An Extension of the Euclid-Euler Theorem to Certain α-Perfect Numbers

Paulo J. Almeida^{1} and Gabriel Cardoso^{2} University of Aveiro Department of Mathematics Campus Universitário de Santiago 3810-193 Aveiro Portugal

                    palmeida@ua.pt
                 gabriel.cardoso@ua.pt

Abstract

In a posthumously published work, Euler proved that all even perfect numbers are of the form 2^{p−1}(2^{p} −1), where 2^{p} −1 is a prime number. In this article, we extend Euler’s method for certain α-perfect numbers for which Euler’s result can be generalized. In particular, we use Euler’s method to prove that if N is a 3-perfect number divisible by 6; then either 2 ∥ N or 3 ∥ N. As well, we prove that if N is a \frac{5}{2}-perfect number divisible by 5, then 2^{4} ∥ N, 5^{2} ∥ N, and 31^{2} | N. Finally, for p ∈ {17, 257, 65537}, we prove that there are no \frac{2p}{p−1}-perfect numbers divisible by p.

1 Introduction

A perfect number is equal to the sum of its proper divisors. These numbers have been studied since the ancient Greek times. An old known result is Proposition 36 of Book IX in the Elements, where Euclid showed that if 2^{p} − 1 is a prime number then 2^{p−1}(2^{p} − 1) is perfect. Around two thousand years later, L. Euler [2, p. 630] proved that every even perfect number is of Euclid’s form. In the same page, there is Euler’s formula for an odd perfect number. If σ(N) = 2N and N is odd, then

1The first author received support from the Center for Research and Development in Mathematics and Applications (CIDMA) through the Portuguese Foundation for Science and Technology (FCT—Fundação para a Ciência e Tecnologia), reference UIDB/04106/2020.

2The second author received support from the Portuguese Foundation for Science and Technology (FCT— Fundação para a Ciência e Tecnologia, reference PD/BD/150533/2019).

N = p^{e}k^{2},  (1)

where p is a prime number, (p, k) = 1, and p ≡ e ≡ 1 (mod 4).

Let σ to be the sum of divisors function and

I(N) = \frac{σ(N)}{N}

to be the index function of N. Clearly, I(N) is multiplicative. For p a prime number and a a non-negative integer, the index function I(p^{a}) is monotonically increasing in a but monotonically decreasing in p. Also, for any prime number p and any positive integer a, we have

\frac{p + 1}{p} ≤ I(p^{a}) < \frac{p}{p − 1}.  (2)

Moreover, if I(N) = \frac{c}{d} and gcd(c, d) = 1, we have

if p is a prime number and p | d, then p | N;  (3)

and

if p is a prime number and p | c, then p | σ(N).  (4)

For any rational number α, a positive integer N is an α-perfect number if I(N) = α. We say α is an abundancy index if α ∈ Im(I). The case α = 2 corresponds to a perfect number and, when α is an integer, N is a multiperfect number. More than 5700 multiperfect numbers have been found (see Achim Flammenkamp’s The Multiply Perfect Numbers Page [3]). Many other rational values for α were studied, some of them with connections with odd perfect numbers [7, 11]. When a rational α is not in the image of I(N), α is called an abundancy outlaw. Holdener and Stanton [5] gave a list of abundancy outlaws as well as rational numbers for which their abundancy status is unknown. Given a prime p, we say that a rational α is a p-abundancy outlaw if there are no α-perfect numbers divisible by p.

We will present a generalization of Euler’s result on perfect numbers that can be applied to certain α-perfect numbers. In particular, it can be applied to the 3-perfect numbers and \frac{5}{2}-perfect numbers. For p ∈ \{17, 257, 65537\}, we also prove that \frac{2p}{p−1} is a p-abundancy outlaw.

The only known 3-perfect numbers are the following

2^{3} · 3 · 5, 2^{5} · 3 · 7, 2^{9} · 3 · 11 · 31, 2^{13} · 3 · 11 · 43 · 127,
2^{8} · 5 · 7 · 19 · 37 · 73, and 2^{14} · 5 · 7 · 19 · 31 · 151.

Dickson [1, p. 33] gives a historical account of the 3-perfect numbers. It is conjectured the numbers above are the only 3-perfect numbers. Clearly, N is a 3-perfect number with 2 ∥ N if and only if \frac{N}{2} is an odd perfect number. Therefore, if the above list of 3-perfect numbers is complete, then there are no odd perfect numbers. Also, it is easy to see that \frac{5}{2} is not an abundancy outlaw, since I(24) = \frac{5}{2}.

In order to apply our generalization of the Euclid-Euler theorem we will need to find all solutions of certain two and three-variable exponential diophantine equations. We will use the general method developed by Styer [10] which formalizes and extends a method used by Guy, Lacampagne, and Selfridge [4] (also see the paper [8] for similar results).

2 Generalizing Euler’s method

Let α be a rational number and N > 1 be an α-perfect number, then I(N) = α. Using (2) and the multiplicativity of I(N), there exist positive integers r and m, prime numbers p_{i}, and positive integers a_{i}, with 1 ≤ i ≤ r; such that

N = m ∏_{i=1}^{r} p_{i}^{a_{i}},  (5)

α ∏_{i=1}^{r} \frac{p_{i} − 1}{p_{i}} ≤ 1,  (6)

and

gcd (m, ∏_{i=1}^{r} p_{i}^{a_{i}}) = 1.

Therefore,

αm ∏_{i=1}^{r} p_{i}^{a_{i}} = σ(N) = σ(m) ∏_{i=1}^{r} \frac{p_{i}^{a_{i}+1} − 1}{p_{i} − 1}.

Hence, for some integers k and d, we have

σ(m) = \frac{βk}{d} ∏_{i=1}^{r} p_{i}^{a_{i}+1}  (7)

and

m = \frac{k}{d} ∏_{i=1}^{r} (p_{i}^{a_{i}+1} − 1) ,  (8)

where

β = α ∏_{i=1}^{r} \frac{p_{i} − 1}{p_{i}} ≤ 1

and

d = gcd (β ∏_{i=1}^{r} p_{i}^{a_{i}+1}, ∏_{i=1}^{r} (p_{i}^{a_{i}+1} − 1)) .

We will try to find a lower bound for σ(m), by summing divisors of m that are explicitly indicated in Eq. (8). A comparison of the lower bound with Eq. (7), will give us contradictions or conditions on the form of N. From now on, we will always consider β = 1.

3 The Euclid-Euler theorem: r = 1 and α = \frac{p}{p−1}

Suppose N = p^{a}m is an α-perfect number, where p is a prime number and α = \frac{p}{p−1} satisfies Condition (6) with equality. Then I(N) = α.

By (3), if p ≠ 2 then 2 | N. Therefore,

I(N) ≥ \frac{3}{2} \frac{p + 1}{p} > \frac{p}{p − 1} = α.

Hence, p = 2 and we have the case studied by Euler [2]. The Euler theorem is a well known result, but we write its proof here to give context for our results.

Theorem 1 (Euler). If N is an even perfect number, then N = 2^{p−1}(2^{p} − 1), where p and 2^{p} − 1 are prime numbers.

Proof. Suppose N = 2^{a}m and σ(N) = 2N. Then

2^{a+1}m = σ(N) = (2^{a+1} − 1)σ(m).

Since gcd(2^{a+1} −1, 2^{a+1}) = 1, then there exists a positive integer k such that m = k(2^{a+1} −1) and σ(m) = k2^{a+1}. Since k and m divide m, and σ(m) = k + m, then 2^{a+1} − 1 is a prime number and k = 1. Now, 2^{a+1} − 1 can only be a prime number if a + 1 is a prime number p. Therefore, N = 2^{p−1}(2^{p} − 1).

4 3-perfect numbers divisible by 6: r = 2 and α = 3

In this section, we will consider r = 2, p_{1} = 2, and p_{2} = 3 in Eq. (5). We will prove that if a 3-perfect number N is divisible by 6, then either 2 ∥ N or 3 ∥ N. Although Steuerwald [9] prove this result in 1954, we will use the method introduced in Section 2. Before we proceed, we need results about the difference between powers of 2 and 3. These results are special cases of Pillai’s conjecture and Mihǎilescu’s theorem [6], which proves the Catalan conjecture. It is worth noting that the solutions of equation |2^{a} − 3^{b}| = 1 were first determined by Levi ben Gershon, also known as Gersonides (1288–1344), in his treatise De Numeris Harmonicis.

Lemma 2. The only solutions of the diophantine equation

2^{a} − 3^{b} = −1  (9)

are (1, 1) and (3, 2). Also, the only solutions of the diophantine equation

2^{a} − 3^{b} = 2^{c} − 1,  (10)

are (2, 1, 1), (4, 2, 3), and (a, 0, a), for all a ∈ ℤ.

Proof. We start by solving the equation |2^{a} − 3^{b}| = 1. Since 2 is a primitive root of 3^{b}, for any b ≥ 1; then the solutions of 2^{x} ≡ ±1 (mod 3^{b}) are x = 3^{b−1}k, for k ≥ 0. But, for b ≥ 3 we have

2^{3^{b−1}} > 2 · 3^{b}.

Therefore, b ∈ {0, 1, 2} and so we obtain the solutions (a, 0, a) and (2, 1, 1) of Eq. (10). Also, we obtain the solutions (1, 1) and (3, 2) of Eq. (9).

Now, suppose c ≥ 2. If c is even, then 3 | 2^{c} − 1 and we obtain a contradiction. If c is odd, we have 2^{a} ≡ 1 (mod 3) so a is even. Therefore, there exists a positive integer a′ such that a = 2a′. As c = 2c′ + 1, for some positive integer c′, we have 2^{2a′} − 3^{b} = 2^{2c′+1} − 1. So (−1)^{b+1} ≡ −1 (mod 4). Hence, b is even and we have b = 2b′, for some non-negative integer b′. Then we have

2^{2a′} − 3^{2b′} = 2^{2c′+1} − 1.

Therefore,

(−1)^{a′} − (−1)^{b′} ≡ 2 · (−1)^{c′} − 1 (mod 5).

If a′ and b′ have the same parity we have a contradiction. If a′ is odd and b′ is even we have a contradiction as well. Therefore, a′ is even and b′ is odd. So we have 4 | a and b ≡ 2 (mod 4), which implies 1 − (−1) ≡ 2 · (−1)^{c′} − 1 (mod 5). Hence, c′ is odd and c ≡ 3 (mod 4). Since c > 1, then a > c. Therefore, we have 2^{c}(2^{a−c} − 1) = 3^{b} − 1. Hence, 2^{c} ∥ 3^{b} − 1.

Since b ≡ 2 (mod 4), then 2^{3} ∥ 3^{b} − 1. Hence, c = 3 and we can rewrite Eq. (10) as

(2^{\frac{a}{2}} − 3^{\frac{b}{2}} ) (2^{\frac{a}{2}} + 3^{\frac{b}{2}} ) = 7.

Therefore, a = 4 and b = 2. Hence, if c ≥ 2, the only solution of Eq. (10) is (4, 2, 3). Therefore, we obtain the stated result,

We can now state and give a new proof of Steuerwald’s theorem on 3-perfect numbers.

Theorem 3. Suppose N is a 3-perfect number of the form N = 2^{a} 3^{b}m, where a, b ≥ 1 and gcd(6, m) = 1. Then a = 1 and b ≠ 1, or a ≠ 1 and b = 1.

Proof. Let N = 2^{a} 3^{b}m such that a, b ≥ 1, gcd(6, m) = 1, and σ(N) = 3N. Then

3N = 2^{a} 3^{b+1}m = σ(N) = (2^{a+1} − 1) \frac{3^{b+1} − 1}{2} σ(m).

Therefore,

\frac{σ(m)}{m} = \frac{2^{a+1} 3^{b+1}}{(2^{a+1} − 1)(3^{b+1} − 1)}.

Let d = gcd (2^{a+1} 3^{b+1}, (2^{a+1} − 1)(3^{b+1} − 1)). It is easy to see that d = 2^{s} 3^{t}, where 1 ≤ s ≤ a + 1 and 0 ≤ t ≤ b + 1. Since

gcd (\frac{2^{a+1}3^{b+1}}{2^{s} 3^{t}} , \frac{(2^{a+1} − 1)(3^{b+1} − 1)}{2^{s} 3^{t}}) = 1,

then

σ(m) = \frac{2^{a+1} 3^{b+1}}{2^{s} 3^{t}} k and m = \frac{2^{a+1} − 1}{3^{t}} \frac{3^{b+1} − 1}{2^{s}}k,

for some positive integer k.

Let us consider the following three cases, which will establish the claim.

Case A: Suppose that t ≠ 0 and let

M = max (\frac{2^{a+1} − 1}{3^{t}}, \frac{3^{b+1} − 1}{2^{s}}) .

Then we have

\frac{σ(m)}{k} = \frac{2^{a+1}3^{b+1}}{2^{s} 3^{t}}

= \frac{((2^{a+1} − 1) + 1) ((3^{b+1} − 1) + 1)}{2^{s} 3^{t}}

= \frac{2^{a+1} − 1}{3^{t}} \frac{3^{b+1} − 1}{2^{s}} + \frac{2^{a+1} − 1}{3^{t}} \frac{1}{2^{s}} + \frac{3^{b+1} − 1}{2^{s}} \frac{1}{3^{t}} + \frac{1}{2^{s} 3^{t}}

< \frac{m}{k} + \frac{M}{2} + \frac{M}{3} + 1

< \frac{m}{k} + M + 1.

Therefore,

σ(m) < m + Mk + k.  (11)

By definition of M, we have m = uMk, with u ∈ \{\frac{2^{a+1}−1}{3^{t}}, \frac{3^{b+1}−1}{2^{s}}\}. By definition of s and t, we have u ∈ ℤ. Then Mk | m. If Mk ≠ m and M ≠ 1, then m, Mk, and k are different divisors of m. Thus,

σ(m) ≥ m + Mk + k.  (12)

The combination of inequalities (11) and (12), give us a contradiction.

If Mk = m or M = 1, then

\frac{2^{a+1} − 1}{3^{t}} = 1 or \frac{3^{b+1} − 1}{2^{s}} = 1.

Therefore, 2^{a+1} − 3^{t} = 1 or 3^{b+1} − 2^{s} = 1. By Lemma 2, we have a = 1 or b = 1.

Case B: Suppose that t = 0 and

2^{a+1} − 1 ≠ \frac{3^{b+1} − 1}{2^{s}}.

Let

M′ = min (2^{a+1} − 1, \frac{3^{b+1} − 1}{2^{s}}) .

If M′ = 1 then

2^{a+1} − 1 = 1 or \frac{3^{b+1} − 1}{2^{s}} = 1.

Therefore, 2^{a+1} = 2 or 3^{b+1} − 2^{s} = 1. As a, b ≥ 1, by Lemma 2, we conclude that b = 1. If M′ ≠ 1 then

m, (2^{a+1} − 1) k, \frac{3^{b+1} − 1}{2^{s}}k, and k,

are different divisors m. Therefore,

σ(m) ≥ m + (2^{a+1} − 1)k + \frac{3^{b+1} − 1}{2^{s}}k + k

> (2^{a+1} − 1) \frac{3^{b+1} − 1}{2^{s}}k + \frac{2^{a+1} − 1}{2^{s}}k + \frac{3^{b+1} − 1}{2^{s}}k + \frac{k}{2^{s}}

= σ(m),

where the strict inequality results from s ≥ 1. So, we obtain a contradiction.

Case C: Suppose that t = 0 and

2^{a+1} − 1 = \frac{3^{b+1} − 1}{2^{s}}.

Then 2^{a+1+s} − 3^{b+1} = 2^{s} − 1. By Lemma 2, we have (a, b) ∈ {(0, 0), (0, 1)}. Since a, b ≥ 1 we obtain a contradiction.

Hence, we must have a = 1 or b = 1.

Next, we prove that we cannot have a = 1 and b = 1, simultaneously.

If a = 1, let N = 2N′, where N′ is odd. Then

6N′ = 3N = σ(N) = 3σ(N′)

Therefore, σ(N′) = 2N′. So N′ is an odd perfect number. By Euler’s formula (1), if 3 | N′ then 3^{2} | N′. Hence, we cannot have b = 1. Thus, we have a = 1 and b ≠ 1, or a ≠ 1 and b = 1.

5 Fermat prime numbers: r = 2 and α = \frac{2p}{p−1}

In Eq. (5), if r = 2, p_{1} = 2, and p_{2} ≠ 3, then α is not an integer. Moreover, if p_{2} is not a Fermat number, then we must have a prime p ∉ \{p_{1}, p_{2}\} such that p | \frac{p_{2}−1}{2}. Using (3), we obtain p | N. For this reason, and in order to keep r = 2 without any other known prime divisors of N, we will only consider the cases where p_{2} ≠ 3 and p_{2} is a Fermat prime number. Let F_{n} = 2^{2^{n}} + 1 denote the n-th Fermat number, for any non-negative integer n. The next result states some properties of Fermat numbers that will be used later.

Lemma 4. Let n be a non-negative integer and F_{n} be the n-th Fermat number. Then

1. F_{n−1} | F_{n}^{2^{n}} − 1, for n ≥ 1;

2. ord_{2^{2^{n}+s}}(F_{n}) = 2^{s}, for n ≥ 1 and s a non-negative integer.

Proof. Let n ≥ 1. Clearly, F_{n} = (F_{n−1} − 1)^{2} + 1. So there exists an integer t, such that

F_{n}^{2^{n}} − 1 = ((F_{n−1} − 1)^{2} + 1)^{2^{n}} − 1

= −1 + ((F_{n−1}^{2} − 2F_{n−1}) + 2)^{2^{n}}

= −1 + ∑_{j=0}^{2^{n}} \binom{2^{n}}{j} (F_{n−1}^{2} − 2F_{n−1})^{j} 2^{2^{n}−j}

= −1 + tF_{n−1} + 2^{2^{n}} = tF_{n−1} + F_{0}F_{1} ⋯ F_{n−1}.

Hence, F_{n−1} | F_{n}^{2^{n}} − 1.

The statement 2. is clearly true for s = 0. We prove by induction that

F_{n}^{2^{s−1}} ≡ 1 + 2^{2^{n}+s−1} (mod 2^{2^{n}+s}),  (13)

for any s ≥ 1.

Since F_{n} ≡ F_{n} (mod 2^{2^{n}+1}), the congruence (13) is valid for s = 1. Now, suppose (13) is valid for s. Then there exists an integer t such that

F_{n}^{2^{s}} = (F_{n}^{2^{s−1}})^{2}
= (1 + (2^{2^{n}+s−1} + 2^{2^{n}+s}t))^{2}
= 1 + 2^{2^{n}+s} + 2^{2^{n}+s+1}t + 2^{2^{n+1}+2s−2} + 2^{2^{n+1}+2s}t^{2} + 2^{2^{n+1}+2s}t
≡ 1 + 2^{2^{n}+s} (mod 2^{2^{n}+s+1}).

Therefore, Eq. (13) is true for any s ≥ 1. So we have

ord_{2^{2^{n}+s}}(F_{n}) ≠ 2^{s−1}.

Since ord_{2^{2^{n}+s}}(F_{n}) is a power of 2 and

F_{n}^{2^{s}} ≡ 1 (mod 2^{2^{n}+s}),

then statement 2. is obtained.

Before applying Euler’s method we still need two technical results about some exponential diophantine equations. Our proof of these results is inspired by Styer’s work [10].

Lemma 5. Let n ≥ 1 and F_{n} be the n-th Fermat number. Suppose there exists a non-negative integer s and a prime number q_{n}, such that

q_{n} | F_{n}^{2^{s}} − 1  (14)

and

ord_{F_{n}}(2) | ord_{q_{n}}(2).  (15)

Then the exponential diophantine equation

2^{a} − F_{n}^{b} = 2^{c} − 1,  (16)

has no solutions (a, b, c) with b ≥ 1 and c ≥ 2^{n} + s.

Proof. Suppose there exists a non-negative integer s and a prime number q_{n} satisfying Cond. (14) and (15). Also, suppose that exists a solution (a, b, c) of Eq. (16) with b ≥ 1 and c ≥ 2^{n} + s. Since b ≥ 1, we have a > c. Therefore, we have

F_{n}^{b} ≡ 1 (mod 2^{2^{n}+s}).

By Lemma 4, we have ord_{2^{2^{n}+s}}(F_{n}) = 2^{s}. Hence, 2^{s} | b. Now, Cond. (14) and Eq. (16) imply 2^{a} − 1 ≡ 2^{c} − 1 (mod q_{n}). Let t = ord_{q_{n}}(2). Then a ≡ c (mod t). Hence, exists a′ such that a = c + a′t. But, by Eq. (15), ord_{F_{n}}(2) | t, which implies

2^{c} − 1 ≡ 2^{a} ≡ 2^{c} (2^{t})^{a′} ≡ 2^{c} (mod F_{n}).

Since the previous equation has no solutions, we obtain a contradiction. Thus, we obtain the stated result.

Lemma 6. Let F_{n} be the n-th Fermat number and consider the following exponential diophantine equations

2^{a} − F_{n}^{b} = −1  (17)

and

2^{a} − F_{n}^{b} = 2^{c} − 1.  (18)

Then

(a) when n ∈ {1, 2, 3, 4}, Equation (17) only holds for (a, b) = (2^{n}, 1);

(b) when n ∈ {2, 3, 4}, Equation (18) only holds for

(a, b, c) ∈ {(a, 0, a) | a ∈ ℤ} ∪ {(2^{n} + 1, 1, 2^{n})};

(c) when n = 1, Equation (18) only holds for

(a, b, c) ∈ {(a, 0, a) | a ∈ ℤ} ∪ {(3, 1, 2), (7, 3, 2), (5, 2, 3)}.

Proof. Suppose n ∈ {1, 2, 3, 4}, 2^{a} − F_{n}^{b} = −1, and a ≥ 2^{n} + 1. Therefore, 2^{2^{n}+1} | F_{n}^{b} − 1. So b is even and we have b = 2b′, for some positive integer b′. But, since 3 | 5^{2} − 1, 9 | 17^{2} − 1, 43 | 257^{2} − 1, and 11 | 65537^{2} − 1, we always obtain a contradiction. Therefore, (2^{n}, 1) is the only solution of Eq. (17), and we obtain statement (a).

Now, we consider Eq. (18) with n ∈ {1, 2, 3, 4}. Notice that we always have a ≥ c. If b = 0 then a = c and so we have (a, b, c) = (a, 0, a). Since 2^{c}(2^{a−c} − 1) = F_{n}^{b} − 1 and 2^{2^{n}} | F_{n}^{b} − 1, for any positive integer b, we have c ≥ 2^{n}. Therefore, if b = 1 then c = 2^{n} and a = 2^{n} + 1.

From now on, we assume b ≥ 2. Since F_{n} is a Fermat number, we have a ≥ 2^{n+1} + 1. As 2^{n} + n < 2^{n+1}, then Eq. (18) implies

F_{n}^{b} ≡ 1 − 2^{c} (mod 2^{2^{n}+n}).  (19)

By Lemma 4, we have ord_{2^{2^{n}+n}}(F_{n}) = 2^{n}. Then, for each c, exist integers b_{c} and b′_{c} such that b = b_{c} + 2^{n}b′_{c}. Notice that b_{c} and b′_{c} also depend on which n we are considering. By Lemma 4, we have F_{n−1} | F_{n}^{2^{n}} − 1. Hence,

2^{a} − F_{n}^{b_{c}} ≡ 2^{c} − 1 (mod F_{n−1}).  (20)

As ord_{F_{n−1}}(2) = 2^{n}, then exist integers a_{c} and a′_{c} such that a = a_{c} + 2^{n}a′_{c}. But then

2^{a_{c}}(−1)^{a′_{c}} ≡ 2^{c} − 1 (mod F_{n}).  (21)

We now analyze Eq. (18) for each n ∈ {1, 2, 3, 4}.

Suppose n = 1.

Since s = 2 and q_{1} = 13 satisfy the conditions of Lemma 5, then Eq. (18) has no solutions for c ≥ 4. Therefore, c = 2 or c = 3.

Clearly, (7, 3, 2) is the only solution of Eq. (18), with c = 2 and a ≤ 7. If a ≥ 8 and c = 2, then 5^{b} ≡ −3 (mod 256). Therefore, b ≡ 35 (mod 64), i.e., there exists a non-negative integer b′ such that b = 35 + 64b′. Then

2^{a} − 5^{35} · 5^{64b′} = 3.

As 641 | 5^{64} − 1 and 5^{35} ≡ 516 (mod 641), then

2^{a} ≡ 519 (mod 641).

But this congruence has no solutions.

If c = 3 and a ≤ 5, then (5, 2, 3) is the only solution of Eq. (18). Suppose c = 3 and a ≥ 6, then 5^{b} ≡ −7 (mod 64). Therefore, we have b ≡ 10 (mod 16); i.e., there exists a non-negative integer b′ such that b = 10 + 16b′. Then

2^{a} − 5^{10} · 5^{16b′} = 7.

As 13 | 5^{16} − 1 and 5^{10} ≡ −1 (mod 13), then

2^{a} ≡ 6 (mod 13).

Therefore, a ≡ 5 (mod 12). On the other hand, since 5^{10} ≡ 9 (mod 17) and 17 | 5^{16} − 1, then

2^{a} ≡ −1 (mod 17).

Hence, we have a ≡ 4 (mod 8), which gives us a contradiction. Thus, we obtain statement (c).

Let us now consider n = 2.

Since s = 4 and q_{1} = 18913 satisfy the conditions of Lemma 5, then Eq. (18) has no solutions for c ≥ 8.

For each c ∈ {4, 5, 6, 7}, consider b = b_{c} + 2^{n}b′_{c} to be the solutions of Eq. (19) and a = a_{c} + 2^{n}a′_{c} to be the solutions of Eq. (20) (when these equations have solution). The values of b_{c} and a_{c} are given in Table 1.

c b_{c}  a_{c}
4 3  3
5 2 No solutions
6 0  2
7 0  2

Table 1: Solutions of Eq. (19) and Eq. (20) for n = 2.

Hence, when c = 5, Eq. (20) has no solutions. Also, when c ∈ {4, 6, 7}, Eq. (21) has no solutions.

Next, we consider n = 3.

Since s = 4 and q_{3} = 193 satisfy the conditions of Lemma 5, then Eq. (18) has no solutions for c ≥ 12.

For each c ∈ {8, 9, 10, 11}, consider b = b_{c} + 2^{n}b′_{c} to be the solutions of Eq. (19) and a = a_{c} + 2^{n}a′_{c} to be the solutions of Eq. (20) (when these equations have solution). The values b_{c} and a_{c} are given in table 2.

c b_{c}  a_{c}
8 7  7
9 6 No solutions
10 4  1
11 0  3

Table 2: Solutions of Eq. (19) and Eq. (20) for n = 3.

Hence, when c = 9, Eq. (20) has no solutions. Also, when c ∈ {8, 10, 11}, Eq. (21) has no solutions.

Finally, consider n = 4.

Since s = 5 and q_{4} = 38899171806337 satisfy the conditions of Lemma 5, then Eq. (18) has no solutions for c ≥ 21.

For each c ∈ {16, 17, 18, 19, 20}, consider b = b_{c} + 2^{n}b′_{c} to be the solutions of Eq. (19) and a = a_{c} + 2^{n}a′_{c} to be the solutions of Eq. (20) (when these equations have solution). The values b_{c} and a_{c} are given in table 3.

c b_{c}  a_{c}
16 15  15
17 14 No solutions
18 12 No solutions
19 8 No solutions
20 0  4

Table 3: Solutions of Eq. (19) and Eq. (20) for n = 4.

Hence, when c ∈ {17, 18, 19}, Eq. (20) has no solutions. When c ∈ {16, 20}, Eq. (21) has no solutions.

Thus, statement (b) is obtained.

We are now in conditions to prove a generalization of the Euclid-Euler theorem for \frac{2F_{n}}{F_{n}−1}- perfect numbers, when n ∈ {1, 2, 3, 4}. The proof of the next theorem follows the same steps as the proof of Theorem 3.

Theorem 7. Let F_{n} be the n-th Fermat number. Then

1. if exists N such that \frac{σ(N)}{N} = \frac{F_{1}}{2} and F_{1} | N, then 2^{4} ∥ N, F_{1}^{2} ∥ N, and 31^{2} | N.

2. if n ∈ \{2, 3, 4\} then \frac{2F_{n}}{F_{n}−1} is a F_{n}-abundancy outlaw.

Proof. Let n ∈ {1, 2, 3, 4} and F_{n} = 2^{2^{n}} + 1. We can write N = 2^{a}F_{n}^{b}m such that a, b ≥ 1, gcd(2F_{n}, m) = 1, and

σ(N) = \frac{2F_{n}}{F_{n} − 1}N.

Then

\frac{2F_{n}}{F_{n} − 1} = \frac{σ(N)}{N} = \frac{2^{a+1} − 1}{2^{a}} \frac{F_{n}^{b+1} − 1}{F_{n}^{b}(F_{n} − 1)} \frac{σ(m)}{m}.

Therefore,

\frac{σ(m)}{m} = \frac{2^{a+1} F_{n}^{b+1}}{(2^{a+1} − 1)(F_{n}^{b+1} − 1)}.

Let

d = gcd (2^{a+1} F_{n}^{b+1}, (2^{a+1} − 1)(F_{n}^{b+1} − 1)) .

As F_{n} − 1 | F_{n}^{b+1} − 1, then d = 2^{s} F_{n}^{t}, where 2^{n} ≤ s ≤ a + 1 and 0 ≤ t ≤ b + 1. Since

gcd (\frac{2^{a+1}F_{n}^{b+1}}{2^{s} F_{n}^{t}} , \frac{(2^{a+1} − 1)(F_{n}^{b+1} − 1)}{2^{s} F_{n}^{t}}) = 1,

then

σ(m) = \frac{2^{a+1}F_{n}^{b+1}}{2^{s} F_{n}^{t}} k and m = \frac{2^{a+1} − 1}{F_{n}^{t}} \frac{F_{n}^{b+1} − 1}{2^{s}}k,

for some positive integer k.

Let us consider the following three cases, which will establish the claim.

Case A: Suppose that t ≠ 0 and let

M = max (\frac{2^{a+1} − 1}{F_{n}^{t}}, \frac{F_{n}^{b+1} − 1}{2^{s}}) .

Then we have that

\frac{σ(m)}{k} = \frac{2^{a+1}F_{n}^{b+1}}{2^{s} F_{n}^{t}}

= \frac{2^{a+1} − 1}{F_{n}^{t}} \frac{F_{n}^{b+1} − 1}{2^{s}} + \frac{2^{a+1} − 1}{F_{n}^{t}} \frac{1}{2^{s}} + \frac{F_{n}^{b+1} − 1}{2^{s}} \frac{1}{F_{n}^{t}} + \frac{1}{2^{s} F_{n}^{t}}

< \frac{m}{k} + \frac{M}{F_{n} − 1} + \frac{M}{F_{n}} + 1

< \frac{m}{k} + M + 1.

Therefore,

σ(m) < m + Mk + k.  (22)

If Mk ≠ m and M ≠ 1, then m has at least the divisors m, Mk, and k. Thus,

σ(m) ≥ m + Mk + k.  (23)

The combination of inequalities (22) and (23), give us a contradiction.

If Mk = m or M = 1, then

\frac{2^{a+1} − 1}{F_{n}^{t}} = 1 or \frac{F_{n}^{b+1} − 1}{2^{s}} = 1.

Thus 2^{a+1} − F_{n}^{t} = 1 or F_{n}^{b+1} − 2^{s} = 1. By Lemma 6, the only solutions of these equations are (a, t) = (0, 0) and (b, s) = (0, 2^{n}). Hence, we obtain a contradiction because a, b ≥ 1.

Case B: Suppose t = 0 and

2^{a+1} − 1 ≠ \frac{F_{n}^{b+1} − 1}{2^{s}}.

Let

M′ = min (2^{a+1} − 1, \frac{F_{n}^{b+1} − 1}{2^{s}}) .

If M′ = 1 then

2^{a+1} − 1 = 1 or \frac{F_{n}^{b+1} − 1}{2^{s}} = 1.

Thus 2^{a+1} = 2 or F_{n}^{b+1} − 2^{s} = 1. As a, b ≥ 1, by Lemma 6, we have a contradiction. If M′ ≠ 1 then m has at least the divisors

m, (2^{a+1} − 1) k, \frac{F_{n}^{b+1} − 1}{2^{s}}k and k.

Therefore,

σ(m) ≥ m + (2^{a+1} − 1)k + \frac{F_{n}^{b+1} − 1}{2^{s}}k + k

> (2^{a+1} − 1) \frac{F_{n}^{b+1} − 1}{2^{s}}k + \frac{2^{a+1} − 1}{2^{s}}k + \frac{F_{n}^{b+1} − 1}{2^{s}}k + \frac{k}{2^{s}}

= σ(m),

where the strict inequality results from s ≥ 2^{n}. Hence, we obtain a contradiction.

Case C: Suppose t = 0 and

2^{a+1} − 1 = \frac{F_{n}^{b+1} − 1}{2^{s}}.

Then 2^{a+1+s} − F_{n}^{b+1} = 2^{s} − 1. By Lemma 6, we have

\begin{matrix}(F_{n}, a, b, s) ∈ \{(5, −1, −1, 1), (5, 0, 0, 2), (5, 1, & 1, 3), (5, 4, 2, 2), (17, 0, 0, 4), \\ (257, 0, 0, 8), (65537, 0, 0, 16)\}. \end{matrix}

Since a, b ≥ 1, we only have solutions for F_{n} = 5 and then (F_{n}, a, b) = (5, 1, 1) or (F_{n}, a, b) = (5, 4, 2).

If (F_{n}, a, b) = (5, 1, 1) then

\frac{σ(N)}{N} = \frac{5}{2} = \frac{3}{2} · \frac{6}{5} · \frac{σ(m)}{m}.

Therefore, 9 | m. But then

\frac{σ(N)}{N} = \frac{5}{2} ≥ \frac{3}{2} · \frac{6}{5} · \frac{13}{9} > \frac{5}{2}.

Hence, (F_{n}, a, b) = (5, 4, 2) and so 31^{2} | N.

6 Acknowledgment

This work was supported by The Center for Research and Development in Mathematics and Applications (CIDMA) through the Portuguese Foundation for Science and Technology (FCT—Fundação para a Ciência e Tecnologia), reference UIDB/04106/2020 and reference PD/BD/150533/2019.

References

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2020 Mathematics Subject Classification: Primary 11A25; Secondary 11N25, 11B83. Keywords: Euclid-Euler theorem, α-perfect number, Fermat number, abundancy index, abundancy outlaw.

(Concerned with sequences \underline{A005820}, \underline{A019434}, \underline{A214409}, and \underline{A214413}.)

Received July 2 2022; revised version received July 3 2022; October 5 2022; October 6 2022; October 12 2022. Published in Journal of Integer Sequences, October 14 2022.

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