Divisibility of Divisor Functions of Even Perfect Numbers

Hùng Vi{ê \atop .}t Chu

Department of Mathematics University of Illinois at Urbana-Champaign Champaign, IL 61820 USA

                     hungchu2@illinois.edu

Abstract

Let k > 2 be a prime such that 2^{k} − 1 is a Mersenne prime. Let n = 2^{α−1}p, where α > 1 and p < 3 · 2^{α−1} − 1 is an odd prime. Define σ_{k}(n) to be the sum of the kth powers of the positive divisors of n. Continuing the work of Cai et al. and Jiang, we prove that n | σ_{k}(n) if and only if n is an even perfect number other than 2^{k−1}(2^{k} − 1). Furthermore, if n = 2^{α−1}p^{β−1} for some β > 1, then n | σ_{5}(n) if and only if n is an even perfect number other than 496.

1 Introduction and main results

For a positive integer n, let σ(n) be the sum of the positive divisors of n. We call n perfect if σ(n) = 2n (sequence \underline{A000396} in the On-Line Encyclopedia of Integer Sequences (OEIS) [11]). Due to the work of Euclid and Euler, it is well-known that an even integer n is perfect if and only if n = 2^{p−1}(2^{p} − 1), where both p and 2^{p} − 1 are primes. A prime of the form 2^{p} − 1 is called a Mersenne prime. Up to now, fewer than 60 Mersenne primes are known. Two questions are still open: whether there are infinitely many even perfect numbers and whether there exists an odd perfect number, though various progress has been made. For example, Pomerance [6] showed that an odd perfect number must have at least 7 distinct prime factors. Nielsen improved the result by proving that an odd perfect number must have at least 9 distinct prime factors. For related results, see [7, 8].

Meanwhile, mathematicians have generalized the concept of perfect numbers. Pollack and Shevelev [5] introduced k-near-perfect numbers. For k ≥ 1, a k-near-perfect number n is the sum of all of its proper divisors with at most k exceptions. A positive integer n is called near-perfect if n is the sum of all but exactly one of its proper divisors (\underline{A181595}). Pollack and Shevelev showed how to construct near-perfect numbers and established an upper bound of x^{5/6+o(1)} for the number of near-perfect numbers in [1, x] as x → ∞. Li and Liao [4] gave two equivalent conditions of all even near-perfect numbers of the form 2^{α}p_{1}p_{2} and 2^{α}p_{1}^{2}p_{2}, where α > 0 and p_{1}, p_{2} are distinct primes. In 2013, Ren and Chen [10] found all near-perfect numbers with two distinct prime factors. Continuing the work, Tang et al.

[14] showed that there is no odd near-perfect number with three distinct prime divisors. For other beautiful results on near-perfect numbers and deficient-perfect numbers (\underline{A271816}, \underline{A341475}), see [12, 13].

The present paper focuses on another generalization of perfect numbers by connecting an even perfect number n with the divisibility of σ_{k}(n), where k ≥ 1 and

σ_{k}(n) := ∑_{d|n} d^{k}.

In 2006, Luca and Ferdinands [3] proved that for k ≥ 2, there are infinitely many n such that n | σ_{k}(n). In 2015, Cai et al. [1] proved the following theorem.

Theorem 1. Let n = 2^{α−1}p, where α > 1 is an integer and p is an odd prime. If n | σ_{3}(n), then n is an even perfect number. The converse is also true for n ≠ 28.

Three years later, Jiang [2] improved the theorem as follows.

Theorem 2. Let n = 2^{α−1}p^{β−1}, where α, β > 1 are integers and p is an odd prime. Then n | σ_{3}(n) if and only if n is an even perfect number ≠ 28.

These theorems show a beautiful relationship between an even perfect number n and σ_{3}(n). A natural extension is to consider σ_{k}(n) for some other values of k. Unfortunately, Theorem 1 does not hold when k = 5 or 7, for example. A quick computer search gives σ_{5}(22) ≡ 0 (mod 22) and σ_{7}(86) ≡ 0 (mod 86). However, if we add one more restriction on p, the following theorem holds.

Theorem 3. Let k > 2 be a prime such that 2^{k} − 1 is a Mersenne prime. If n = 2^{α−1}p, where α > 1 and p < 3 · 2^{α−1} − 1 is an odd prime. Then n | σ_{k}(n) if and only if n is an even perfect number ≠ 2^{k−1}(2^{k} − 1).

Theorem 3 can be considered a generalization of Theorem 1 as we have a wider range of k with the new restriction on p as a compensation. Interestingly, when k = 5, we can generalize Theorem 3 the same way as Jiang generalized Theorem 1.

Theorem 4. If n = 2^{α−1}p^{β−1}, where α, β > 1 and p < 3 · 2^{α−1} − 1 is an odd prime. Then n | σ_{5}(n) if and only if n is an even perfect number ≠ 496.

Unfortunately, our method is not applicable to other values of k even though computation supports the following conjecture.

Conjecture 5. Let k > 2 be a prime such that 2^{k} − 1 is a Mersenne prime. If n = 2^{α−1}p^{β−1}, where α, β > 1 and p < 3 · 2^{α−1} − 1 is an odd prime. Then n | σ_{k}(n) if and only if n is an even perfect number ≠ 2^{k−1}(2^{k} − 1).

Our paper is structured as follows. Section 2 provides several preliminary results that are used repeatedly throughout the paper, Section 3 proves Theorem 3 and Section 4 proves Theorem 4. Since the proof of several claims made in Section 3 and Section 4 are quite technical, we move them to the Appendix for the ease of reading.

2 Preliminaries

Let n = 2^{α−1}p^{β−1}, where α, β > 1 are integers and p < 3 · 2^{α−1} − 1 is an odd prime. Let k > 2 be a prime such that 2^{k} − 1 is a Mersenne prime. We will stick with these notation throughout the paper. If n | σ_{k}(n), then

2^{α−1}p^{β−1} | σ_{k}(2^{α−1})σ_{k}(p^{β−1}) = (1 + 2^{k} + ⋯ + 2^{(α−1)k})(1 + p^{k} + ⋯ + p^{(β−1)k})

= \frac{2^{αk} − 1}{2^{k} − 1} · \frac{p^{βk} − 1}{p^{k} − 1}.

Because (2, 2^{αk} − 1) = 1 and (p, p^{βk} − 1) = 1, it follows that

2^{α−1} divides \frac{p^{βk} − 1}{p^{k} − 1}, so 2^{α} divides p^{βk} − 1,  (1)
p^{β−1} divides \frac{2^{αk} − 1}{2^{k} − 1}.  (2)

Furthermore, rewrite (1) as

2^{α−1} | \frac{p^{βk} − 1}{p^{k} − 1} = \frac{(p^{k} − 1)(p^{k(β−1)} + p^{k(β−2)} + ⋯ + 1)}{p^{k} − 1} = ∑_{i=0}^{β−1} p^{ki}.

Since each term is odd and the summation is divisible by 2, we know that 2 | β. The following lemma is the key ingredient in the proof of Theorem 3.

Lemma 6. Let n = 2^{α−1}(2^{k} − 1)^{β−1}, where α, β > 1 are integers. Then n ∤ σ_{k}(n).

Proof. We use proof by contradiction. Suppose n | σ_{k}(n). By (1) and (2), we have

2^{α} | (2^{k} − 1)^{βk} − 1,  (3)
(2^{k} − 1)^{β} | (2^{αk} − 1) = (2^{k} − 1)((2^{k})^{α−1} + ⋯ + 1).  (4)

Write α = (2^{k} − 1)^{u}α_{1} and β = 2^{v}β_{1}, where u ≥ 0, v ≥ 1 and (2^{k} − 1, α_{1}) = (2, β_{1}) = 1. By Lemma 13, α ≤ v + k.

If u = 0, we get α = α_{1}. From (4), β = 1, which contradicts the fact that 2 | β.

If u ≥ 1, Remark 15 implies that β ≤ u + 2^{k} − 1. We have

2^{(2^{k}−1)^{u}−k} ≤ 2^{α−k}β_{1} ≤ 2^{v}β_{1} = β ≤ u + 2^{k} − 1.

Since for all u ≥ 1 and k ≥ 3,

2^{(2^{k}−1)^{u}−k} > u + 2^{k} − 1

by Lemma 11, we have a contradiction. This finishes our proof.

3 Proof of Theorem 3

For the forward implication, we prove that if n = 2^{α−1}p and n | σ_{k}(n), then α is prime and p = 2^{α} − 1. By Lemma 6, n ≠ 2^{k−1}(2^{k} − 1). We have

σ_{k}(n) = σ_{k}(2^{α−1}p) = σ_{k}(2^{α−1})σ_{k}(p)

= (1 + 2^{k} + ⋯ + 2^{k(α−1)})(1 + p^{k})

= (1 + 2^{k} + ⋯ + 2^{k(α−1)})(1 + p) ∑_{i=1}^{k} p^{k−i}(−1)^{i+1}.

So, 2^{α−1}p | σ_{k}(n) implies that 2^{α−1} | 1+p and p | 1+2^{k} +⋯+2^{k(α−1)}. There exist k_{1}, k_{2} ∈ ℕ such that p = k_{1}2^{α−1} − 1 and 1 + 2^{k} + ⋯ + 2^{k(α−1)} = \frac{2^{kα}−1}{2^{k}−1} = k_{2}p. So,

(2^{α} − 1) ∑_{i=0}^{k−1} 2^{iα} = 2^{kα} − 1 = k_{3}(k_{1}2^{α−1} − 1),  (5)

where k_{3} = (2^{k} − 1)k_{2}.

Suppose that k_{1} = 1. Then p = 2^{α−1} − 1 and (5) implies that either 2^{α−1} − 1 | (2^{α} − 1) or 2^{α−1} − 1 | ∑_{i=0}^{k−1} 2^{iα}. If the former, we write

1 = 2^{α} − 1 − 2(2^{α−1} − 1) ≡ 0 (mod 2^{α−1} − 1),

which is impossible. Suppose the latter. Because 2^{α} ≡ 2 (mod p), we have

∑_{i=0}^{k−1} 2^{iα} ≡ ∑_{i=0}^{k−1} 2^{i} ≡ 2^{k} − 1 (mod p),

which implies that p divides 2^{k} − 1. Hence, p = 2^{k} − 1. However, Lemma 6 implies that n ∤ σ_{k}(n), which contradicts our assumption. So, k_{1} ≥ 2; however, k_{1} < 3 by assumption.

So, k_{1} = 2; we have p = 2^{α} − 1 and α is a prime. Therefore, n is an even perfect number ≠ 2^{k−1}(2^{k} − 1).

For the backward implication, write n = 2^{q−1}(2^{q} − 1), where q ≠ k and 2^{q} − 1 are primes. We have

σ_{k}(n) = (1 + 2^{k} + 2^{2k} + ⋯ + 2^{(q−1)k})(1 + (2^{q} − 1)^{k})

= \frac{2^{qk} − 1}{2^{k} − 1}(1 + (2^{q} − 1)^{k}).

Clearly, 2^{q−1} divides 1 + (2^{q} − 1)^{k}. It suffices to show that 2^{q} − 1 divides \frac{2^{qk}−1}{2^{k}−1}. The fact n ≠ 2^{k−1}(2^{k} −1) implies that 2^{q} −1 and 2^{k} −1 are two distinct primes. So, (2^{q} −1, 2^{k} −1) = 1. Because 2^{q} − 1 | 2^{qk} − 1, 2^{q} − 1 divides \frac{2^{qk}−1}{2^{k}−1}. Therefore, n | σ_{k}(n).

4 Proof of Theorem 4

4.1 Preliminary results

We provide lemmas that give useful bounds used in the proof of Theorem 4.

Lemma 7. Let n = 2^{α−1}p^{3}, where α > 1, p ≡ 3 (mod 4) and p < 3 · 2^{α−1} − 1. Then n ∤ σ_{5}(n).

Proof. We prove by contradiction. Suppose that n | σ_{5}(n). We have

σ_{5}(2^{α−1}p^{3}) = (1 + 2^{5} + ⋯ + 2^{5(α−1)})(1 + p^{5} + p^{10} + p^{15})
= (1 + 2^{5} + ⋯ + 2^{5(α−1)})(p^{10} + 1)(p + 1)(p^{4} − p^{3} + p^{2} − p + 1).

So,

2^{α−1} | (p^{10} + 1)(p + 1)  (6)
p^{3} | 1 + 2^{5} + ⋯ + 2^{5(α−1)} = \frac{2^{5α} − 1}{2^{5} − 1}.  (7)

Because p^{10} + 1 ≡ 2 (mod 4), we know that 2^{α−2} | p + 1. Hence, p = k_{1}2^{α−2} − 1 for some k_{1} ∈ ℕ. Combining with p < 3 · 2^{α−1} − 1, we get 1 ≤ k_{1} ≤ 5. By (7), write 2^{5α} − 1 = 31k_{2}p^{3} for some k_{2} ∈ ℕ. Therefore,

31k_{2}(k_{1}2^{α−2} − 1)^{3} = (2^{α} − 1)(2^{4α} + 2^{3α} + 2^{2α} + 2^{α} + 1).  (8)

Suppose that p divides both 2^{α} − 1 and ∑_{i=0}^{4} 2^{iα}. Then 2^{α} ≡ 1 (mod p) and so, ∑_{i=0}^{4} 2^{iα} ≡ 5 (mod p). Hence, p = 5, which contradicts the congruence p ≡ 3 (mod 4). It must be that either p^{3} | ∑_{i=0}^{4} 2^{iα} or p^{3} | 2^{α} − 1. We consider two corresponding cases.

Case 1: (k_{1}2^{α−2} − 1)^{3} | 2^{α} − 1. So, (k_{1}2^{α−2} − 1)^{3} ≤ 2^{α} − 1. In order that the inequality is true for some α ≥ 2, we must have 1 ≤ k_{1} ≤ 2. Otherwise,

(k_{1}2^{α−2} − 1)^{3} ≥ (3 · 2^{α−2} − 1)^{3} > 2^{α} − 1,

for all α ≥ 2. We consider two cases.

(i) k_{1} = 1. Then 2^{α−2} − 1 | 2^{α} − 1. Because

3 = (2^{α} − 1) − 4(2^{α−2} − 1) ≡ 0 (mod 2^{α−2} − 1),

p = 2^{α−2} − 1 = 3. So, α = 4 and n = 2^{3}3^{3}, a contradiction as 2^{3}3^{3} |̸ σ_{5}(2^{3}3^{3}).

(ii) k_{1} = 2. Then 2^{α−1} − 1 | 2^{α} − 1. Because

1 = (2^{α} − 1) − 2(2^{α−1} − 1) ≡ 0 (mod 2^{α−1} − 1),

p = 2^{α−1} − 1 = 1, a contradiction.

Case 2: (k_{1}2^{α−2} − 1)^{3} | ∑_{i=0}^{4} 2^{iα}. Let x_{0} = 2^{α}. Let f(x) = x^{4} + x^{3} + x^{2} + x + 1.

(i) If k_{1} = 1, we have 2^{α} ≡ 4 (mod p). So, f(x_{0}) ≡ 341 (mod p). Because p divides f(x_{0}), it follows that p divides 341 and so, p = 11 or 31. Since p = 2^{α−2} − 1, p = 31, and α = 7. However, n = 2^{6}31^{3} ∤ σ_{5}(n).

(ii) If k_{1} = 2, we have 2^{α} ≡ 2 (mod p). So, f(x_{0}) ≡ 31 (mod p). Because p divides f(x_{0}), it follows that p = 31 and α = 6. However, n = 2^{5}31^{3} ∤ σ_{5}(n).

(iii) If k_{1} = 3, we have 3x_{0} ≡ 4 (mod p). So, 3^{4}f(x_{0}) ≡ 781 (mod p). Because p divides f(x_{0}), it follows that p | 781 and so, p ∈ {11, 71}. Since p = 3 · 2^{α−2} − 1, we know p = 11, α = 4, and n = 2^{3}11^{3}. However, n = 2^{3}11^{3} ∤ σ_{5}(n).

(iv) If k_{1} = 4, we have x_{0} ≡ 1 (mod p). So, f(x_{0}) ≡ 5 (mod p). It follows that p = 5, which contradicts the congruence p ≡ 3 (mod 4).

(v) If k_{1} = 5, we have 5x_{0} ≡ 4 (mod p). So, 5^{4}f(x_{0}) ≡ 2101 (mod p). It follows that p divides 2101, so p = 11 or 191. Both cases are impossible.

This completes our proof.

Lemma 8. Let n = 2^{α−1}p^{β−1}, p ≡ 1 (mod 4) and n | σ_{k}(n). Write β = 2^{v}β_{1}, where v ≥ 1 and (2, β_{1}) = 1. Then

p^{2^{v}−1} ≤ \frac{2^{k(v+1)} − 1}{2^{k} − 1}.  (9)

Proof. Let p − 1 = 2^{t}p_{1}, where t ≥ 2 and 2 ∤ p_{1}. Because

p^{k} − 1 = (p − 1) ∑_{i=1}^{k} p^{k−i} = 2^{t}p_{1} ∑_{i=1}^{k} p^{k−i},  (10)

we have 2^{t} || (p^{k} − 1). By Lemma 16, 2^{t+v} || p^{kβ} − 1. Hence,

2^{v} || \frac{p^{kβ} − 1}{p^{k} − 1}.

By (1),

α ≤ v + 1.  (11)

Hence, we have

p^{2^{v}−1} ≤ p^{β−1} ≤ \frac{2^{kα} − 1}{2^{k} − 1} ≤ \frac{2^{k(v+1)} − 1}{2^{k} − 1}.

Lemma 9. Let n = 2^{α−1}p^{β−1}, p ≡ 3 (mod 4) and n | σ_{k}(n). Write β = 2^{v}β_{1}, where v ≥ 1 and (2, β_{1}) = 1. Then

p^{2^{v}−2k−1} < \frac{2^{k(v−1)}}{2^{k} − 1} .  (12)

Proof. Let p^{2} − 1 = 2^{s}p_{2}, where 2 ∤ p_{2}. Then s ≥ 3. By (10), 2 || p^{k} − 1 and by Lemma 17, 2^{v+s−1} || p^{kβ} − 1. Hence,

2^{v+s−2} || \frac{p^{kβ} − 1}{p^{k} − 1}.

By (1),

α ≤ v + s − 1.  (13)

We have

p^{2^{v}−1} ≤ p^{β−1} ≤ \frac{2^{kα} − 1}{2^{k} − 1} ≤ \frac{2^{k(v+s−1)} − 1}{2^{k} − 1}

= \frac{2^{ks}2^{k(v−1)} − 1}{2^{k} − 1} < \frac{p^{2k}2^{k(v−1)} − 1}{2^{k} − 1} because p^{2} > 2^{s}.

Therefore,

p^{2^{v}−2k−1} < \frac{2^{k(v−1)} − 1/p^{2k}}{2^{k} − 1} < \frac{2^{k(v−1)}}{2^{k} − 1} .

Lemma 10. Let n = 2^{α−1}p^{β−1}, p ≡ 3 (mod 4) and n | σ_{k}(n). Write β = 2^{v}β_{1} and p + 1 = 2^{λ}p_{1}, where (2, β_{1}) = (2, p_{1}) = 1. Then one of the following must hold (a)

p = k,

(b)

(2^{λ} − 1)^{β−1} ≤ 2^{λ+v} − 1,

(c)

(2^{λ} − 1)^{β−1} ≤ ∑_{i=0}^{k−1} 2^{i(λ+v)}.

Proof. From (1) and (2), we have

2^{α} | p^{β} − 1 and p^{β−1} | 2^{kα} − 1 = (2^{α} − 1) ∑_{i=0}^{k−1} 2^{iα}.

By Lemma 18, 2^{λ+v} || p^{β} − 1. So, α ≤ λ + v. Case 1: p | 2^{α} − 1 and p | ∑_{i=0}^{k−1} 2^{iα}. The fact that 2^{α} ≡ 1 (mod p) implies that ∑_{i=0}^{k−1} 2^{iα} ≡ k (mod p). Because p | ∑_{i=0}^{k−1} 2^{iα} and k is prime, it must be that p = k. We have scenario (a). Case 2: p | 2^{α} − 1 and p ∤ ∑_{i=0}^{k−1} 2^{iα}. So,

2^{α} | p^{β} − 1 and p^{β−1} | 2^{α} − 1.

We have

(2^{λ} − 1)^{β−1} ≤ p^{β−1} ≤ 2^{α} − 1 ≤ 2^{λ+v} − 1.

We have scenario (b). Case 3: p |̸ 2^{α} − 1 and p | ∑_{i=0}^{k−1} 2^{iα}. So,

2^{α} | p^{β} − 1 and p^{β−1} | ∑_{i=0}^{k−1} 2^{iα}.

We have

(2^{λ} − 1)^{β−1} ≤ p^{β−1} ≤ ∑_{i=0}^{k−1} 2^{iα} ≤ ∑_{i=0}^{k−1} 2^{i(λ+v)}.

We have scenario (c).

4.2 Proof of Theorem 4

We now bring together all preliminary results and prove Theorem 4 by case analysis.

Proof. The backward implication follows from Theorem 3. We prove the forward implication. Let n = 2^{α−1}p^{β−1}, where α, β > 1 and p < 3·2^{α−1}−1 is an odd prime. Suppose that n | σ_{5}(n). Computation shows that n ≠ 496.

Case 1: p ≡ 1 (mod 4). By (9),

5^{2^{v}−1} ≤ p^{2^{v}−1} ≤ \frac{2^{5(v+1)} − 1}{2^{5} − 1},  (14)

which only holds if 1 ≤ v ≤ 2 by Lemma 19.

(i) v = 1. By (11), α = 2 then by (2), p | 33, which contradicts the congruence p ≡ 1 (mod 4).

(ii) v = 2. By (14), p ≤ 10 and so p = 5. By (11), 2 ≤ α ≤ 3. However, neither value of α satisfies (2).

Case 2: p ≡ 3 (mod 4). Note that because k = 5, we can ignore scenario (a) of Lemma 10. By (12),

3^{2^{v}−11} ≤ p^{2^{v}−11} < \frac{2^{5(v−1)}}{2^{5} − 1} ,  (15)

which implies 1 ≤ v ≤ 4 by Lemma 20.

(i) v = 4. By (15), p = 3. So, in (13), s = 3 and 2 ≤ α ≤ 6. If α ≤ 5, (2) gives

3^{15} | 3^{16β_{1}−1} ≤ \frac{2^{25} − 1}{31}, a contradiction.

If α = 6, (2) does not hold.

(ii) v = 3. Then β ≥ 8. By Lemma 10, either (2^{λ} − 1)^{β−1} ≤ 2^{λ+3} − 1 or (2^{λ} − 1)^{β−1} ≤ ∑_{i=0}^{4} 2^{i(λ+3)}.

(a) If (2^{λ} − 1)^{β−1} ≤ 2^{λ+3} − 1, then λ < 2 because β ≥ 8, a contradiction.

(b) If (2^{λ} − 1)^{β−1} ≤ ∑_{i=0}^{4} 2^{i(λ+3)}, then β ≤ 15 in order that λ ≥ 2. Since 8 | β, we know β = 8. Plugging β = 8 into (2^{λ} − 1)^{β−1} ≤ ∑_{i=0}^{4} 2^{i(λ+3)}, we have 2 ≤ λ ≤ 4 and so 2 ≤ s ≤ 5. By (13), 2 ≤ α ≤ 7 and by (2), we acquire

p^{7} | \frac{2^{5α} − 1}{31} ≤ \frac{2^{35} − 1}{31}.

Hence, p ∈ {3, 7, 11, 19}. Computation shows that for each pair (α, p), (2) does not hold.

(iii) v = 2. Then 4 |β. By Lemma 10, either (2^{λ} − 1)^{β−1} ≤ 2^{λ+2} − 1 or (2^{λ} − 1)^{β−1} ≤ ∑_{i=0}^{4} 2^{i(λ+2)}. Since β ≥ 4 and λ ≥ 2, the former does not hold. If the later, since λ ≥ 2, it must be that β < 12 and so β ∈ {4, 8}.

(a) β = 4. Lemma 7 rejects this case.

(b) β = 8. Plugging β = 8 into (2^{λ} − 1)^{β−1} ≤ ∑_{i=0}^{4} 2^{i(λ+2)}, we have 2 ≤ λ ≤ 3 and so 2 ≤ s ≤ 4. By (13), 2 ≤ α ≤ 5. We are back to item (ii) part (b).

(iv) v = 1. By Lemma 10, either (2^{λ} − 1)^{β−1} ≤ 2^{λ+1} − 1 or (2^{λ} − 1)^{β−1} ≤ ∑_{i=0}^{4} 2^{i(λ+1)}. If the former, β = 2 and n = 2^{α−1}p. By Theorem 3, n is an even perfect number. If the latter, since λ ≥ 2, it must be that β ≤ 9 and so β ∈ {2, 6}.

(a) If β = 2, Theorem 3 guarantees that n is an even perfect number.

(b) If β = 6, then 2 ≤ λ ≤ 4 and so 2 ≤ s ≤ 5. By (13), 2 ≤ α ≤ 5 and by (2), we acquire

p^{5} | \frac{2^{5α} − 1}{31} ≤ \frac{2^{25} − 1}{31}.

Hence, p ∈ {3, 7, 11}. Computation shows that for each pair (α, p), (2) does not hold.

We have finished the proof.

5 Acknowledgments

The author would like to thank the anonymous referee for helpful suggestions that improve the exposition of this paper.

A Technical proofs used for Lemma 6

We provide proofs of claim(s) made in the proof of Lemma 6. Notation from Lemma 6 is retained here.

Lemma 11. For all u ≥ 1 and k ≥ 3, we have

2^{(2^{k}−1)^{u}−k} > u + 2^{k} − 1.

Proof. We prove by induction on u. For u = 1, it is clear that 2^{2^{k}−1−k} > 2^{k} for all k ≥ 3. Assume that the inequality holds for u = n ≥ 1 and for all k ≥ 3. We want to show that the inequality holds for u = n + 1 and for all k ≥ 3. Fixing k ≥ 3, we have

2^{(2^{k}−1)^{n+1}−k} = 2^{−k}(2^{(2^{k}−1)^{n}})^{2^{k}−1} > 2^{−k}(2^{k}(n + 2^{k} − 1))^{2^{k}−1}

by the inductive hypothesis. Hence, it suffices to show that 2^{−k}(2^{k}(n + 2^{k} − 1))^{2^{k}−1} ≥ n + 2^{k}; equivalently, 2^{k(2^{k}−1)}(n + 2^{k} − 1)^{2^{k}−1} ≥ 2^{k}(n + 2^{k}). Because 2^{k(2^{k}−1)} ≥ 2^{k}, it remains to show

(n + 2^{k} − 1)^{2^{k}−1} ≥ n + 2^{k}.

Let ℓ = 2^{k} − 1. The above inequality becomes

(n + ℓ)^{ℓ} − 1 ≥ n + ℓ.

Equivalently,

(n + ℓ)^{ℓ−1} − \frac{1}{n + ℓ} ≥ 1,

which is true because (n + ℓ)^{ℓ−1} ≥ 2.

Lemma 12. For all odd k ≥ 3, we have 2^{k+1} || (2^{k} − 1)^{2k} − 1.

Proof. Write

(2^{k} − 1)^{2k} − 1 = ∑_{i=0}^{2k} \binom{2k}{i}(2^{k})^{2k−i}(−1)^{i} − 1 = ∑_{i=0}^{2k−1} \binom{2k}{i}(2^{k})^{2k−i}(−1)^{i}.

When i = 2k − 1, we have the term −2k · 2^{k} = −k2^{k+1}. Because k is odd, 2^{k+1} || k2^{k+1}. This finishes our proof.

Lemma 13. The following holds

2^{v+k} || (2^{k} − 1)^{βk} − 1.

Proof. We prove by induction on v. When v = 1, write

(2^{k} − 1)^{βk} − 1 = (2^{k} − 1)^{2kβ_{1}} − 1 = ((2^{k} − 1)^{2k} − 1) ∑_{i=1}^{β_{1}}(2^{k} − 1)^{2k(β_{1}−i)}.

Because the summation is 1 mod 2 and by Lemma 12, 2^{k+1} || (2^{k} − 1)^{2k} − 1, our claim holds for v = 1. Inductive hypothesis: suppose that there exists z ≥ 1 such that the claim holds for all v ∈ [1, z]. We show that the claim holds for v = z + 1. We have

(2^{k} − 1)^{2^{z+1}β_{1}k} − 1 = ((2^{k} − 1)^{2^{z}β_{1}k} − 1)((2^{k} − 1)^{2^{z}β_{1}k} + 1).

By the inductive hypothesis, 2^{z+k} || (2^{k} − 1)^{2^{z}β_{1}k} − 1, so it suffices to show that 2 || (2^{k} − 1)^{2^{z}β_{1}k} + 1. Observe that

(2^{k} − 1)^{2^{z}β_{1}k} + 1 = (4^{k} − 2^{k+1} + 1)^{2^{z−1}β_{1}k} + 1 ≡ 2 (mod 4).

Hence, 2 || (2^{k} − 1)^{2^{z}β_{1}k} + 1, as desired. This completes our proof.

Lemma 14. Let m be chosen such that (2^{k} − 1)^{m} || 2^{(2^{k}−1)k} − 1. Then for all u ≥ 0,

(2^{k} − 1)^{u+m} || 2^{(2^{k}−1)^{u+1}kα_{1}} − 1.

Proof. First, we claim that m ≥ 2. To prove this, write

2^{(2^{k}−1)k} − 1 = (2^{k} − 1) ∑_{i=2}^{2^{k}}(2^{k})^{(2^{k}−i)}.

Since each term in the summation is congruent to 1 mod 2^{k} − 1 and there are 2^{k} − 1 terms, the summation is divisible by 2^{k} − 1. Therefore, (2^{k} − 1)^{2} | 2^{(2^{k}−1)k} − 1.

We are ready to prove the lemma. We proceed by induction. Recall that in the proof of Lemma 6, we define α_{1} := α/(2^{k} − 1)^{u}. For u = 0, write

2^{(2^{k}−1)kα_{1}} − 1 = (2^{(2^{k}−1)k} − 1)(2^{(2^{k}−1)k(α_{1}−1)} + 2^{(2^{k}−1)k(α_{1}−2)} + ⋯ + 1)

= (2^{(2^{k}−1)k} − 1) ∑_{i=1}^{α_{1}}(2^{k})^{(2^{k}−1)(α_{1}−i)}.

By assumption, (2^{k} − 1)^{m} || 2^{(2^{k}−1)k} − 1. Each term in the summation ∑_{i=1}^{α_{1}}(2^{k})^{(2^{k}−1)(α_{1}−i)} is congruent to 1 mod 2^{k} − 1, so the summation is congruent to α_{1} mod 2^{k} − 1. Hence, our lemma holds for u = 0. Inductive hypothesis: suppose that there exists z ≥ 0 such that our lemma holds for all u ≤ z. We show that our lemma holds for u = z + 1. Write

2^{(2^{k}−1)^{z+2}kα_{1}} − 1 = (2^{(2^{k}−1)^{z+1}kα_{1}} − 1)·

(2^{(2^{k}−1)^{z+1}kα_{1}(2^{k}−2)} + 2^{(2^{k}−1)^{z+1}kα_{1}(2^{k}−3)} + ⋯ + 1)

= (2^{(2^{k}−1)^{z+1}kα_{1}} − 1) ∑_{i=2}^{2^{k}} 2^{(2^{k}−1)^{z+1}kα_{1}(2^{k}−i)}.

By the inductive hypothesis, (2^{k} − 1)^{z+m} || 2^{(2^{k}−1)^{z+1}kα_{1}} − 1. Each term in the summation is congruent to 1 mod (2^{k} − 1)^{m}. Since there are 2^{k} − 1 terms, the summation is congruent to (2^{k} − 1) mod (2^{k} − 1)^{m}. Because m ≥ 2, (2^{k} − 1) exactly divides the summation. So,

(2^{k} − 1)^{z+m+1} exactly divides 2^{(2^{k}−1)^{z+2}kα_{1}} − 1,

as desired. This completes our proof.

Remark 15. Note that for all k ≥ 3, in order that (2^{k} − 1)^{m} ≤ 2^{(2^{k}−1)k} − 1, we must have m < 2^{k}. By Lemma 14, (2^{k} − 1)^{u+2^{k}} does not divide 2^{(2^{k}−1)^{u+1}kα_{1}} − 1 for all u ≥ 0.

B Technical proofs used for Lemma 8

We provide proofs of claim(s) made in the proof of Lemma 8. Notation from Lemma 8 is retained here.

Lemma 16. With notation as in Lemma 8, the following holds

2^{t+v} || p^{2^{v}β_{1}k} − 1.

Proof. We prove by induction on v. When v = 1, write

p^{2kβ_{1}} − 1 = (p^{2k} − 1)(p^{2k(β_{1}−1)} + p^{2k(β_{1}−2)} + ⋯ + 1)

= (p^{k} − 1)(p^{k} + 1) ∑_{i=1}^{β_{1}} p^{2k(β_{1}−i)}

= (p^{k} − 1)(p + 1) (∑_{i=1}^{k} p^{k−i}(−1)^{i+1}) ∑_{i=1}^{β_{1}} p^{2k(β_{1}−i)}.  (16)

Since p + 1 ≡ 2 (mod 4), 2 || (p + 1). We showed that 2^{t}||(p^{k} − 1) in the proof of Lemma 8. Also, the two summations are odd. Therefore, 2^{t+1} || p^{2kβ_{1}} − 1.

Inductive hypothesis: suppose that there exists z ≥ 1 such that our claim holds for all v ∈ [1, z]. We show that our claim holds for v = z + 1. We have

p^{2^{z+1}kβ_{1}} − 1 = p^{(2^{z}kβ_{1})·2} − 1 = (p^{2^{z}kβ_{1}} + 1)(p^{2^{z}kβ_{1}} − 1).

By the inductive hypothesis, 2^{z+t} || p^{2^{z}kβ_{1}} − 1. Also, p ≡ 1 (mod 4) implies that p^{2^{z}kβ_{1}} + 1 ≡ 2 (mod 4). So, 2 || p^{2^{z}kβ_{1}} +1. Therefore, 2^{z+t+1} || p^{2^{z+1}kβ_{1}} −1. We have finished our proof.

C Technical proofs used for Lemma 9

We provide proofs of claim(s) made in the proof of Lemma 9. Notation from Lemma 9 is retained here.

Lemma 17. With notation as in Lemma 9, the following holds

2^{v+s−1} || p^{k2^{v}β_{1}} − 1.

Proof. We prove by induction on v. When v = 1, by (16), we only consider (p + 1)(p^{k} − 1). We showed that 2 || p^{k} − 1 in the proof of Lemma 9. Since 2^{s} || (p − 1)(p + 1) and 2 || p − 1, it follows that 2^{s−1} || p + 1. Therefore, 2^{s} || p^{k2β_{1}} − 1.

Inductive hypothesis: suppose that there exists z ≥ 1 such that for all v ∈ [1, z], our claim holds. We show that our claim also holds for v = z + 1. We have

p^{2^{z+1}kβ_{1}} − 1 = p^{(2^{z}kβ_{1})·2} − 1 = (p^{2^{z}kβ_{1}} + 1)(p^{2^{z}kβ_{1}} − 1).

By the inductive hypothesis, 2^{z+s−1} || p^{2^{z}kβ_{1}}−1. Also, p^{2} ≡ 1 (mod 4) implies that p^{2^{z}kβ_{1}}+1 ≡ 2 (mod 4). So, 2 || p^{2^{z}kβ_{1}} + 1. Therefore, 2^{z+s} || p^{2^{z+1}kβ_{1}} − 1. We have finished our proof.

D Technical proofs used for Lemma 10

We provide proofs of claim(s) made in the proof of Lemma 10. Notation from Lemma 10 is retained here.

Lemma 18. With notation as in Lemma 10, the following holds

2^{λ+v} || (2^{λ}p_{1} − 1)^{2^{v}β_{1}} − 1.

Proof. We prove by induction on v. Observe that

(2^{λ}p_{1} − 1)^{2β_{1}} − 1 = ∑_{i=0}^{2β_{1}} \binom{2β_{1}}{i}(2^{λ}p_{1})^{2β_{1}−i}(−1)^{i} − 1

= ∑_{i=0}^{2β_{1}−1} \binom{2β_{1}}{i}(2^{λ}p_{1})^{2β_{1}−i}(−1)^{i},

which clearly indicates that 2^{λ+1} || (2^{λ}p_{1} −1)^{2β_{1}} −1. So, the claim holds for v = 1. Inductive hypothesis: suppose that there exists z ≥ 1 such that for all v ∈ [1, z], the claim holds. We prove that the claim holds for v = z + 1. We have

(2^{λ}p_{1} − 1)^{2^{z+1}β_{1}} − 1 = ((2^{λ}p_{1} − 1)^{2^{z}β_{1}} − 1)((2^{λ}p_{1} − 1)^{2^{z}β_{1}} + 1).

By the inductive hypothesis, 2^{λ+z} || (2^{λ}p_{1} − 1)^{2^{z}β_{1}} − 1. Also, (2^{λ}p_{1} − 1)^{2^{z}β_{1}} + 1 ≡ 2 (mod 4) since λ ≥ 2. Hence, 2^{λ+z+1} || (2^{λ}p_{1} − 1)^{2^{z+1}β_{1}} − 1, as desired.

E Technical proofs used for Theorem 4

Lemma 19. If v ≥ 1 and 5^{2^{v−1}} ≤ \frac{2^{5(v+1)}−1}{31}, then 1 ≤ v ≤ 2.

Proof. The inequality 5^{2^{v−1}} ≤ \frac{2^{5(v+1)}−1}{31} implies that

5^{2^{v}−1} ≤ 5^{5(v+1)},

which is equivalent to 2^{v} −1 ≤ 5(v +1). Clearly, we have 1 ≤ v ≤ 4. However, the inequality 5^{2^{v−1}} ≤ \frac{2^{5(v+1)}−1}{31} does not hold when v ∈ \{3, 4\}. We conclude that 1 ≤ v ≤ 2.

Lemma 20. If v ≥ 1 and 3^{2^{v}−11} < \frac{2^{5(v−1)}}{31} , then 1 ≤ v ≤ 4.

Proof. The inequality 3^{2^{v}−11} < \frac{2^{5(v−1)}}{31} implies that 3^{2^{v}−11} < 3^{5(v−1)}. Hence, 2^{v} < 5v + 6, which holds only if 1 ≤ v ≤ 4.

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2010 Mathematics Subject Classification: 11A25. Keywords: perfect number, divisor function.

(Concerned with sequences \underline{A000396}, \underline{A181595}, \underline{A271816}, and \underline{A341475}.) Received October 20 2020; revised version received February 13 2021. Published in Journal of Integer Sequences, February 13 2021. Return to Journal of Integer Sequences home page.