Unitary super perfect numbers^{∗}^{†}

Tomohiro Yamada

Abstract

We shall show that 9, 165 are all of the odd unitary super perfect numbers.

1 Introduction

We denote by σ(N) the sum of divisors of N. N is called to be perfect if σ(N) = 2N. It is a well-known unsolved problem whether or not an odd perfect number exists. Interest to this problem has produced many analogous notions.

D. Suryanarayana [10] called N to be super perfect if σ(σ(N)) = 2N. It is asked in this paper and still unsolved whether there were odd super perfect numbers.

A special class of divisors is the class of unitary divisors defined by Cohen [2]. A divisor d of n is called a unitary divisor if (d, n/d) = 1. Then we write d || n. We denote by σ^{∗}(N) the sum of unitary divisors of N. Replacing σ by σ^{∗}, Subbarao and Warren [9] introduced the notion of a unitary perfect number. N is called to be unitary perfect if σ^{∗}(N) = 2N. They proved that there are no odd unitary perfect numbers. Moreover, Subbarao [8] conjectured that there are only finitely many unitary perfect numbers.

Combining these two notions, Sitaramaiah and Subbarao [6] studied unitary super perfect (USP) numbers, integers N satisfying σ^{∗}(σ^{∗}(N)) = 2N. They found all unitary super perfect numbers below 10^{8}. The first ones are 2, 9, 165, 238. Thus there are both even and odd USPs. They proved that another odd USP must have at least four distinct prime factors and conjectured that there are only finitely many odd USPs.

∗2000 Mathematics Subject Classification: 11A05, 11A25.

†Key words: Odd perfect numbers. Super perfect numbers. Unitary divisors.

The purpose of this paper is to prove this conjecture. Indeed, we show that the known two USPs are all.

Theorem 1.1. If N is an odd USP, then N = 9 or N = 165.

Our proof is completely elementary. The key point of our proof is the fact that if N is an odd USP, then σ^{∗}(N) must be of the form 2^{f_{1}}q^{f_{2}}, where q is an odd prime. This yields that if p^{e} is an unitary divisor of N, then p^{e} + 1 must be of the form 2^{a}q^{b}. Moreover, elementary theory of cyclotomic polynomials and quadratic residues gives that a ≤ 2 or b = 0. Hence p^{e} belongs a to very thin set. Using this fact, we deduce that q must be small. For each small primes q, we show that σ^{∗}(σ^{∗}(N))/N < 2 and therefore N cannot be an USP unless N = 9, 165, with the aid of the fact that f_{1}, f_{2} must be fairly large. We sometimes use facts already stated in [6] but we shall present proofs of these facts when proofs are omitted in [6].

Our method does not seem to work to find all odd super perfect numbers. Since σ(σ(N)) = 2N does not seem to imply that ω(σ(N)) ≤ 2. Even assuming that ω(σ(N)) ≤ 2, the property of σ that σ(p^{e})/p^{e} > 1 + 1/p prevents us from showing that σ(σ(N)) < 2. Nevertheless, with the aid of a theory of exponential diophantine equations, we can show that for any given k, there are only finitely many odd super perfect numbers N with ω(σ(N)) ≤ k.

2 Preliminary Lemmas

Let us denote by v_{p}(n) the solution e of p^{e}||n. For distinct primes p and q, we denote by o_{q}(p) the exponent of p mod q and we define a_{q}(p) = v_{q}(p^{d} − 1), where d = o_{q}(p). Clearly o_{q}(p) divides q − 1 and a_{q}(p) is a positive integer. Now we quote some elementary properties of v_{q}(σ(p^{x})). Lemmas 2.1 is wellknown. Lemma 2.1 has been proved by Zsigmondy[12] and rediscovered by many authors such as Dickson[3] and Kanold[4]. See also Theorem 6.4A.1 in [5].

Lemma 2.1. If a > b ≥ 1 are coprime integers, then a^{n} − b^{n} has a prime factor which does not divide a^{m} − b^{m} for any m < n, unless (a, b, n) = (2, 1, 6) or a − b = n = 1, or n = 2 and a + b is a power of 2.

By Lemma 2.1, we obtain the following lemmas.

Lemma 2.2. Let p, q be odd primes and e be a positive integer. If p^{e} + 1 = 2^{a}q^{b} for some integers a and b, then one of the following holds:

a)e = 1.

b)e is even and q ≡ 1 (mod 2e).

c)p is a Mersenne prime and q ≡ 1 (mod 2e).

Proof. We first show that if a) does not hold, then either b) or c) must hold. Since (p, e) ≠ (2, 3) and e ≠ 1, it follows from Lemma 2.1 that p^{2e} − 1 has a prime factor r which does not divide p^{m} − 1 for any m < 2e. Since the order of p (mod r) is 2e, r ≡ 1 (mod 2e). Since r is odd and does not divide p^{e} − 1, r divides p^{e} + 1 and therefore q = r.

If e is even, then b) holds. Assume that e is odd. If p + 1 has an odd prime factor, then this cannot be equal to q and must be a prime factor of p^{e} + 1 = 2^{a}q^{b}, which is contradiction. Thus p is a Mersenne prime and c) follows. ♢

Lemma 2.3. Let p be an odd prime and e be a positive integer. If p^{e} + 1 = 2^{a}3^{b} for some integers a and b, then e = 1.

Proof. By Lemma 2.2, e = 1 or 3 ≡ 1 (mod 2e). The latter is equivalent to e = 1. ♢

Lemma 2.4. Let p be an odd prime and e, x be positive integers. If p^{e} +1 = 2^{x}, then e = 1.

Proof. If e > 1, then by Lemma 2.1, p^{2e} − 1 has a prime factor which does not divide p^{m} −1 for any m < 2e. This prime factor must be odd and divide p^{e} + 1, which violates the condition p^{e} + 1 = 2^{x}. ♢

Lemma 2.5. Let p be an odd prime and e, x be positive integers. If 2^{x} +1 = 3^{e}, then (e, x) = (1, 1) or (2, 3).

Proof. We apply Lemma 2.1 with (a, b, n) = (3, 1, e). If e > 2, then 3^{e} − 1 has a prime factor which does not divide 3 − 1 = 2. ♢

Lemma 2.6. If a prime p divides 2^{a} + 1 for some integer a, then p is congruent to 1, 3 or 5 (mod 8).

Proof. If a is even, then it is well known that p ≡ 1 (mod 4). If a is odd, then p divides 2x^{2} + 1 with x = 2^{(a−1)/2}. We have (−2/p) = 1 and therefore p ≡ 1 or 3 (mod 8). ♢ Lemma 2.7. Let p and q be odd primes and b be a positive integer. If a prime p divides q^{b} + 1 and 4 does not divide q^{b} + 1, then 4q does not divide p + 1.

Proof. If b is even, then p ≡ 1 (mod 4) and clearly 4q does not divide p + 1.

If b is odd, then we have (−q/p) = 1 and q ≡ 1 (mod 4). Assume that q divides p + 1. Since q ≡ 1 (mod 4), we have, by the reciprocity law, (−q/p) = (−1/p)(q/p) = (−1/p)(p/q) = (−1/p)(−1/q) = (−1/p). Thus (−1/p) = 1 and p ≡ 1 (mod 4) and therefore 4 does not divide p + 1. ♢

3 Basic properties of odd USPs

In this section, we shall show some basic properties of odd USPs.

We write N = p_{1}^{e_{1}}p_{2}^{e_{2}} … p_{k}^{e_{k}}, where p_{1}, p_{2}, …, p_{k} are distinct primes. Moreover, we denote by C the constant

(3.1)  ∏_{p,2^{p}−1 is prime} \frac{2^{p}}{2^{p} − 1} < 1.6131008.

This upper bound follows from the following estimate:

∏_{p,2^{p}−1 is prime} \frac{2^{p}}{2^{p} − 1} < \frac{4}{3} · ( ∏_{n≥3,n is odd} \frac{2^{n}}{2^{n} − 1})

< \frac{4}{3} · exp ( ∑_{n≥3,n is odd} \frac{1}{2^{n} − 1})

< \frac{4}{3} · exp (\frac{1}{7} ∑_{n≥0} \frac{1}{4^{n}} )

= \frac{4}{3} · exp (\frac{4}{21}) = 1.631007 · · · .

(3.2)

Lemma 3.1. If N is an odd USP, then σ^{∗}(N) = 2^{f_{1}}q^{f_{2}} for some odd prime q and positive integers f_{1}, f_{2}. Moreover, q^{f_{2}} + 1 is not divisible by 4.

Proof. Since N is odd, σ^{∗}(N) must be even. Moreover, since σ^{∗}(σ^{∗}(N)) = 2N with N odd, σ^{∗}(N) has exactly one odd prime factor. Hence σ^{∗}(N) = 2^{f_{1}}q^{f_{2}} for some odd prime q and positive integers f_{1}, f_{2}. Since σ^{∗}(q^{f_{2}}) = q^{f_{2}} + 1 divides σ^{∗}(σ^{∗}(N)) = 2N, 4 does not divide q^{f_{2}} + 1. ♢

4 Q CANNOT BE 3 5

Henceforth, we let N ≠ 9, 165 be an odd USP and write σ^{∗}(N) = 2^{f_{1}}q^{f_{2}} as allowed by Lemma 3.1.

Lemma 3.2. Unless p_{i} is a Mersenne prime and e_{i} is odd, we have p_{i}^{e_{i}} = 2^{a_{i}}q^{b_{i}} − 1 for some positive integers a_{i} and b_{i} with a_{i} ≤ 2. Moreover, f_{1} = ∑_{i=1}^{k} a_{i} and f_{2} = ∑_{i=1}^{k} b_{i}.

Proof. Since σ^{∗}(p_{i}^{e_{i}}+1) divides σ^{∗}(N) = 2^{f_{1}}q^{f_{2}}, we can write p_{i}^{e_{i}}+1 = 2^{a_{i}}q^{b_{i}} with some nonnegative integers a_{i} and b_{i}. Since p_{i} is odd and non-Mersenne, a_{i} and b_{i} are positive by Lemma 2.4.

If e_{i} is even, then p_{i}^{e_{i}} + 1 ≡ 2 (mod 4). Hence a_{i} = 1.

Assume that p_{i} is not a Mersenne prime and e_{i} is odd. By Lemma 2.2, we have e_{i} = 1 and therefore p_{i} = p_{i}^{e_{i}} = 2^{a_{i}}q^{b_{i}}−1. By Lemma 2.6 and 2.7, we have a_{i} ≤ 2 since q^{f_{2}} + 1 is not divisible by 4. This shows a_{i} ≤ 2. The latter part of the lemma immediately follows from 2^{f_{1}}q^{f_{2}} = σ^{∗}(N) = ∏(p_{i}^{e_{i}} + 1). ♢

Lemma 3.3. ω(N) ≥ 3.

Proof. First we assume that N = p_{1}^{e_{1}}. Since we have σ^{∗}(N)/N = 1 + 1/N and σ^{∗}(σ^{∗}(N))/σ^{∗}(N) ≤ (1 + 1/2)(1 + 2/N) by Lemma 3.1, we have N ≤ 9. We can easily confirm that N = 9 is the sole odd USP with N ≤ 9.

Next we assume that N = p_{1}^{e_{1}}p_{2}^{e_{2}}. Since we have σ^{∗}(N)/N ≤ (1+1/3)(1+ 3/N) and σ^{∗}(σ^{∗}(N))/σ^{∗}(N) ≤ (1 + 1/4)(1 + 4/N), we have N < 37. We can easily confirm that there is no odd USP N with N < 37 and ω(N) = 2.

Another proof of impossibility of ω(N) = 1 unless N = 2, 9 (whether N is even or odd) can be found in [6, Theorem 3.2] and impossibility of ω(N) = 2 (again, N may be even) is stated in [6, Theorem 3.3] with their proof presented only in the case N is even. ♢

4 q cannot be 3

In this section, we show that q ≠ 3. There are two cases: the case 3 | N and the case 3 ∤ N.

Proposition 4.1. If 3 ∤ N and 3 | σ^{∗}(N), then f_{1} and f_{2} are even, p_{i} has the form 2 · 3^{b_{i}} − 1 with positive integers b_{i}.

Proof. We have e_{i} = 1 by Lemma 2.3. Thus any p_{i} must be of the form 2^{a_{i}} · 3^{b_{i}} − 1 with nonnegative integers a_{i}, b_{i}. Since 3^{f_{2}} + 1 is not divisible by 4, f_{2} must be even. Since 3 does not divide 2^{f_{1}} + 1, f_{1} must also be even. By Lemma 2.6, any prime factor of N is congruent to 1 (mod 4) and therefore a_{i} must be odd. By Lemma 3.2, we have a_{i} = 1. ♢

Hence we have p_{i} ∈ {5, 17, 53, 4373, …}.

Lemma 4.2. If 3 | σ^{∗}(N), then 3 | N.

Proof. Suppose 3 | σ^{∗}(N) and 3 ∤ N. By Proposition 4.1, we have

(4.1)  \frac{σ^{∗}(N)}{N} ≤ \frac{6}{5} · \frac{18}{17} · \frac{54}{53} · (∏_{i=7}^{∞} \frac{2 · 3^{i}}{2 · 3^{i} − 1}) .

Since

(4.2)  ∏_{i=7}^{∞} \frac{2 · 3^{i}}{2 · 3^{i} − 1} ≤ exp ∑_{i=7}^{∞} \frac{1}{2 · 3^{i} − 1} ≤ exp (\frac{1}{2 · 3^{7} − 1} ∑_{i=0}^{∞} 3^{−i}) ,

we have

(4.3)  \frac{σ^{∗}(N)}{N} < \frac{6}{5} · \frac{18}{17} · \frac{54}{53} · exp (\frac{3}{8744}) .

Since k ≥ 3 by Lemma 3.3, we have f_{1} = k ≥ 3 and f_{2} ≥ 3 + 2 + 1 = 6. Thus we obtain

(4.4)  \frac{σ^{∗}(σ^{∗}(N))}{σ^{∗}(N)} ≤ \frac{9}{8} · \frac{730}{729}.

Multiplying (4.3) and (4.4), we obtain

(4.5) 2 = \frac{σ^{∗}(σ^{∗}(N))}{N} < \frac{9}{8} · \frac{730}{729} · \frac{6}{5} · \frac{18}{17} · \frac{54}{53} · exp (\frac{3}{8744}) = 1.4588 ⋯ < 2,

which is contradiction. ♢

Lemma 4.3. It is impossible that 3 | N and 3 | σ^{∗}(N).

Proof. Suppose 3 | N and 3 | σ^{∗}(N). We have e_{i} = 1 by Lemma 2.3. By Lemma 2.6, 2^{a_{i}} + 1 is divisible by no Mersenne prime other than 3. Since 3^{b_{i}} + 1 cannot be divisible by 4, b_{i} must be odd and therefore 3^{b_{i}} + 1 is divisible by no Mersenne prime. Hence it follows from Lemma 3.2 that any p_{i} must be of the form 2^{a_{i}} · 3^{b_{i}} − 1, where a_{i} ≤ 2 and b_{i} are positive integers. Hence p_{i} ∈ {5, 11, 17, 53, 107, 971, 4373, …}.

Thus we obtain

(4.6) \frac{σ^{∗}(N)}{N} ≤ \frac{4}{3} · \frac{6}{5} · \frac{12}{11} · \frac{18}{17} · \frac{54}{53} · \frac{108}{107} · (∏_{i=7}^{∞} \frac{2 · 3^{i}}{2 · 3^{i} − 1}) · (∏_{i=5}^{∞} \frac{4 · 3^{i}}{4 · 3^{i} − 1}) .

As in the proof of the previous lemma, substituting the inequality

(4.7)  ∏_{i=5}^{∞} \frac{4 · 3^{i}}{4 · 3^{i} − 1} ≤ exp (\frac{1}{4 · 3^{5} − 1} ∑_{i=0}^{∞} 3^{−i})

we have

(4.8)  \frac{σ^{∗}(N)}{N} ≤ \frac{4}{3} · \frac{6}{5} · \frac{12}{11} · \frac{18}{17} · \frac{54}{53} · \frac{108}{107} · exp (\frac{3}{8744} + \frac{3}{1942}) .

Since k ≥ 46 by [6, Theorem 3.4], we have

(4.9)  \frac{σ^{∗}(σ^{∗}(N))}{σ^{∗}(N)} ≤ \frac{2^{46} + 1}{2^{46}} · \frac{3^{45} + 1}{3^{45}}.

Multiplying (4.8) and (4.9), we obtain

\begin{aligned}2 & = \frac{σ^{∗}(σ^{∗}(N))}{N} \\ & ≤ \frac{2^{46} + 1}{2^{46}} · \frac{3^{45} + 1}{3^{45}} · \frac{4}{3} · \frac{6}{5} · \frac{12}{11} · \frac{18}{17} · \frac{54}{53} · \frac{108}{107} · exp (\frac{3}{8744} + \frac{3}{1942}) \\ & ≤ 1.9041 ⋯ < 2, \end{aligned} (4.10)

which is contradiction. ♢

It immediately follows from these two lemmas that q ≠ 3.

5 The remaining part

The remaining case is the case 3 ∤ σ^{∗}(N), i.e., q ≠ 3.

Lemma 5.1. Suppose p_{i} is not a Mersenne prime. Then p_{i}^{e_{i}} has the form 2^{a_{i}} · q^{b_{i}} − 1 with positive integers a_{i} ≤ 2 and b_{i}. Moreover, for any integer b, at most one of the pairs (1, b) and (2, b) appear in (a_{i}, b_{i})’s.

Proof. The former part follows from Lemma 3.2. Since q ≠ 3, 3 divides at least one of 2 · q^{b} − 1 and 4 · q^{b} − 1. If both pairs (a_{i}, b_{i}) = (1, b) and (a_{j}, b_{j}) = (2, b) appear, then at least one of p_{i}^{e_{i}} and p_{j}^{e_{j}} must be a power of three, which violates the condition that p_{i} and p_{j} are not Mersenne. ♢ Lemma 5.2. q ≤ 13. Furthermore, provided f_{2} ≥ 2, we have q = 5 or q = 7.

Proof. By Lemma 5.1, we have

(5.1)  \frac{σ^{∗}(N)}{N} ≤ C · (∏_{a=1}^{∞} \frac{2 · q^{a}}{2 · q^{a} − 1}) .

Since ∏_{a=1}^{∞} 2 · q^{a}/(2 · q^{a} − 1) ≤ exp (q/ {(q − 1)(2q − 1)}), we have

(5.2)  \frac{σ^{∗}(N)}{N} ≤ C · exp (\frac{q}{(q − 1)(2q − 1)}) .

By Lemma 3.3, we have

(5.3)  \frac{σ^{∗}(σ^{∗}(N))}{σ^{∗}(N)} ≤ \frac{2^{f_{1}} + 1}{2^{f_{1}}} · \frac{q^{f_{2}} + 1}{q^{f_{2}}} ≤ \frac{2^{3} + 1}{2^{3}} · \frac{q^{f_{2}} + 1}{q^{f_{2}}}.

Combining these inequalities, we obtain

(5.4) 2 ≤ \frac{σ^{∗}(σ^{∗}(N))}{N} ≤ \frac{2^{3} + 1}{2^{3}} · C · \frac{q^{f_{2}} + 1}{q^{f_{2}}} · exp (\frac{q}{(q − 1)(2q − 1)}) .

Hence

(5.5)  \frac{q^{f_{2}} + 1}{q^{f_{2}}} · exp (\frac{q}{(q − 1)(2q − 1)}) ≥ \frac{16}{9C} ≥ 1.102087.

This yields q ≤ 13. If f_{2} ≥ 2, then this inequality yields q ≤ 7. ♢

Theorem 5.3. q ≠ 5.

Proof. Suppose that q = 5. Then we have p_{i}^{e_{i}} = 2 · 5^{b_{i}} − 1 or p_{i}^{e_{i}} = 4·5^{b_{i}}−1 or p_{i} is Mersenne. Hence p_{i}^{e_{i}} ∈ {19, 499, 7812499, …, 9, 49, 1249, …, 3, 7, 31, 127, 8191, …}. We note that 9 = 3^{2} and 49 = 7^{2}.

Let us assume that 19 | N. Then f_{1} ≡ 9 (mod 18) and hence 3^{3} | N. By (3.2), we have

(5.6)  \frac{σ^{∗}(N)}{N} ≤ \frac{3}{4} · \frac{28}{27} · C · exp (\frac{5}{36}) .

Since f_{1} ≥ 9, we have

(5.7)  \frac{σ^{∗}(σ^{∗}(N))}{N} ≤ \frac{2^{9} + 1}{2^{9}} · \frac{6}{5} · \frac{7}{9} · C · exp (\frac{5}{36}) = 1.7332 ⋯ < 2,

which is contradiction. Thus 19 cannot divide N. From this we deduce that if p_{i}^{e_{i}} = 2 · 5^{b_{i}} − 1 or p_{i}^{e_{i}} = 4 · 5^{b_{i}} − 1, then b_{i} ≥ 3.

It is impossible that 7 | N since 7 does not divide 2^{x} + 1 or 5^{x} + 1 for any integer x.

Hence, by Lemma 3.3 we have

(5.8)  \frac{σ^{∗}(σ^{∗}(N))}{N} ≤ \frac{7}{8} · C · exp (\frac{5}{4} · \frac{250}{249}) · \frac{6}{5} · \frac{9}{8} < 1.9150 ⋯ < 2.

So that, we cannot have q = 5. ♢

Theorem 5.4. q ≠ 7, 11, 13.

Proof. Suppose q = 7. Observing that 4·7^{b} −1 is divisible by 3, we deduce from Lemma 3.2 that, for any i, p_{i} is a Mersenne prime or p_{i}^{e_{i}} = 2 · 7^{b_{i}} − 1. By Lemma 2.6, (2^{f_{1}} + 1)(7^{f_{2}} + 1) is not divisible by 7. Hence

\begin{aligned}\frac{σ^{∗}(N)}{N} & ≤ \frac{4}{3} · (∏_{i=2}^{∞} \frac{2^{2i+1}}{2^{2i+1} − 1}) · (∏_{i=1}^{∞} \frac{2 · 7^{i}}{2 · 7^{i} − 1}) \\ & ≤ \frac{4}{3} · exp (\frac{1}{31} · \frac{4}{3} + \frac{1}{13} · \frac{8}{7}) . \end{aligned} ; (5.9)

By Lemma 3.3, we have k ≥ 3. We deduce from Lemma 3.2 that we can take an integer s with 1 ≤ s ≤ 3 for which the following statement holds: there is at least 3 − s indices i such that p_{i} is a Mersenne prime and e_{i} is odd, and there is at least s indices i such that p_{i}^{e_{i}} = 2 · 7^{b_{i}} − 1. If s = 1, then f_{1} ≥ 6 and f_{2} ≥ 1. If s = 2, then f_{1} ≥ 4 and f_{2} ≥ 3. If s = 3, then f_{1} ≥ 3 and f_{2} ≥ 6.

\begin{aligned}\frac{σ^{∗}(σ^{∗}(N))}{σ^{∗}(N)} & ≤ max \{\frac{2^{6} + 1}{2^{6}} · \frac{8}{7} · \frac{2^{4} + 1}{2^{4}} · \frac{7^{3} + 1}{7^{3}}, \frac{2^{3} + 1}{2^{3}} · \frac{7^{6} + 1}{7^{6}}\} \\ & ≤ \frac{65}{56}. \end{aligned} ; (5.10)

Combining two inequalities (5.9) and (5.10), we have

(5.11) \frac{σ^{∗}(σ^{∗}(N))}{N} ≤ \frac{65}{56} · \frac{4}{3} · exp (\frac{1}{31} · \frac{4}{3} + \frac{1}{13} · \frac{8}{7}) = 1.7604 ⋯ < 2,

which is contradiction.

Suppose q = 11. Observing that 2 · 11^{2b+1} − 1 and 4 · 11^{2b} − 1 is divisible by 3, we deduce from Lemma 3.2 that, for any i, p_{i} is a Mersenne prime or p_{i}^{e_{i}} = 2^{a_{i}} · 7^{b_{i}} − 1 with a_{i} + b_{i} odd.

\begin{aligned}\frac{σ^{∗}(N)}{N} & ≤ \frac{4}{3} · (∏_{i=2}^{∞} \frac{2^{2i+1}}{2^{2i+1} − 1}) · (∏_{i=1}^{∞} \frac{2 · 11^{i}}{2 · 11^{i} − 1}) \\ & ≤ \frac{4}{3} · exp (\frac{1}{31} · \frac{4}{3} + \frac{1}{21} · \frac{12}{11}) . \end{aligned} ; (5.12)

In a similar way to derive (5.10), we obtain

(5.13)  \frac{σ^{∗}(σ^{∗}(N))}{σ^{∗}(N)} ≤ \frac{2^{3} + 1}{2^{3}} · \frac{11^{6} + 1}{11^{6}}.

Combining these inequalities, we have

\begin{aligned}2 = \frac{σ^{∗}(σ^{∗}(N))}{N} & ≤ \frac{2^{3} + 1}{2^{3}} · \frac{11^{6} + 1}{11^{6}} · \frac{4}{3} · \frac{8}{7} · exp (\frac{1}{31} · \frac{4}{3} + \frac{1}{21} · \frac{12}{11}) \\ & ≤ 1.8850 ⋯ < 2, \end{aligned} ; (5.14)

which is contradiction.

Suppose q = 13. 3 | N and 3 ∤ (q^{f_{2}} +1) since q = 13 ≡ 1 (mod 3). Hence f_{1} must be odd. Moreover, f_{2} = 1 by Lemma 5.2. Hence σ^{∗}(N) = 2^{f_{1}} · 13 and N = 7(2^{f_{1}} +1). There is exactly one index j such that p_{j}^{e_{j}} is of the form 2^{a}13^{b} − 1 for some positive integers a, b. By Lemma 3.2, we have a ≤ 2. Moreover, we have b = 1 since b ≤ f_{2} = 1. Hence p_{j}^{e_{j}} = 25 = 5^{2}. Since 13^{f_{2}} + 1 = 2 · 7, 2^{f_{1}} + 1 must be divisible by 5. But this is impossible since f_{1} is odd. ♢

Now Theorem 1.1 is clear. By Lemma 5.2, q must be one of 3, 5, 7, 11, 13. In the previous section, it is shown that q ≠ 3. Theorem 5.3 shows that q ≠ 5. Theorem 5.4 eliminates the remaining possibilities.

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Tomohiro Yamada Department of Mathematics Faculty of Science Kyoto University Kyoto, 606-8502 Japan

e-mail: tyamada@math.kyoto-u.ac.jp