shall be proven now that repunits with different odd prime bases and exponents are
not equal and there always exists a prime which is not a common divisor.
4
q
2αi +1
q
−1
2αj +1
−1
Theorem 1. The pair of integers i qi −1 and j qj −1 , qi > qj do not have identical sets of prime divisors if the bases and exponents are odd primes.
Proof.
If αi 6= αj ,
qi2αi +1 − 1 =
2αi
Y
k
(qi − ω2α
)
i +1
k=0
2αj
2α +1
qj j
−1=
Y
(2.11)
0
k
(qj − ω2α
)
j +1
k0 =0
ωn = e
2πi
n
2α +1
Quotients of products of linear factors in qi2αi +1 −1 and qj j −1 generally will not involve
cancellation of integer divisors because products of powers of the different units of unity
cannot be equal unless the exponents are a multiples of the primes 2αi + 1 or 2αj + 1. The
sumsets which give rise to equal exponents can be enumerated by determining the integers
s1 , ..., st such that the sums are either congruent to 0 modulo 2αi + 1 and 2αj + 1. It is
apparent that the entire integer sets {1, 2, ..., 2αi } and {1, 2, ..., 2αj } cannot be used for
unequal repunits. The number of sequences satisfying the congruence relations for each
P
P
0
repunit must equal P tk =2αi +1 (2tk − 1) and P t0 =2αj +1 (2tl − 1), with sequences of
l
integers which sum to a non-zero value giving rise to cancellation of complex numbers in
the product being included.
Q t0 k
Q tk
s0n
sm
). Upper and lower
(qj − ω2α
) and n=1
(qi − ω2α
Consider two products m=1
j +1
i +1
bounds for these products are
(qi − 1)
tk
<
tk
Y
sm
(qi − ω2α
) < (qi + 1)tk
i +1
m=1
(2.12)
0
0
(qj − 1)tl <
tl
Y
s0n
0
(qj − ω2αj +1 ) < (qj + 1)tl
n=1
Since the minimum difference between primes is 2, one of the inequalities (qi + 1)tk ≤
0
0
0
0
(qj − 1)tl , (qj + 1)tl ≤ (qi − 1)tk , (qi + 1)tk > (qj − 1)tl or (qj + 1)tl > (qi − 1)tk will be
satisfied when t0l 6= tk . The first two inequalities imply that the factors of the repunits
Qt0k
Qtk
s0n
sm
defined by the products m=1
(qi − ω2α
)
and
(q
−
ω
j
n=1
2αj +1 ) are distinct.
i +1
If either of the next two inequalities hold, the factors possibly could be equal when
0
0
(qi − 1)tk = (qj − 1)ti or (qi + 1)tk = (qj + 1)tl . In the first case, t0l = nij tk , qi − 1 =
5
P
P 0
(qj − 1)nij , nij > 2. This relation holds for all tk and t0l so that
k tk = nij
l tl .
However, 2αi + 1 and 2αj + 1 are prime, so that this equation is not valid. When qi + 1 =
P
P
(qj + 1)nij , nij ≥ 2, the relation k tk = nij l t0l again leads to a contradiction.
The identification of the products in Eq.(2.11) for each corresponding k, l is necessary, since any additional product would give rise either to a new prime factor after appropriate rescaling, or a different power of the same number, which would would prevent a cancellation of an extra prime, described later in the proof. The inequalities
0
0
(qj − 1)tl < (qi − 1)tk < (qj + 1)tl imply that
t0l
ln(qj + 1)
ln(qj − 1)
< tk < t0l
ln(qi − 1)
ln(qi − 1)
(2.13)
If these inequalities hold for two pairs of exponents (tk1 , t0l1 ), (tk2 , t0l2 ), the interval
h
i
t0
t0
t
ln(q −1)
ln(q +1)
t0l2 ln(qji −1) , t0l2 ln(qji −1) contains t0l2 tk1 . The fractions tkkm cannot be equal to tl0m for all
m, n as this would imply that ttk0 =
l
prime fraction when αi 6= αj .
Then
Then tk2
n
l1
2αi +1
2αj +1 ,
which is not possible since
2αi +1
2αj +1
ln
is an irreducible
¯
¯
¯t
tk2 ¯¯
1
¯ k1
¯ 0 − 0 ¯≥ 0 0
¯ t l1
tl2 ¯ tl1 tl2
(2.14)
h
i
0 ln(qj −1) 0 ln(qj +1)
6∈ tl2 ln(qi −1) , tl2 ln(qi −1) if
µ
ln
qj + 1
qj − 1
¶
<
ln(qi − 1)
t0l1 t0l2
(2.15)
As t0l ≤ 2αj + 1, this inequality is valid when
qj · ln(qi − 1) > 3(2αj + 1)2
Since the arithmetic primitive factor of
or
Φn (q)
p ,
q n −1
q−1 ,
n 6= 6 is Φn (q) =
where p is the largest prime factor of both
prime, which is a primitive divisor of
2α +1
qi i −1
(2.16)
n
gcd(n,3)
Q
1≤k≤n
gcd(k,n)=1
(qi − e
2πik
n
)
and Φn (q) [1][2][3][15][16], any
, divides Φ2αi +1 (qi ). It follows that these
Qtk
sm
(qi − ω2α
)
primes can be obtained by appropriate multiplication of the products m=1
i +1
Qt0k
s0n
and n=1 (qj − ω2αj +1 ). Although these products may not be the primitive divisors, they
Q t1
sm
are real, and multiplication of m=1
(qi − ω2α
) by κ1 ∈ R can be compensated by
i +1
0
Q t2
sm
−1
to obtain the integer factor. Since products of two complex
m0 =1 (qi − ω2αi +1 by κ1
qi −1
6
2πik1
linear factors (qi − e 2αi +1 )(qi − e
−2πik1
2αi +1
) are real, let
2πik1
− 2α
2πik2
− 2α
x1 = (qi − e 2αi +1 )(qi − e
x2 = (qi − e 2αi +1 )(qi − e
y1 = (qj − e
2πik3
2αj +1
2πik4
)(qj − e
y2 = (qj − e 2αj +1 )(qj − e
2πik1
i +1
)
2πik2
i +1
)
2πik3
− 2α +1
j
2πik4
j +1
− 2α
(2.17)
)
)
The occurrence of the same prime divisors in the two repunits
relations of the form
κ1 x1 = P1`1 = P1h1 κ01 y1
2αi +1
qi
−1
qi −1
2αj +1
and
qj
−1
qj −1
`2
h2 0−1
κ−1
1 x2 = P2 = P2 κ1 y2
yields
(2.18)
for two prime factors implies
P1`2 h1 −h2 `1
x`12 −h2 x`22 −h2
=
y1`2 y2`2
which cannot be satisfied by the prime P1 . This ratio equals
³
´
³
´
(`2 −h2 )
(`2 −h2 )
2πk1
2πk2
(`2 −h2 )
2
2
2
2
[q
−
2q
cos
+
1]
[q
−
2q
cos
i
i
i
i
2αi +1
2αi +1 + 1]
(x1 x2 ) 2
³
´
³
´
=
`2
`2
`2
2πk3
2πk4
2 [q 2 − 2q cos
2
(y1 y2 ) 2
[qj2 − 2qj cos 2α
+
1]
+
1]
j
j
+1
2α
+1
j
j
(2.19)
(2.20)
2
2
In general,
the
³
´ quadratic expressions are not rational. However, as (qi − 1) < [qi −
2πk1
2qi cos 2α
+1] < (qi +1)2 , a power of this expression which is integer cannot be equated
i +1
³
´
2α +1
qi i −1
2πk1
2
to a prime p prime unless p = qi . Since qi 6 |
qi −1 , the factors [qi − 2qi cos 2αi +1 +
1]
(`2 −h2 )
2
, for example, are not fractional powers of only one prime. Therefore, the ratio
` −h
` −h
x12 2 x22 2
`
`
y12 y22
must contain at least two primes. While the product of two quadratic factors
may be a surd, it is possible to choose a set of quadratic factors such that the product
is rational and even an integer. Although a further rescaling of this product might be
required to obtain a prime divisor, it has been established that every prime factor arises in
the cyclotomic polynomials which are constructed from the linear and quadratic factors.
It follows that the rescaling follows a definite pattern to introduce the prime factors.
The occurrence of the same three prime divisors in the two repunits would imply that
κ1 x1 = P1`1 = P1h1 κ01 y1
`2
h2 0−1 0
κ−1
1 κ2 x2 = P2 = P2 κ1 κ2 y2
`3
h3 0−1
κ−1
2 x3 = P3 = P2 κ2 y3
7
(2.21)
yielding the equation
P1`2 h1 −h2 `1 P3`2 h3 −h2 `3 =
(x1 x2 )`2 −h2
(y1 y2 )`2
However,
(x1 x2 x3 )`2 −h2
(y1 y2 y3 )`2
(2.22)
introduces a minimum of two different primes, whereas
` −h2
x32
tains at least one more prime divisor. Alternatively, this fraction equals
(x1 x2 )
(`2 −h2 )
2
`2
(y1 y2 ) 2
·
(x2 x3 )
(`2 −h2 )
2
`2
(y2 y3 ) 2
·
(x3 x1 )
`
y32
(`2 −h2 )
2
con-
(2.23)
`2
(y3 y1 ) 2
Each term in the product contains n primes, n ≥ 2, and cancellation between the each pair
of fractions leaves at least two distinct primes, since cancellation of all of the primes would
imply that the ratio of a power of one term and the other term is a prime power, which is
not possible based on the form of the linear complex factors. In addition to the cancellation
between pairs of fractions, the rescaling for each subsequent ratio yields an additional prime
divisor, except possibly for the last ratio, by analogy with Eq.(2.21). This follows if the
products of the quadratic expressions are rescaled to the cyclotomic polynomials, each with
a different index, since Φn (qi ) introduces a new primitive divisor for each n and odd prime
qi . If the rescaled quantities equal the same cyclotomic polynomial, the coefficients would
be different for different products, implying that cancellation requires distinct prime factors
2α +1
to be introduced elsewhere in decomposition of qi2αi +1 − 1 or qj j − 1. Consequently,
the product (2.23) introduces at least three distinct primes. Continuing this factorization
for products of more than three prime powers, it follows that a product of ` ratios yields a
minimum of ` distinct prime divisors. From this property, it follows that the two repunits
generally could not have identical sets of prime factors.
An alternative proof of this result can be given as follows. First, let 2αi + 1 = 2αj + 1
P
P
and tk1 = tk2 . However, k tk = ` t0` , which implies that some of the elements of {tkm }
0
are less than t` , while other tkn are greater than t` . The inequalities (qj −1)t` < (qi −1)tk <
Qt
0
sm
(qj + 1)t` cannot be satisfied for all k, ` implying that the products skm =1 (qi − ω2α
)
i +1
0
Qt0`
sn
and s0n =1 (qj − ω2α
) are not equal for some k, `. The rescaling then cannot be done
j +1
without introducing distinct prime divisors in one of the repunits.
If αi 6= αj , the exponents t0`1 and t0`2 can be set equal, because the products of linear
tkj
t`0
factors in Eq.(2.11) cam
i to magnitude. Then (qj −1) < (qj −1) <
h be paired according
ln(q −1)
0
ln(q +1)
(qj + 1)t` and tk1 ∈ t0` ln(qji −1) , t0` ln(qji −1) . For some tk2 ,
tk 1
t0`
irreducible prime fraction, and |tk1 − tk2 | ≥ 1. However, tk2
³
´
t0` ln
qj +1
qj −1
< 1, and it is then
ln(qi −1)
0
Qt
s0n
2
with s`0n =1 (qj − ω2α
).
j +1
not possible to equate the product
2
8
tk2
t0`
, since
2αi +1
2αj +1
is an
i
0 ln(qj −1) 0 ln(qj +1)
6
∈
t` ln(qi −1) , t` ln(qi −1) if
1
6=
h
2
Qtk2
sm2 −1 (qi
s
m2
− ω2α
)
i +1
Furthermore, the exponents t0l1 and t0l2 can be set equal to 2, provided that products are
multiplied by the power of a prime ish included to match the intervals.i Then, P h (qj − 1)2 <
(qi − 1)tk1 < P h (qj + 1)2 and tk1 ∈
k2 ,
tk1
t0l
1
h
6=
tk 2
t0l
2
2ln(qj −1)+h lnP
ln(qi −1)
if αi 6= αj , and |tk1 − tk2 | ≥ 1. The
2ln(qj +1)+h lnP
ln(qi −1)
inequalities P h (qj
,
− 1)2 < (qi − 1)tk2 <
2πikl
2πikl
2
. Again, for some k1 ,
P (qj +1) arise because each of the terms yl = (qj −e 2αj +1 )(qj −e 2αj +1 ), which is matched
Qt
sm
with a corresponding product of linear factors skm =1 (qi − ω2α
), has the same upper
i +1
³
´
+1
h
i 2ln qqjj −1
2ln(qj −1)+h lnP 2ln(qj +1)+h lnP
and lower bound. However, tk2 6∈
,
if ln(qi −1) < 1 which
ln(qi −1)
ln(qi −1)
implies qi > 5.
2ln(qj +1)+h lnP
only
ln(qi −1)
Then t0l = h0 tk , and if
The equality is also valid because tk2 =
0
if P h =
t
(qi −1) k2
(qj +1)2
which
implies that qi − 1 = (qj + 1)h , h0 ≥ 1.
h0 > 1, summation over
k, l gives 2αj + 1 = h0 (2αi + 1) which is not possible as 2αi + 1 and 2αj + 1 are prime. If
h0 = 1, tk = t0l and 2αi + 1 must be set equal to 2αj + 1. Setting qj = 3 and qi = 5, it
2αi +1
2αi +1
can be proven by induction that one of the repunits 3 2 −1 and 5 4 −1 has a distinct
3
5
3
5
−1
prime divisor. First, 3 2−1 = 13, 3 2−1 = 11 · 11, 55−1
= 31 and 5 4−1 = 11 · 71. Secondly,
when 2αi + 1, there are only primitive divisors. A distinct prime divisor arises in one of
the repunits
2α +1
j
−1
−1 qj
,
qi −1
qj −1
2αi +1
qi
for qi ≥ 5 and therefore for all odd primes qi , qj , qi > qj .
n
m
−1
−1
For the known solutions to the equation xx−1
= yy−1
, one of the prime bases is 2.
Since qj − 1 = 1, the theorem is circumvented because the bounds (2.12) are satisfied for a
larger set of primes qi and exponents tk . Furthermore, the integers sets {1, 2, ..., 2αi } and
{1, 2, ..., 2αj } are used entirely in the products to give the same integer, which must then
Q t0
0
s0n
) < 3t`
be a prime. The existence of solutions to the inequalities 1 < s`n =1 (qj − ω2α
j +1
implies that these bounds do not exclude the equality of
2αj + 1, qi , 2αi + 1.
22αj +1 −1
qj −1
and
2αi +1
−1
qi
qi −1
for some
If the primes which divide only qi2αi +1 − 1, are factors of qi − 1 or qj − 1, they would be
partially cancelled in a comparison of
2αi +1
qi
−1
qi −1
2αj +1
and
qj
−1
qj −1 .
Any divisor of qi − 1 which
2α +1
qi i −1
is a factor of qi −1 also must divide 2αi + 1. When 2αi + 1 and 2αj + 1 are prime, this
divisor would have to be 2αi + 1 or 2αj + 1.
A potential counterexample to the result concerning the occurrence of a distinct prime
9
factor in one of the repunits arises if equations of the form
2αj +1
qj
−1
qi2αi +1 − 1
= (2αi + 1)ν
qi − 1
qj − 1
(2αj + 1)
2α +1
2αi +1
ν qi
(2αj + 1)ν1
qj j − 1
−1
=
qi − 1
qj − 1
(2.24)
qi2αi +1 − 1
q 2αi +1 − 1
= (2αi + 1)ν2 i
qi − 1
qi − 1
¯ 2αj +1
¯q
−1
are satisfied, as the set of prime divisors of the repunits would be identical if 2αi +1¯¯ j qj −1
¯ 2α +1
¯ 2αj +1
¯ q i −1
¯q
−1
in the first relation, 2αj + 1¯¯ i qi −1 in the second condition and 2αi + 1¯¯ j qj −1 , 2αj +
¯ 2α +1
¯ q i −1
1¯¯ i qi −1 in the third relation. The conditions in Eq. (2.24) follow if different powers of
the other primes do not arise. However, qip − 1 ≡ 0 (mod pν ) only if qip − 1 ≡ 0 (mod p),
q p −1
and since qip − q ≡ 0 (mod p), this is possible when p|qi − 1. If p|qi − 1, then qii −1 ≡ p and
qip −1
qi −1
6≡ 0 (mod pν ), ν ≥ 2. Setting ν = 1 gives the three conditions
2αj +1
qj
−1
qi2αi +1 − 1
= (2αi + 1)
qi − 1
qj − 1
2α +1
qj j − 1
qi2αi +1 − 1
(2αj + 1)
=
qi − 1
qj − 1
(2.25)
qi2αi +1 − 1
qi2αi +1 − 1
= (2αi + 1)
(2αj + 1)
qi − 1
qi − 1
q
2αj +1
−1
q
2αi +1
−1
with 2αi + 1 6 | j qj −1 and 2αj + 1 6 | i qi −1 . It follows that either 2αi + 1 or 2αj + 1 is
a prime which divides only one of the repunits.
Furthermore, qi − 1 and qj − 1 are integers which are not rescaled, so that distinct
prime divisors arise in the products of the remaining linear factors. If the exponents
2αi + 1 and 2αj + 1 are equal, it would be necessary for nij to be equal to one, which is
q
2αi +1
−1
q
2αj +1
−1
not feasible since qi 6= qj . It follows also that i qi −1 and j qj −1 cannot be equal, so
that the original assumption of their inequality is valid The existence of a prime divisor
not common to both repunits can be deduced for all odd prime bases and exponents
(qi , 2αi + 1, qj , 2αj + 1), qi 6= qj .
3. Formulation of the Condition for Perfect Numbers in terms of the
Coefficients of Repunits
10
Let N = (4k + 1)4m+1
Q`
i=1
qi2αi [8] and the coefficients {ai } and {bi } be defined by
qi2αi +1 − 1
(4k + 1)4m+2 − 1
ai
= bi
qi − 1
4k
If
σ(N )
N
gcd(ai , bi ) = 1
(3.1)
6= 2,
p
"
2(4k + 1)
q12α1 +1
−1
q1 − 1
− 1 q`2α` +1 − 1 (4k + 1)4m+2 − 1
...
q2 − 1
q` − 1
4k
q22α2 +1
1
p
(b1 ...b` ) 2
= 2(4k + 1)
1 ·
(a1 ...a` ) 2
6= 2(4k + 1)2m+1
Ỳ
µ
(4k + 1)4m+2 − 1
4k
# 12
¶ (`+1)
2
(3.2)
qiαi
i=1
or
¶`
Ỳ q 2αi +1 − 1 µ
b1 ...b`
4k
i
A
=
·
a1 ...a`
qi − 1
(4k + 1)4m+2 − 1
i=1
·
¸`+1 `−1
2α` +1
Y
−1
4k
2αi q`
4m+1
6= 2(4k + 1)
q
·
i
4m+2
(4k + 1)
−1
q` − 1
i=1
(3.3)
When ` > 5 is odd, there exists an odd integer `o and an even integer `e such that
` = 3`0 + 2`e , so that
µ
¶µ
¶ µ
¶µ
¶
b1 ...b`
b13 a2
b46 a5
b3`o −2,3`o a3`o −1
b3`o +1 b3`o +2
=
...
a1 ...a`
a13 b2
a46 b5
a3`o −2,3`o b3`o −1
a3`o +1 a3`o +2
(3.4)
¶
µ
b`−1 b`
·¤
...
a`−1 a`
It has been proven that
b3ī−2,3ī a3ī−1
a3ī−2,3ī b3ī−1
6= 2(4k + 1) · ¤ any choice of a3ī−2 , a3ī−1 , a3ī and
b3ī−2 , b3ī−1 , b3ī consistent with equation (1.1) [7]. Similarly,
b`
a`
b3ī−2,3ī a3ī−1
ρ̄
≡ 2(4k + 1) 3ī−2 · ¤
a3ī−2,3ī b3ī−1
χ̄3ī−2
b3j̄+1 b3j̄+2
ρ̂3j̄+2
= 2(4k + 1)
·¤
a3j̄+1 a3j̄+2
χ̂3j̄+2
6= 2(4k + 1) · ¤ so that
(3.5)
where the fractions are square-free and gcd(ρ̄3ī−2 , χ̄3ī−2 ) = 1, gcd(ρ̂3j̄+2 , χ̂3j̄+2 ) = 1. *
Then,
b1 ...b`
ρ̄1 ρ̄3`o −2 ρ̂`−2`e +2,2 ρ̂`2
= 2(4k + 1) ·
...
...
·¤
(3.6)
a1 ...a`
χ̄1 χ̄3`o −2 χ̂`−2`e +2,2 χ̂`2
* The notation has been changed from that of reference [7] with a different choice of
index for ρ̄, χ̄.
11
When ` = 2`o + 3`e > 4 for odd `0 and even `e ,
¶
µ
b1 ...b`
4k + 1)4m+2 − 1
ρ̂22 ρ̂2`o ,2 ρ̄`−3`e +1 ρ̄`−2
= 2(4k + 1)
·
...
...
·¤
a1 ...a`
4k
χ̂22 χ̂2`o ,2 χ̄`−3`e +1 χ̄`−2
If the products
¶ `e µ
`o µ
Y
ρ̂3`
ρ̄3ī−2 Y
¶
o +2j̄,2
ī=1
for odd ` and
χ̄3ī−2
(3.8)
χ̂3`o +2j̄,2
j̄=1
¶ `e −1 µ
`o µ
Y
ρ̂2ī,2 Y
ρ̄2`
¶
o +3j̄+1
ī=1
χ̂2ī,2
j=1
(3.7)
(3.9)
χ̄2`o +3j̄+1
for even ` are not the squares of rational numbers,
b1 ...b`
6 2(4k + 1) · ¤
=
` is odd
a1 ...a`
µ
¶
b1 ...b`
4k + 1)4m+2 − 1
6 2(4k + 1)
=
·¤
a1 ...a`
4k
(3.10)
` is even
which would imply the inequality (3.3) and the non-existence of an odd perfect number.
4. Examples of Integers satisfying the Rationality Condition
Q`
qi2αi which satisfy the condition of rationality of
# 21
2α` +1
2α1 +1
4m+2
− 1 (4k + 1)
q
− 1 q`
−1
2(4k + 1) 1
...
q1 − 1
q` − 1
4k
Integers N = (4k + 1)4m+2
"
i=1
(4.1)
are given in the following list:
375 32 292 792 832 1372 2832 3132
375 32 292 672 792 832 1372 2832
375 32 72 112 292 792 832 1372 1912 2832
375 32 52 72 112 132 292 472 792 3132
375 32 112 132 472 792 2112
375 32 792 832 1372 2112 2832 3132
375 32 52 72 112 792 832 1372 2112 2832 3132
375 32 112 132 292 472 672 792
375 72 132 292 472 792 1912 3132
375 32 72 112 672 792 832 1372 1912 2112 2832 3132
375 32 52 112 472 672 792 2112 3132
375 32 52 472 792 1912 2112 3132
12
(4.2)
None of these integers satisfy the condition
the first integer is
σ(N )
N
= 2. For example, the sum of divisors of
σ(375 32 292 792 832 1372 2832 3132 ) = 2·37·34 ·74 ·132 ·192 ·312 ·432 ·672 ·732 ·1812 ·3672 (4.3)
so that new prime factors 7, 13, 19, 31, 43, 67, 73, 181, 367 are introduced. The inclusion of
these prime factors in the integer N leads to yet additional primes, and then the lack of
closure of the set of prime factors renders it impossible for them to be paired to give even
powers in σ(N ) with the exception of 2(4k + 1).
It may be noted that the decompositions of repunits with prime bases of comparable
magnitude and exponent 6 include factors that are too large and cannot be matched easily
with factors of other repunits. It also can be established that repunits with prime bases
and other exponents do not have square-free factors that can be easily matched to provide
a closed sequence of such pairings. This can be ascertained from the partial list
55 − 1
U5 (6, 5) =
= 781
4
U6 (6, 5) = 3906 = 2 · 7 · 31 · 32
U5 (8, 7) = 2801
U5 (12, 11) = 16105
U5 (14, 13) = 30941
(4.4)
U5 (18, 17) = 88741
U7 (4, 3) = 1093
U7 (6, 5) = 19531
U9 (4, 3) = 9841
U3 (32, 31) = 993 = 3 · 331
The first prime of the form 4k + 1 which arises as a coefficient D in the equality
q3 − 1
= Dy 2
q−1
q prime
(4.5)
4m+2
is 3541. This prime is too large to be the basis for the factor (4k+1)4k −1 , since it
would give rise to many other unmatched prime factors in the product of repunits and
the rationality condition would not be satisfied. Amongst the coefficients D which are
composite, the least integer with a prime divisor of the form 4k + 1 is 183, obtained when
q = 13. However, the repunit with base 61 still gives rise to factors which cannot be
matched since
616 − 1
= 858672906 = 2 · 3 · 7 · 13 · 31 · 97 · 523
(4.6)
60
13
5. A Proof of the Odd Perfect Number Conjecture
Theorem 2. Minimization of the number of primes in the relations between the repunits
requires the introduction of new prime divisors in σ(N ). If the minimization condition is
imposed for each relation, the sequence of additional prime factors is infinite.
Proof. The exponents will be assumed to be prime, since repunits with composite exponents have at least three prime divisors [7]. By Theorem 1, if the bases qi , qj and the
exponents 2αi + 1, 2αj + 1 are odd primes, there exists a prime divisor which is not a
2αj +1
2αi +1
qi
qj
−1
qi −1
−1
common factor of the repunits
and qj −1 . Consequently, with the condition
(2.1) on the prime bases and exponents, there will be no odd prime pairs (qi , qj ) satisfying
the exponential Diophantine equation
2αi +1
qi
−1
qi −1
2αj +1
=
qj
−1
qj −1 .
Additionally, it has been established that there are at most two solutions for fixed x
y−x
x−1
and y if y ≥ 7 [4]. If a = y−1
δ , b = δ , c = δ , δ = gcd(x − 1, y − 1) and s is the least
integer such that xs ≡ 1 (mod by n1 ), where (x, y, m1 , n1 ) is a presumed solution of the
Diophantine equation, then m2 − m1 is a multiple of s if (x, y, m2 , n2 ) is another solution
with the same bases x, y [4]. As x < y, x can be identified with qj , y with with qi and
δ with gcd(qj − 1, qi − 1). Given that the first exponents are 2αi + 1 and
³³ 2αj´+ 1, the
´
q −1
j
constraint on a second exponent of qj then would be m2 − (2αj + 1) = ιϕ
qi2αi +1
δ
where ι ∈ Q, with ι ∈ Z when qj 6 |qi − 1. This constraint can be extended to m1 = n1 = 1,
since it follows from the equations (y − 1)xmi − (x − 1)y ni = y − x, axmi − by ni = c,
which hold also for these values of the exponents. Based on the trivial solution to these
equations for an arbitrary pair of odd primes qi , qj , the first non-trivial
³
´ solution for the
exponent of qj would be greater than or equal to than 1+ι(qi −1)ϕ
¡¡ qj −1 ¢¢
1+ι(qi −1)ϕ
qj
δ
qj −1
δ
. This exponent
−1
introduces new prime divisors which are factors of
and therefore does
qj −1
not minimize the number of prime factors in the product of repunits.
Imprimitive prime divisors can be introduced into equations generically relating the
two unequal repunits and minimizing the number of prime factors in the product
2αj +1
2α +1
Q`
−1
qi i −1 qj
2αi
4m+1
the least number of
.
Given
an
odd
integer
N
=
(4k
+
1)
i=1 qi
qi −1
qj −1
unmatched prime divisors in σ(N ) will be attained if there are pairs (qi , qj ) satisfying one
14
of the three relations
2α +1
qj j − 1
qi2αi +1 − 1
= (2αi + 1) ·
qi − 1
qj − 1
2α +1
qj j − 1
qi2αi +1 − 1
(2αj + 1) ·
=
qi − 1
qj − 1
(5.1)
2α +1
qj j − 1
qi2αi +1 − 1
(2αj + 1) ·
= (2αi + 1) ·
qi − 1
qj − 1
2αj +1
qj
−1
As qj −1 would not introduce any additional prime divisors if the first relation in
Eq.(5.1) is satisfied, the product of two pairs of repunits of this kind, with prime bases
(qi , qj ), (qk , qk0 ), yields a minimum of three distinct prime factors, if two of the repunits
are primes. In the second relation, (qi , 2αi + 1) and (qj , 2αj + 1) are interchanged, whereas
in the the third relation, an additional prime divisor is introduced when 2αi + 1 6= 2αj + 1.
Equality of σ(N ) and 2N is possible only when the prime divisors of the product of repunits
· 2αj +1 ¸ k1
1
q
−1
also arise in the decomposition of N. If the the repunits have roots j qj −1
and
· 2α 0 +1 ¸ k1
2
qk 0 k
−1
are prime, they will be bases for new repunits
q 0 −1
k
·
2αj +1
qj
·
·
−1
qj −1
2αj +1
qj
1
−1
qj −1
2α 0 +1
−1
qk 0 k
qk0 −1
·
¸ 2αnk1 +1
2α 0 +1
−1
=
¸ k1
1
−1
(qj − 1)
2αj +1
(qj
1
k1
1
1
− 1) k1 − (qj − 1) k1
¸ 2αnk2 +1
−1
qk 0 k
qk0 −1
2
−1
=
¸ k1
2
−1
(qk0 − 1)
2α +1
(qk0 k0
1
k2
1
1
− 1) k2 − (qk0 − 1) k2


" 2αj +1
# 2αnk1 +1
1
−1
 qj

·
− 1
qj − 1


" 2α +1
# 2αnk2 +1
0
2
−1
 q 0k

· k
− 1
qk 0 − 1
(5.2)
where k1 < 2αj + 1 and k2 < 2αk0 + 1. To minimize the number of prime factors in
the product of repunits, these new quotients must be either powers of additional prime
divisors or satisfy relations of the form
·
2αj +1
qj
·
−1
qj −1
2αj +1
qj
¸ 2αnk1 +1
−1
qj −1
1
−1
= (2αn1
¸ k1
1
2α
−1
15
+1
q t1 t 1 − 1
+ 1)
q t1 − 1
(5.3)
Furthermore, 2αn1 + 1 is the factor of lesser magnitude that can be absorbed into a power
with exponent 2αn1 + 1,
qt2αt +1 − 1
2α
> qj n1
(5.4)
qt − 1
q
2αt +1
−1
The number of prime divisors of N is again minimized if the repunit t qt −1 is the power
of a prime. However, it is then the basis of another repunit which satisfies the inequality
"
h
2α +1
qt t −1
#2αn3 +1
i k10
h
2αt +1
qt
−1
qt −1
³
−1
1
qt −1
i k10
1
2αn1
> qj
2αn
3
k0
1
´
(5.5)
−1
h
Continuing this procedure yields a repunit larger qj
2αt +1 2αt +1
2α
+1
1
2
... tkm
0
k0
k0
m
1
2
i
2αn2m+1
. There-
2αt` +1
k`0
fore, an infinite sequence of repunits is generated, and when
has a lower bound
greater than 1, this product will diverge as m → ∞. There do not exist of odd integers N
with prime factors satisfying either of the first two relations in Eq. (5.1).
If the third relation in Eq.(5.1) holds,
·
(2αt + 1)
2αj +1
qj
·
−1
qj −1
2αj +1
qj
¸ 2αnk1 +1
−1
qj −1
1
−1
= (2αn1
¸ k1
1
2α
−1
+1
q t1 − 1
+ 1) t1
q t1 − 1
(5.6)
an inequality similar to Eq.(5.4) is satisfied, and again, repunits of increasing magnitude
are introduced in the sum of divisors.
2αt +1
`
−1
1
qt
If the repunits qt −1 are not powers of a single prime, then the product of the four
`
repunits with bases qi , qj , qk , qk0 would have a minimum of five different prime divisors.
2α +1
4m+2
Q`
q i −1
The product of (4k+1)4k −1 , with at least two distinct prime divisors, and i=1 i qi −1
then possesses a minimum of `+3 different prime factors and the perfect number condition
cannot be satisfied.
Even when none of the three relations hold in Eq.(5.1) hold, by Theorem 1, there exists
a primitive divisor which is not a common divisor of both repunits. It can be assumed
q
2α` +1
−1
that the last repunit ` q` −1 has a prime exponent since it will introduce at least three
prime divisors if the exponent is composite. The new prime divisor may be denoted qj` if
it does not equal 4k + 1, and interchanging q` with qī and 4k + 1, it can be deduced that
Q`
U4m+2 (4k+2, 4k+1) i=1 U2αi +1 (qi +1, qi ) will not be divisible by jī ∈ 1, ..., `, jī 6= ī, with
i6=ī
16
the exception of one value i0 . The prime qj0 will not be a factor of
Then
2α +1
qī ī − 1
hj
= qjī bari
ī 6= i0
qī − 1
2α
Q`
i=1
U2αi +1 (qi + 1, qi ).
+1
qi0 i0 − 1
= (4k + 1)4m+1
q i0 − 1
hj
(4k + 1)4m+2 − 1
= 2qji i0
0
4k
(5.7)
The last equation is equivalent to
(4k + 1)2m+1 − 1
= y12
4k
(4k + 1)2m+1 + 1
= y22
2
(5.8)
hj
i0
2
ji0
y1 y2 = q
which has no integer solutions for k and m. There are no prime sets {4k + 1; q1 , ..., q` }
which satisfy these conditions with hjī and hji0 even. Thus, there are no odd integers
Q`
N = (4k + 1)4m+1 i=1 qi2αi such that the primes and exponents {qi , 2αi + 1} satisfy
none of the relations in Eq.(5.1) and equality between σ(N ) and 2N . For all odd integers,
σ(N )
N 6= 2.
6. On the Non-Existence of Coefficients of Repunits satisfying the
Perfect Number Condition
Theorem 3. There does not exist any set of primes {4k + 1; q1 , ..., q` } such that the
coefficients {ai } and {bi } satisfy the equation
·
¸`+1 Ỳ
b1 ...b`
4k
qi2αi
= 2(4k + 1)
a1 ...a`
(4k + 1)4m+2 − 1
i=1
There are no odd perfect numbers
Proof. It has been observed that ab11 6= 2(4k + 1) · ¤ by the non-existence of multiply
perfect numbers with less than four prime factors [5][6],
·
¸3
b1 b2
4k
6= 2(4k + 1)
q12α1 q22α2
a1 a2
(4k + 1)4m+2 − 1
17
(6.1)
by the non-existence of perfect numbers with three prime divisors [10], and proven that
b1 b2 b3
a1 a2 a3 6= 2(4k + 1) · ¤ so that the inequality is valid for ` = 1, 2, 3.
Suppose that there are no odd integers N of the form (4k + 1)4m+1
σ(N )
N = 2 so that
·
¸` `−1
Y
b1 ...b`−1
4k
6= 2(4k + 1)
q 2αi
a1 ...a`−1
(4k + 1)4m+2 − 1 i=1 i
Q`−1
i=1
qi2αi with
(6.2)
...b`
If there exists a perfect number with prime factors {4k +1, q1 , ..., q` }, then ab11 ...a
must have
`
´
³
Q
(4k+1)4m+2 −1
− τ (U4m+2 (4k + 2, 4k + 1), i=1 qi ) distinct prime factors, where
1+`+τ
4k
4m+2
Q
U4m+2 (4k+2, 4k+1) is the Lucas number (4k+1)4k −1 and τ (U4m+2 (4k+2, 4k+1), i=1 qi )
denotes the number of common divisors of the two integers. However, equality σ(N )
³
´
4m+2
Q` q2αi +1 −1
and 2N also implies that i=1 i qi −1 has at least ` + 2 − τ (4k+1)4k −1 different
prime divisors and a maximum of ` + 1 prime factors. If there is no cancellation between
Q`
Q`
Q` qi2αi +1 −1
2αi
4k
and
by
q
,
multiplication
of
U
(q
+
1,
q
)
=
i
i=1 i
i=1 2αi +1 i
i=1
(4k+1)4m+2 −1
qi −1
³
´`
³
´
U4m+2 (4k + 2, 4k + 1) = (4k+1)4k
introduces τ (4k+1)4k
new prime divisors
4m+2 −1
4m+2 −1
...b`
and ab11 ...a
would have ` + 2 distinct prime factors. However, if gcd(U4m+2 (4k + 2, 4k +
`
1), U2αi +1 (qi + 1, qi )) = 1 for all i, the repunit U4m+2 (4k + 2, 4k + 1) must introduce
additional prime divisors. It follows that
(4k + 1)4m+2 − 1
2α
= qi0 i0
(6.3)
8k
must be valid for some index i0 . There are no integer solutions to
xn − 1
= 2y 2
x ≡ 1(mod 4), n ≡ 2(mod 4)
(6.4)
x−1
and there is no odd integer satisfying these conditions. A variant of this proof has been
·
³
´¸ 12
2α +1
Q`
qi i −1
8k(4k+1)
obtained by demonstrating the irrationality of
· (4k+1)4m+2 −1
when
i=1
qi −1
gcd(U2α+1 (qi + 1, qi ), U4m+2 (4k + 2, 4k + 1)) = 1 [6].
2α +1
Q`
q i −1
When the number of prime factors of i=1 i qi −1 is less than ` + 1, there must be
at least two divisors of U4m+2 (4k + 2, 4k + 1) which do not arise in the decomposition of
this product. While one of the factors is 2, 4k + 1 is not a divisor, implying that any
other divisor must be qj̄ for some j̄. It has been assumed that there are no prime sets
`−1
{4k + 1; q1 , ..., q`−1 } satisfying the perfect number condition. Either ab11 ...b
...a`−1 does not have
Q`−1 q2αi +1 −1
` + 1 factors, or if it does have ` + 1 factors, then i=1 i qi −1 has ` − 1 factors but
`−1
Y
i=1
Y
qi2αi +1 − 1
6= (4k + 1)4m+1
qi2αi −ti
qi − 1
i6=ī,j̄
18
(6.5)
for some prime qī , where qiti k U4m+2 (4k + 2, 4k + 1). Multiplication by U2α` +1 (q` + 1, q` )
4m+2
2α
must contain the factor qī ī , because (4k+1)4k −1 only would introduce the two primes
Q`−1
2, qj̄ and not qī , since i=1 U2αi +1 (qi + 1, qi ) contains ` − 1 primes, including 4k + 1 and
excluding qī . Interchanging the roles of the primes in the set {qi , i = 1, ..., `}, it follows
that the prime equations
qi2αi +1 − 1
2αj
= (4k + 1)hi qji i
qi − 1
2α` +1
q`
−1
2α
= (4k + 1)h` qī ī
q` − 1
4m+2
(4k + 1)
−1
2α
= 2qj̄ j̄
4k
(6.6)
must hold, where h` 6= 0. Since the third equation has no integer solution, k ≥ 1 [7], it
`−1
follows that ab11 ...b
...a`−1 must not have ` + 1 prime factors.
There cannot be less than ` + 1 different factors of
2α +1
qi i −1
b1 ...b`−1
a1 ...a`−1 ,
as each new repunit
introduces at least one distinct prime divisor by Theorem 1 and the factorization of
qi −1
4m+2
(4k+1)
−1
`−1
contains at least two new primes. Consequently, this would imply ab11 ...b
4k
...a`−1 has
`
at least `+2 prime factors and ab11 ...b
...a` has a minimum of `+3 prime divisors, which is larger
Q`
than the number necessary for equality between U4m+2 (4k+2, 4k+1) i=1 U2αi +1 (qi +1, qi )
Q`
and 2(4k + 1) i=1 qi2αi .
Q`
If the number of prime factors of i=1 U2αi +1 (qi + 1, qi ) does equal ` + 1, then the
only additional prime divisor arising from multiplication with U4m+2 (4k + 2, 4k + 1) is
2. However, since both 2 and 2k + 1 divide the repunit ³
with base 4k +
´ 1, this property
(4k+1)4m+2 −1
1
does not hold unless the prime factors of 2k + 1 and 2k+1
can be included
4k
in the set {qi , i = 1..., `}. There are then a total of ` + 2 prime factors in
h
i`+1
Q`
8k
can be absorbed into i=1 qi2αi .
4m+2
(4k+1)
−1
b1 ...b`
a1 ...a`
only if
The repunit then can be expressed as
Y t
(4k + 1)4m+2 − 1
=2
qj j
4k
(6.7)
j∈{K}
where {K} ⊆ {1, ..., q` }. By the non-existence of odd perfect numbers with prime factors
Q`−1
4k + 1, qi , i = 1..., ` − 1, either U4m+2 (4k + 2, 4k + 1) i=1 U2αi +1 (qi + 1, qi ) has less than
` + 1 factors or
`−1
Y
i=1
qi2αi +1 − 1
6= (4k + 1)4m+1
qi − 1
Y
i∈{K}−{qī }
19
qi2αi
Y
{K}−{qī }
2αj −tj
qj
(6.8)
for some ī ∈ {1, ..., `}. If the product of repunits for the prime basis {4k + 1, q1 , ..., q`−1 }
q
2α` +1
−1
has less than ` + 1 factors, ` q` −1 introduces at least two new prime factors. Even if one
of these divisors is 4k + 1, the other factor must be qī for some ī 6= `. Interchange of the
primes qi , i = 1, ..., ` again yields the relations in Eq.(6.6).
2α
If the product (6.8) includes ` primes, and not qī , then qī ī must be a factor of
Q
Interchanging the primes, it follows that the product (6.8) includes i6=ī qi2αi .
2α` +1
q`
−1
q` −1 .
The prime divisors in U2α` +1 (q` +1, q` ) either can be labelled qj` for some j` ∈ {1, ..., `−
1} or equals 4k + 1. while 4k + 1 does not divide U4m+2 (4k + 2, 4k + 1), it can occur in the
other repunits
4m+2
(4k+1)
4k
−1
2αi +1
qi
−1
qi −1
for i = 1, ..., ` − 1. The prime qj` = qī also may not be a divisor of
.
One choice for q` is a prime divisor of 2k + 1. If 2k + 1 is prime, the factors of
4m+2
and (4k+1)4k −1 can be compared. Since the latter quotient is equal to
(2k+1)2α` +1 −1
2k
(4k+1)2m+1 −1
4k
· [(4k + 1)2m+1 + 1], consider setting m equal to α` . By Theorem 1, the
primitive divisor P divides
both repunits if 4k = nP + 2k or 2k = nP implying n = 2,
¯
¯ (2k+1)2α` +1 −1
P = k. However, if k ¯¯
, 1 + (2k + 1) + ... + (2k + 1)2α` ≡ 2α` + 1 ≡ 0 (mod k)
2k
k
Q
−1
and k = 2α` + 1 for prime exponents 2α` + 1. Suppose (2k+1)
= k · i Pimi and
2k
Q
nj
(4k+1)k −1
=
k
·
j P̃j . If a primitive divisor rk + 1, r ∈ Z, divides both repunits,
4k
(2k + 1)k − 1 = (rk + 1)(x − 1), x ∈ Z, (4k + 1)k − 1 = (rk + 1)(x0 − 1), x0 ∈ Z, so that
(2k + 1)k − 1
(4k + 1)k − 1
=
x−1
x0 − 1
(6.9)
x0 (2k + 1)k − (2k + 1)k − x0 = x(4k + 1)k − (4k + 1)k − x
(6.10)
This relation would imply
and therefore x − 1 = b(4k + 1)k , x0 − 1 = a(4k + 1)k . However, the equation is then
a(4k + 1)k (2k + 1)k − a(4k + 1)k = b(4k + 1)k (2k + 1)k − b(2k + 1)k
(6.11)
which cannot be satisfied by any integers a, b. Not all primitive divisors of the two repunits
2α` +1
are identical. For the exception, k = 1, 2α` + 1 = 5, 4m + 2 = 10n, (2k+1)2k −1 = 112 ,
(4k+1)4m+2 −1
4k
has other prime factors in addition to 11.
If m 6= α` , it can be demonstrated that there is a primitive divisor of
4m+2
(2k+1)2α` +1 −1
2k
which is not a factor of (4k+1)4k −1 by using the comparison of the linear factors in the
decomposition of each numerator in Theorem 1, since 4k 6= (2k)n , n ≥ 2 and 4k + 2 6=
(2k + 2)n , n ≥ 2 for any k > 1.
20
In general, either there exists a prime factor of
2α` +1
−1
q`
4m+2
which does not divide (4k+1)4k −1 or there are more than ` + 2 primes in the
decomposition of the product of repunits. The existence of a distinct prime divisor requires
a separate demonstration for composite exponents. Let p1 (4m + 2), p2 (4m + 2) be two
q` −1
2α` +1
−1
has a prime factor different from the divisors of
q` −1
p2 (4m+2)
p1 (4m+2)
(4k+1)
−1
−1
and (4k+1) 4k
by Theorem 1. While the union of the sets of prime
4k
p1 (4m+2)p2 (4m+2)
−1
divisors of the two repunits is contained in the factorization of (4k+1)
,
4k
prime divisors of 4m + 2. Then
q`
the repunit with the composite exponent will have a primitive divisor, which does not
q
2α` +1
−1
belong to the union of the two sets and equals a1 p1 p2 + 1. If this prime divides ` q` −1 ,
a1 p1 p2 + 1 = a2 (2α` + 1) + 1. Either 2α` + 1|4m + 2 or the primitive divisor equals
a12 (2α` + 1)p1 p2 + 1. Suppose
(a12 (2α` + 1) + 1)(x + 1) =
(4k + 1)p1 p2 − 1
≡ p1 p2 (mod 4)
4k
(6.12)
Since p1 , p2 are odd primes, either a12 ≡ 0 (mod 4) or a12 ≡ 2 (mod 4), so that the primitive
divisor equals 4c(2α` + 1)p1 p2 + 1 when p1 p2 ≡ 1 (mod 4) or (4c + 2)(2α` + 1)p1 p2 + 1 if
p1 p2 ≡ 3 (mod 4). Let the primitive divisor P have the form 4c(2α` + 1)p1 p2 + 1 so that
(4c(2α` + 1)p1 p2 + 1)(y + 1) =
q`2α` +1 − 1
≡ 2α` + 1 (mod q` − 1)
q` − 1
(6.13)
which implies that y + 1 = κ2 [2α` + χ2 (q` − 1)] is either the multiple of an imprimitive
divisor, κ02 (2α` + 1) or it is the multiple of a primitive divisor. Consider the equation
q`2α` +1 − 1
(4c(2α` + 1)p1 p2 + 1)κ2 (2α` + 1) =
q` − 1
(6.14)
(4c(2α` + 1)p1 p2 + 1)κ2 (2α` + 1) ≡ 2α` + 1 (mod q` − 1)
(6.15)
with 2α` |q` − 1. If
q` − 1|(4c(2α` + 1)p1 p2 + 1)κ2 − 1 or (4c(2α` + 1)p1 p2 + 1)κ2 − 1 ≡
Since 2α` + 1|κ2 − 1, κ2 = κ3 (2α` + 1) + 1. Let
q` −1
2α` −1
(mod q` − 1).
q`2α` +1 − 1
(z(q` − 1) + 1)(2α` + 1) =
q` − 1
(6.16)
Then
z = 2α` +
q` − 1
(2α` − 1 + (2α` − 2)(q` + 1) + ... + q`2α` −2 + ... + 1)
2α` + 1
21
(6.17)
is integer and z(q` −1) = 4c(2α` +1)2 κ3 p1 p2 +κ3 (2α` +1)+4c(2α` +1)p1 p2 and 2α` +1|κ2 −1
so that κ2 = κ3 (2α` + 1) + 1. Let
q`2α` +1 − 1
(z(q` − 1) + 1)(2α` + 1) =
q` − 1
(6.18)
Then
z = 2α` +
q` − 1
(2α` − 1 + (2α` − 2)(q` + 1) + ... + q`2α` −2 + ... + 1)
2α` + 1
(6.19)
is integer and z(q` − 1) = 4c(2α` + 1)2 κ3 p1 p2 + κ3 (2α` + 1) + 4c(2α` + 1)p1 p2
In the latter case, the product of the primitive divisors takes the form b(2α` + 1) + 1 so
that κ2 [(2α` + 1) + χ2 (q` − 1)] = b(2α` + 1) + 1 since 2α` + 1 is prime. The two congruence
relations
[(b − 1) + (b(2α` + 1)4cp1 p2 ](2α` + 1) + 1 ≡ 0 (mod q` − 1)
(6.20)
(b − κ2 )(2α` + 1) + 1 ≡ 0 (mod q` − 1)
imply
[κ2 − 1 + (b(2α` + 1) + 1)4cp1 p2 ](2α` + 1) ≡ 0 (mod q` − 1)
(6.21)
When 2αi + 1 6 |q` − 1, it follows that q` − 1|κ2 − 1 + (b(2α` + 1) + 1)4cp1 p2 . Since
every primitive divisor is congruent to 1 modulo 2α` + 1, κ2 = κ3 (2α` + 1) + 1 and
q` − 1|κ3 (2α` + 1) + (b(2α` + 1) + 1)4cp1 p2 .
The factorizations
(4k + 1)4m+2 − 1 =
4m+1
Y
k
((4k + 1) − ω4m+2
)
k=0
q`2α` +1
−1=
2αi
Y
(6.22)
0
k
(q` − ω2α
)
i +1
k0 =0
0
yield real factors which can be identified only if ((4k + 1) − 1)tk = (4k)tk = (q` − 1)tk or
0
(4k + 2)tk = (q` + 1)tk . Then, 4k = (q` − 1)n` , tk = n` t0k or 4k + 2 = (q` + 1)n` , tk = n` t0k .
P
P
Since k tk = 4m + 2, k t0k = 2αi + 1, 4m + 2 = n` (2α` + 1). If 4m + 2 is the product
of two primes p1 p2 , 2α` + 1 must equal one of the primes, p2 .
The congruence (6.15) becomes
[(4c(2α` + 1)2 p1 + 1)κ2 − 1](2α` + 1) ≡ 0 (mod q` − 1)
(6.23)
¡
¢
p1
(4k+1)p1 p2 −1
+1)p1 p2 −1
= ((q` −1)
= p1 p2 + p12p2 (q` − 1) + ...
4k
(q` −1)p1
¡
¢
p1 p2 (q` − 1)p1 (p1 p2 −2) + (q` − 1)p1 (p1 p2 −1) . When 2αi + 1|q` − 1, 2αi + 1|p1 p2 + p12p2 (q`
If 4k = (q` − 1)p1 , then
22
+
−
1)+...+p1 p2 (q` −1)p1 (p1 p2 −2) +(q` −1)p1 (p1 p2 −1) . However,
(4k+1)p1 p2 −1
4k
2α` +1
q`
−1
q` −1
≡ 2α` +1 (mod q` −1),
whereas
≡ p1 (2α` + 1) (mod q` − 1), p1 > 1. The repunits have the same
prime divisors only if p1 ≡ 1 (mod q` − 1), which implies that p1 = ρ(q` − 1) + 1. Since
q` − 1 ≥ 2, p1 cannot equal 2, implying that 4m + 2 must be the product of at least three
(ρ(q −1)+1)(2α` +1)
`
((q` −1)ρ(q` −1)+1 +1)
(q` −1)ρ(q` −1)+1
p1 p2
(q` + 1)p1 , then (4k+1)4k −1
p1
p1
prime factors and
−1
includes at least seven new factors.
p1
p1 p2
−1
If 4k + 2 =
≡ (2 −1)
(mod q` − 1). If 2p1 − 1 ≡
2p1 −2
1 (mod q` − 1), 1 + (2 − 1) + ... + (2 − 1)p1 p2 −1 ≡ p1 p2 ≡ 2α` + 1 (mod q` − 1) only if
p1
p1 p2
−1
(2n+1)p1 p2 −1
≡
p1 ≡ 1 (mod q` − 1). When 2p1 − 1 ≡ 2n + 1 (mod q` − 1), (2 −1)
p
1
2 −2
2n
p1 p2
which is not divisible by 2αi + 1 if p2 6| 2n. When 2αi + 1|n, (2n) 2n −1 ≡ p1 p2 (mod 2n)
which would be congruent to 2αi + 1 only if p1 ≡ 1 (mod 2n) so that p1 ≥ 4αi + 3 and
therefore does not equal 2. The exponent must be a product of a minimimum of three
primes and introduce at least seven different divisors in the product of repunits. A similar
conclusion is obtained for all exponents 4m + 2 = p1 ..ps , s ≥ 3. Therefore, there are
4m+2
primes which divide only the repunit with the composite exponent (4k+1)4k −1 and not
2αi +1
−1
qi
qi −1
, and the number of such factors increases with the τ (4m + 2).
4m+2
If each prime divisor of (4k+1)4k
following equations are obtained
−1
is distinct from the factors of
Q`
2αi +1
qi
i=1
−1
qi −1 ,
qi2αi +1 − 1
2αj
= (4k + 1)hi qji i
qi − 1
4m+2
(4k + 1)
−1
=2
4k
X̀
hi = 4m + 1
the
(6.24)
i=1
which only has the solution k = 0.
2αi +1
0
−1
0
qi
4m+2
The number of equal prime divisors in qi −1 and (4k+1)4k −1 can be chosen to be
0
greater than 1, but equality of σ(N ) and 2N implies that each distinct prime divisor qji of
the repunit
2αi +1
qi
−1
qi −1
appears in the factorization with the exponent 2αji . Consequently,
2αi +1
0
it would be inconsistent with the perfect number condition for divisors of
than qji0 to exist.
If the prime divisor qji0 of
2αi +1
0
−1
0
qi
qi0 −1
is a factor of
23
(4k+1)4m+2 −1
,
4k
qi
0
−1
qi0 −1
other
then one formulation
of the perfect number condition for the integer N = (4k + 1)4m+1 qi2αi is
qi2αi +1 − 1
2αj
= (4k + 1)hi qji i
qi − 1
2α
i 6= i0
+1
h0j
qi0 i0 − 1
= qji i0
0
qi0 − 1
2αji −h0j
(4k + 1)4m+2 − 1
i0
= 2qi0 0
4k
(6.25)
(4k+1)4m+2 −1
has at least
4k
(4k+1)2m+1 −1
and its primitive
four different prime divisors, 2,2k + 1, the prime factors of
4k
(4k+1)4m+2 −1
divisors. Furthermore, as the number of prime divisors of
is less than seven,
4k
2αi +1
qi 0 −1
The last relation in Eq.(6.25) has no solutions with k ≥ 1 since
0
qi0 −1
also should have a different prime factor which is contrary to the Eq.(6.25).
Q`
qi2αi , the quotient
2α +1
Q`
qi i −1
(4k+1)2m+1 −1
,
k
≥
1
has
a
distinct
prime
divisor
from
the
factors
of
because
i=1
4k
qi −1
Since k ≥ 1 in the decomposition N = (4k + 1)4m+1
i=1
p1
−1
of its existence in the factorization of (4k+1)
by Theorem 1. There would be then at
4k
least ` + 3 prime factors of σ(N ) implying that N is not an odd perfect number.
This result clearly follows when the number of prime factors of the integer N exceeds the
)
current lower bound derived from the value of σ(N
[11].
N
24
References
[1] A. S. Bang, Taltheoretiske Undersogelser, Tidsskrift for Mathematik, 5 (1886), 265-284
[2] Yu. Bilu, G. Hanrot and P. M. Voutier, Existence of primitive divisors of
Lucas and Lehmer numbers, J. für die reine und angewandte
Mathematik , 539 (2001), 75-122
[3] G. D. Birkhoff and H. S. Vandiver, On the integral of an − bn , Ann. Math., 5 no. 2
(1904), 173-180
[4] Y. Bugeaud and T. N. Shorey, On the Diophantine Equation
Math., 207(1) (2002), 61-75
xm −1
x−1
=
y n −1
y−1 ’,
Pac. J.
[5] R. D. Carmichael, Multiply perfect odd numbers with three prime factors Amer. Math.
Monthly, 13 (1906), 35-36
[6] R. D. Carmichael, Multiply perfect numbers of four different primes, Ann. Math., 8
(1907), 149-158
[7] S. Davis, A Rationality Condition for the Existence of Odd Perfect Numbers,
International Journal of Mathematics and Mathematical Sciences, 2003 No. 20, 12611293
[8] L. Euler, Tractatus de Numerorum Doctrina, Commentationes Arithmeticae, 2 (1849)
503-575, §109 in: Opera Omnia I 5 Auctoritate et Impensis Societatis Scientarum
Naturalium Helveticae, Genevae, MCMXLIV
[9] R. Goormaghtigh, Nombres tel que A = 1 + x + x2 + ... + xm = 1 + y + y 2 + ... + y n ,
L’Intermediare des Mathematiciens, 24 (1917) 88
[10] P. Hagis, Jr., Outline of a Proof that Every Odd Perfect Number has at Least Eight
Prime Factors, Math. Comp., 40 (1983) 399-404
[11] K. G. Hare, ‘More on the Total Number of Prime Factors of an Odd Perfect Number’,
Math. Comp. 74 (2005) 1003-1008
[12] N. Koblitz, p-adic Numbers, p-Adic Analysis and Zeta Functions, (Springer-Verlag,
1977)
[13] Y. Nesterenko and T. N. Shorey, On the Dirophantine Equation
Arith., 83 (1998), 381-399
25
xm −1
x−1
=
y n −1
y−1 ,
Acta
[14] T. N. Shorey, On the Equation axm − by n = k, Indag. Math. 48 (1986) 353-358
[15] C. L. Stewart, On Divisors of Fermat, Fibonacci, Lucas and Lehmer Numbers, Proc.
London Math. Soc., Series 3, 35 (1977), 425-477
[16] K. Zsigmondy, Zur Theorie der Potenzreste, Monatsh. Math. 3 (1892) 265-284
26