On the simultaneous equations $\sigma(2^a)=p^{f_1}q^{g_1},   \sigma(3^b)=p^{f_2}q^{g_2}, \sigma(5^c)=p^{f_3}q^{g_3}$

Tomohiro Yamada

arXiv:1703.10555·math.NT·March 31, 2017

On the simultaneous equations $\sigma(2^a)=p^{f_1}q^{g_1}, \sigma(3^b)=p^{f_2}q^{g_2}, \sigma(5^c)=p^{f_3}q^{g_3}$

Tomohiro Yamada

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TL;DR

This paper investigates solutions to a set of simultaneous equations involving the sum-of-divisors function applied to powers of 2, 3, and 5, expressed as products of two distinct primes, aiming to classify all such solutions.

Contribution

The paper provides a complete solution to the system of equations involving the sum-of-divisors function and two primes, extending understanding of divisor sum equations with prime factorizations.

Findings

01

Classified all solutions to the equations involving σ(2^a), σ(3^b), σ(5^c).

02

Established conditions under which the equations hold for distinct primes p and q.

03

Enhanced the understanding of divisor sum equations with prime power arguments.

Abstract

We shall solve the simultaneous equations $σ (2^{a}) = p^{f_{1}} q^{g_{1}}, σ (3^{b}) = p^{f_{2}} q^{g_{2}}, σ (5^{c}) = p^{f_{3}} q^{g_{3}}$ with $p, q$ distinct primes.

Tables6

Table 1. Table 1: The residual orders of 3 , 5 3 5 3,5 modulo p 𝑝 p for p = 2 e 1 − 1 𝑝 superscript 2 subscript 𝑒 1 1 p=2^{e_{1}}-1

$e_{1}$	$o_{p} (3)$	$o_{p} (5)$
$2$	N/A	$2$
$3$	$6$	$6$
$5$	$30$	$3$
$7$	$126$	$42$
$13$	$910$	$1365$
$17$	$131070$	$65535$
$19$	$524286$	$74898$
$31$	$715827882$	$195225786$
$61$	$(10)$	$(15)$
$89$	$(6)$	$(84)$
$107$	$(6)$	$(6)$
$127$	$(6)$	$(6)$

Table 2. Table 2: The residual orders of 3 , 5 3 5 3,5 modulo p , q 𝑝 𝑞 p,q for p q = 2 e 1 − 1 , p < q formulae-sequence 𝑝 𝑞 superscript 2 subscript 𝑒 1 1 𝑝 𝑞 pq=2^{e_{1}}-1,p<q

$e_{1}$	$2^{e_{1}} - 1 = p q$	$o_{p} (3)$	$o_{q} (3)$	$o_{p} (5)$	$o_{q} (5)$
$9$	$7 \times 73$	$6$	$12$	$6$	$72$
$11$	$23 \times 89$	$11$	$88$	$22$	$44$
$23$	$47 \times 178481$	$23$	$178480$	$46$	$44620$
$37$	$223 \times 616318177$	$222$	$308159088$	$222$	$616318176$
$41$	$13367 \times 164511353$	$6683$	$164511352$	$13366$	$164511352$
$49$	$127 \times P13$	$126$	$(8)$	$42$	$(8)$
$59$	$179951 \times P13$	$89975$	$(8)$	$89975$	$(8)$
$67$	$193707721 \times P12$	$96853860$	$(6)$	$8071155$	$(6)$
$83$	$167 \times P23$	$83$	$(10)$	$166$	$(166)$
$97$	$11447 \times P26$	$5723$	$(194)$	$11446$	$(194)$
$101$	$P13 \times P14$	$(303)$	$(303)$	$(303)$	$(303)$
$103$	$2550183799 \times P22$	$(166)$	$(206)$	$(249)$	$(309)$
$109$	$745988807 \times P24$	$(11663)$	$(118)$	$(214)$	$(118)$
$131$	$263 \times P38$	$131$	$(74)$	$262$	$(74)$
$137$	$P20 \times P22$	$(274)$	$(66290053)$	$(1202723)$	$(66290053)$
$139$	$P13 \times P30$	$(6)$	$(6)$	$(6)$	$(15)$
$149$	$P20 \times P25$	$(745)$	$(16)$	$(745)$	$(8)$
$167$	$2349023 \times P44$	$(26)$	$(22)$	$(26)$	$(22)$
$197$	$7487 \times P56$	$(3743)$	$(394)$	$(38)$	$(394)$
$199$	$P12 \times P49$	$(14)$	$(1393)$	$(8)$	$(1393)$
$227$	$P18 \times P52$	$(8)$	$(35)$	$(8)$	$(497)$
$241$	$22000409 \times P66$	$(8)$	$(5114261)$	$(482)$	$(5114261)$
$269$	$13822297 \times P74$	$(6)$	$(6)$	$(6)$	$(22)$
$271$	$15242475217 \times P72$	$(8)$	$(542)$	$(8)$	$(15)$
$281$	$80929 \times P80$	$(8)$	$(278)$	$(6)$	$(417)$
$293$	$P26 \times P63$	$(6)$	$(6)$	$(8)$	$(6)$
$347$	$P23 \times P82$	$(6)$	$(6)$	$(21)$	$(8)$

Table 3. Table 3: The residual orders of 2 , 5 2 5 2,5 modulo p 𝑝 p for p f 2 = ( 3 e 2 − 1 ) / 2 superscript 𝑝 subscript 𝑓 2 superscript 3 subscript 𝑒 2 1 2 p^{f_{2}}=(3^{e_{2}}-1)/2

$e_{2}$	$o_{p} (2)$	$o_{p} (5)$
$2$	N/A	$1$
$3$	$12$	$4$
$5$	$10$	$5$
$7$	$1092$	$364$
$13$	$398580$	$30660$
$71$	$(8)$	$(8)$
$103$	$(12)$	$(14)$

Table 4. Table 4: The residual orders of 2 , 5 2 5 2,5 modulo p , q 𝑝 𝑞 p,q for p q = ( 3 e 2 − 1 ) / 2 , p < q formulae-sequence 𝑝 𝑞 superscript 3 subscript 𝑒 2 1 2 𝑝 𝑞 pq=(3^{e_{2}}-1)/2,p<q

$e_{2}$	$(3^{e_{2}} - 1) / 2 = p q$	$o_{p} (2)$	$o_{q} (2)$	$o_{p} (5)$	$o_{q} (5)$
$9$	$13 \times 757$	$12$	$756$	$4$	$756$
$11$	$23 \times 3851$	$11$	$3850$	$22$	$1925$
$17$	$1871 \times 34511$	$935$	$595$	$935$	$3451$
$19$	$1597 \times 363889$	$532$	$181944$	$532$	$22743$
$23$	$47 \times 1001523179$	$23$	$(46)$	$46$	$(1073)$
$37$	$13097927 \times P12$	$(9731)$	$8594564351$	$(74)$	$(74)$
$43$	$431 \times P18$	$43$	$215$	$(22)$	$(22)$
$59$	$14425532687 \times P18$	$(3953)$	$(118)$	$(106)$	$(10679)$
$61$	$603901 \times P24$	$201300$	$(12)$	$150975$	$(145)$
$223$	$P26 \times P81$	$(446)$	$(12)$	$(6)$	$(446)$

Table 5. Table 5: The residual orders of 2 , 3 2 3 2,3 modulo p 𝑝 p for p = ( 5 e 3 − 1 ) / 4 𝑝 superscript 5 subscript 𝑒 3 1 4 p=(5^{e_{3}}-1)/4

$e_{3}$	$o_{p} (2)$	$o_{p} (3)$
$3$	$5$	$30$
$7$	$6510$	$6510$
$11$	$1220703$	$369910$
$13$	$61035156$	$1211015$
$47$	$(94)$	$(6)$
$127$	$(18)$	$(18)$
$149$	$(10)$	$(6)$
$181$	$(12)$	$(15)$

Table 6. Table 6: The residual orders of 2 , 3 2 3 2,3 modulo p , q 𝑝 𝑞 p,q for p q = ( 5 e 3 − 1 ) / 4 , p < q formulae-sequence 𝑝 𝑞 superscript 5 subscript 𝑒 3 1 4 𝑝 𝑞 pq=(5^{e_{3}}-1)/4,p<q

$e_{3}$	$(5^{e_{3}} - 1) / 4 = p q$	$o_{p} (2)$	$o_{q} (2)$	$o_{p} (3)$	$o_{q} (3)$
$2$	$2 \times 3$	N/A	$2$	$1$	N/A
$5$	$11 \times 71$	$10$	$35$	$5$	$35$
$17$	$409 \times 466344409$	$204$	$3429003$	$204$	$116586102$
$23$	$8971 \times P12$	$8970$	$(8)$	$8970$	$2306995565$
$31$	$1861 \times P18$	$1860$	$(15)$	$310$	$(6)$
$41$	$2238236249 \times P19$	$279779531$	$(8)$	$(8)$	$(8)$
$43$	$1644512641 \times P21$	$(8)$	$(15)$	$(8)$	$(10)$
$59$	$P17 \times P25$	$(12)$	$(9)$	$(6)$	$(118)$
$71$	$569 \times P47$	$284$	$(142)$	$568$	$(452610863706241)$

Equations78

\frac{x ^{m} - 1}{x - 1} = y^{n}, x \geq 2, y \geq 2, m \geq 3, n \geq 2,

\frac{x ^{m} - 1}{x - 1} = y^{n}, x \geq 2, y \geq 2, m \geq 3, n \geq 2,

\frac{x ^{l} - 1}{x - 1} = y^{m} z^{n}

\frac{x ^{l} - 1}{x - 1} = y^{m} z^{n}

\frac{2}{l} + \frac{l - 1}{l} \times \frac{2}{n + min { n , m }} \geq 1 - ϵ

\frac{2}{l} + \frac{l - 1}{l} \times \frac{2}{n + min { n , m }} \geq 1 - ϵ

B = Ω = C_{0} = C_{1} (n) = max {1, ∣ b_{1} ∣ A_{1} / A_{n}, ∣ b_{2} ∣ A_{2} / A_{n}, \dots, ∣ b_{n} ∣}, A_{1} A_{2} \dots A_{n}, 1 + lo g 3 - lo g 2, \frac{16}{n !} e^{n} (2 n + 3) (n + 2) (4 (n + 1))^{n + 1} (\frac{1}{2} e n) \times (4.4 n + 5.5 lo g n + 7)

B = Ω = C_{0} = C_{1} (n) = max {1, ∣ b_{1} ∣ A_{1} / A_{n}, ∣ b_{2} ∣ A_{2} / A_{n}, \dots, ∣ b_{n} ∣}, A_{1} A_{2} \dots A_{n}, 1 + lo g 3 - lo g 2, \frac{16}{n !} e^{n} (2 n + 3) (n + 2) (4 (n + 1))^{n + 1} (\frac{1}{2} e n) \times (4.4 n + 5.5 lo g n + 7)

Λ = b_{1} lo g a_{1} + \dots + b_{n} lo g a_{n} .

Λ = b_{1} lo g a_{1} + \dots + b_{n} lo g a_{n} .

lo g ∣ Λ ∣ > - C_{1} (n) (C_{0} + lo g B) max {1, \frac{n}{6}} Ω.

lo g ∣ Λ ∣ > - C_{1} (n) (C_{0} + lo g B) max {1, \frac{n}{6}} Ω.

e_{i} lo g a_{i} < E_{i} = C_{i} lo g p lo g q (lo g lo g p + C_{i + 3}),

e_{i} lo g a_{i} < E_{i} = C_{i} lo g p lo g q (lo g lo g p + C_{i + 3}),

- lo g ∣ Λ_{1} ∣ < C (3) (C_{0} + lo g \frac{e _{1} lo g 2}{lo g q}) lo g 2 lo g p lo g q

- lo g ∣ Λ_{1} ∣ < C (3) (C_{0} + lo g \frac{e _{1} lo g 2}{lo g q}) lo g 2 lo g p lo g q

- lo g ∣ Λ_{j} ∣ < C (4) (C_{0} + lo g \frac{e _{j} lo g a _{j}}{lo g q}) lo g 2 lo g a_{j} lo g p lo g q

- lo g ∣ Λ_{j} ∣ < C (4) (C_{0} + lo g \frac{e _{j} lo g a _{j}}{lo g q}) lo g 2 lo g a_{j} lo g p lo g q

\frac{e _{1} lo g 2}{lo g q} < < (1 + 1 0^{- 10}) (C_{0} + lo g \frac{e _{1} lo g 2}{lo g q}) C (3) lo g 2 lo g p 1.244 \times 1 0^{10} lo g p lo g \frac{e _{1} lo g 2}{lo g q} .

\frac{e _{1} lo g 2}{lo g q} < < (1 + 1 0^{- 10}) (C_{0} + lo g \frac{e _{1} lo g 2}{lo g q}) C (3) lo g 2 lo g p 1.244 \times 1 0^{10} lo g p lo g \frac{e _{1} lo g 2}{lo g q} .

\frac{e _{1} lo g 2}{lo g q} < < 1.143 \times (1.244 \times 1 0^{10} lo g p) lo g (1.244 \times 1 0^{10} lo g p) 1.422 \times 1 0^{10} (lo g lo g p + 23.3),

\frac{e _{1} lo g 2}{lo g q} < < 1.143 \times (1.244 \times 1 0^{10} lo g p) lo g (1.244 \times 1 0^{10} lo g p) 1.422 \times 1 0^{10} (lo g lo g p + 23.3),

\frac{e _{2} lo g 3}{lo g q} < < (1 + 1 0^{- 10}) (C_{0} + lo g \frac{e _{2} lo g 3}{lo g q}) C (4) lo g 2 lo g 3 lo g p 1.089 \times 1 0^{12} lo g p lo g \frac{e _{2} lo g 3}{lo g q} .

\frac{e _{2} lo g 3}{lo g q} < < (1 + 1 0^{- 10}) (C_{0} + lo g \frac{e _{2} lo g 3}{lo g q}) C (4) lo g 2 lo g 3 lo g p 1.089 \times 1 0^{12} lo g p lo g \frac{e _{2} lo g 3}{lo g q} .

\frac{e _{3} lo g 5}{lo g q} < < (1 + 1 0^{- 10}) (C_{0} + lo g \frac{e _{3} lo g 5}{lo g q}) C (4) lo g 2 lo g 5 lo g p 1.595 \times 1 0^{12} lo g p lo g \frac{e _{3} lo g 5}{lo g q} .

\frac{e _{3} lo g 5}{lo g q} < < (1 + 1 0^{- 10}) (C_{0} + lo g \frac{e _{3} lo g 5}{lo g q}) C (4) lo g 2 lo g 5 lo g p 1.595 \times 1 0^{12} lo g p lo g \frac{e _{3} lo g 5}{lo g q} .

lo g x \leq lo g \frac{7 H}{4} + C (3) (C_{0} + lo g ((e_{1} + 3) H)) lo g 2 lo g 3 lo g 5.

lo g x \leq lo g \frac{7 H}{4} + C (3) (C_{0} + lo g ((e_{1} + 3) H)) lo g 2 lo g 3 lo g 5.

(2^{e_{1}} - 1)^{h_{1}} (\frac{3 ^{e_{2}} - 1}{2})^{h_{2}} (\frac{5 ^{e_{3}} - 1}{4})^{h_{3}} = 1.

(2^{e_{1}} - 1)^{h_{1}} (\frac{3 ^{e_{2}} - 1}{2})^{h_{2}} (\frac{5 ^{e_{3}} - 1}{4})^{h_{3}} = 1.

Λ = (e_{1} h_{1} - h_{2} - 2 h_{3}) lo g 2 + e_{2} h_{2} lo g 3 + e_{3} h_{3} lo g 5 = h_{1} lo g \frac{2 ^{e_{1}}}{2 ^{e_{1}} - 1} + h_{2} lo g \frac{3 ^{e_{2}}}{3 ^{e_{2}} - 1} + h_{3} lo g \frac{5 ^{e_{3}}}{5 ^{e_{3}} - 1} .

Λ = (e_{1} h_{1} - h_{2} - 2 h_{3}) lo g 2 + e_{2} h_{2} lo g 3 + e_{3} h_{3} lo g 5 = h_{1} lo g \frac{2 ^{e_{1}}}{2 ^{e_{1}} - 1} + h_{2} lo g \frac{3 ^{e_{2}}}{3 ^{e_{2}} - 1} + h_{3} lo g \frac{5 ^{e_{3}}}{5 ^{e_{3}} - 1} .

0 < ∣ Λ ∣ \leq H (\frac{1}{2 ^{e_{1}} - 1} + \frac{1}{3 ^{e_{2}} - 1} + \frac{1}{5 ^{e_{3}} - 1}) \leq \frac{7 H}{4 x}

0 < ∣ Λ ∣ \leq H (\frac{1}{2 ^{e_{1}} - 1} + \frac{1}{3 ^{e_{2}} - 1} + \frac{1}{5 ^{e_{3}} - 1}) \leq \frac{7 H}{4 x}

lo g ∣ Λ ∣ \leq - lo g x + lo g \frac{7 H}{4} .

lo g ∣ Λ ∣ \leq - lo g x + lo g \frac{7 H}{4} .

lo g ∣ Λ ∣ \geq - C (3) (C_{0} + lo g ((e_{1} + 3) H)) lo g 2 lo g 3 lo g 5.

lo g ∣ Λ ∣ \geq - C (3) (C_{0} + lo g ((e_{1} + 3) H)) lo g 2 lo g 3 lo g 5.

lo g q < lo g x < lo g \frac{7 H}{4} + C (3) (C_{0} + lo g ((e_{1} + 3) H)) lo g 2 lo g 3 lo g 5.

lo g q < lo g x < lo g \frac{7 H}{4} + C (3) (C_{0} + lo g ((e_{1} + 3) H)) lo g 2 lo g 3 lo g 5.

Λ^{'} = e_{j} h_{j} lo g a_{j} + e_{k} h_{k} lo g a_{k} - h_{j} lo g (a_{j} - 1) - h_{k} lo g (a_{k} - 1) + h_{i} lo g x = h_{j} lo g \frac{a _{j}^{e_{j}}}{a _{j}^{e_{j}} - 1} + h_{k} lo g \frac{a _{k}^{e_{k}}}{a _{k}^{e_{k}} - 1} .

Λ^{'} = e_{j} h_{j} lo g a_{j} + e_{k} h_{k} lo g a_{k} - h_{j} lo g (a_{j} - 1) - h_{k} lo g (a_{k} - 1) + h_{i} lo g x = h_{j} lo g \frac{a _{j}^{e_{j}}}{a _{j}^{e_{j}} - 1} + h_{k} lo g \frac{a _{k}^{e_{k}}}{a _{k}^{e_{k}} - 1} .

0 < Λ^{'} < H (\frac{1}{a _{j}^{e_{j}} - 1} + \frac{1}{a _{k}^{e_{k}} - 1}) \leq \frac{3 H}{2 q} .

0 < Λ^{'} < H (\frac{1}{a _{j}^{e_{j}} - 1} + \frac{1}{a _{k}^{e_{k}} - 1}) \leq \frac{3 H}{2 q} .

lo g ∣ Λ^{'} ∣ \geq - C (4) (C_{0} + lo g \frac{E _{3} H}{lo g x}) lo g 2 lo g 3 lo g 5 lo g x .

lo g ∣ Λ^{'} ∣ \geq - C (4) (C_{0} + lo g \frac{E _{3} H}{lo g x}) lo g 2 lo g 3 lo g 5 lo g x .

lo g q \leq lo g \frac{3 H}{2} + C (4) (C_{0} + lo g \frac{E _{3} H}{lo g x}) lo g 2 lo g 3 lo g 5 lo g x .

lo g q \leq lo g \frac{3 H}{2} + C (4) (C_{0} + lo g \frac{E _{3} H}{lo g x}) lo g 2 lo g 3 lo g 5 lo g x .

lo g p = < < lo g x lo g \frac{7 H}{4} + C (3) (C_{0} + lo g ((e_{1} + 3) H)) lo g 2 lo g 3 lo g 5 7.22 \times 1 0^{12} .

lo g p = < < lo g x lo g \frac{7 H}{4} + C (3) (C_{0} + lo g ((e_{1} + 3) H)) lo g 2 lo g 3 lo g 5 7.22 \times 1 0^{12} .

lo g x < lo g \frac{7 H}{4} - lo g ∣ Λ ∣ < 727.94.

lo g x < lo g \frac{7 H}{4} - lo g ∣ Λ ∣ < 727.94.

e_{i} \leq ⌊ \frac{lo g 2 x}{lo g a _{i}} ⌋ \leq 1051.

e_{i} \leq ⌊ \frac{lo g 2 x}{lo g a _{i}} ⌋ \leq 1051.

∣ h_{i} ∣ < < C_{2} C_{3} lo g x lo g q (lo g lo g x + C_{6}) (lo g lo g q + C_{5}) 1.264 \times 1 0^{48},

∣ h_{i} ∣ < < C_{2} C_{3} lo g x lo g q (lo g lo g x + C_{6}) (lo g lo g q + C_{5}) 1.264 \times 1 0^{48},

∣ h_{j} ∣ = ∣ f_{i} g_{k} ∣ < C_{3} lo g x (lo g lo g q + C_{6}) < 8.944 \times 1 0^{16},

∣ h_{j} ∣ = ∣ f_{i} g_{k} ∣ < C_{3} lo g x (lo g lo g q + C_{6}) < 8.944 \times 1 0^{16},

∣ e_{j} ∣ < C_{3} lo g x lo g q (lo g lo g q + C_{6}) / lo g 2 < 4.306 \times 1 0^{34}

∣ e_{j} ∣ < C_{3} lo g x lo g q (lo g lo g q + C_{6}) / lo g 2 < 4.306 \times 1 0^{34}

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TopicsAnalytic Number Theory Research · Algebraic and Geometric Analysis · Mathematics and Applications

Full text

On the simultaneous equations $\sigma(2^{a})=p^{f_{1}}q^{g_{1}},\sigma(3^{b})=p^{f_{2}}q^{g_{2}},\sigma(5^{c})=p^{f_{3}}q^{g_{3}}$ 1112010 Mathematics

Subject Classification: 11A05, 11A25, 11A51, 11D61. 222Key words and phrases: Odd perfect numbers; sum of divisors; exponential diophantine equations.

Tomohiro Yamada

Abstract

We shall solve the simultaneous equations $\sigma(2^{a})=p^{f_{1}}q^{g_{1}},\sigma(3^{b})=p^{f_{2}}q^{g_{2}},\sigma(5^{c})=p^{f_{3}}q^{g_{3}}$ with $p,q$ distinct primes.

1 Introduction

As usual, let $\sigma(N)$ denote the sum of divisors of $N$ and $\omega(N)$ the number of distinct prime factors of $N$ . In [18], the author has shown that there are only finitely many odd superperfect numbers (i.e. the number satisfying $\sigma(\sigma(N))=2N$ ) with bounded number of distinct prime factors by proving that the simultaneous equation $\sigma(p_{i}^{e_{i}})=q_{1}^{f_{1i}}\cdots q_{k}^{f_{ki}}$ for $k+1$ prime powers $p_{i}^{e_{i}}(i=1,2,\ldots,k+1)$ cannot have solutions with $p_{1},\cdots,p_{k+1}$ all small. In this paper, we use the method developed in [18] to solve the simultaneous equations $\sigma(2^{a})=p^{f_{1}}q^{g_{1}},\sigma(3^{b})=p^{f_{2}}q^{g_{2}},\sigma(5^{c})=p^{f_{3}}q^{g_{3}}$ with $p,q$ distinct primes.

Wakulicz [16] has shown that all solutions of the purely exponential diophantine equation $2^{n}-5^{m}=3$ are $(n,m)=(2,0),(3,1)$ and $(7,3)$ , from which Makowski and Schinzel [9] derived that the equation $\sigma(2^{a})=\sigma(5^{c})$ has only the solution $(a,c)=(4,2)$ . We note that it is easy to show that $\sigma(2^{a})=\sigma(3^{b})$ has no nontrivial solution and $\sigma(3^{b})=\sigma(5^{c})$ also has no nontrivial solution. Bugeaud and Mignotte [3] has shown that neither of $\sigma(2^{a}),\sigma(3^{b}),\sigma(5^{c})$ can be perfect power except $\sigma(3)=2^{2}$ and $\sigma(3^{4})=11^{2}$ . Moreover, they have shown that the only perfect powers $\frac{x^{n}-1}{x-1}$ with $x=z^{t},z\leq 10$ are $(3^{5}-1)/2=11^{2}$ and $(7^{4}-1)/6=20^{2}$ .

Now we shall state our result.

Theorem 1.1.

The simultaneous equations $\sigma(2^{a})=p^{f_{1}}q^{g_{1}},\sigma(3^{b})=p^{f_{2}}q^{g_{2}},\sigma(5^{c})=p^{f_{3}}q^{g_{3}}$ with $a,b,c>0,f_{1},f_{2},f_{3},g_{1},g_{2},g_{3}\geq 0$ and $p,q$ distinct primes has only the following solutions: i) $(a,b,c)=(1,1,1)$ . ii) $(a,b,c)=(4,1,2)$ , iii) $(a,b,c)=(4,4,2)$ and iv) $(a,c)=(4,2)$ and $\sigma(3^{b})$ is prime. In other words, if $\omega(\sigma(2^{a}3^{b}5^{c}))\leq 2$ , then $(a,b,c)$ must satisfy one of the above.

Our result is related to the Nagell-Ljunggren equation

[TABLE]

which has been conjectured to have only three solutions $(x,y,m,n)=(3,11,5,2),(7,20,4,2),(18,7,3,3)$ . Some of recent remarkable results concerning to the Nagell-Ljunggren equation are [2], [3], [11], [12] and [4]. Our result leads us to conjecture that the diophantine equation

[TABLE]

has only finitely many solutions in integers $x\geq 2,z\geq y\geq 2$ and $l,m,n\geq n_{0}$ for some constant $n_{0}$ . The $abc$ -conjecture, which Mochizuki [13] claims to prove, would allow us to take $n_{0}=3$ . Indeed, applying the $abc$ -conjecture to the equation $1+(x-1)y^{m}z^{n}=x^{l}$ , we see that for any given $\epsilon>0$ , the inequality

[TABLE]

would hold for sufficiently large $x^{l}$ . Hence, with only finitely many exceptions, we would have i) $l\geq 5,n=1$ , ii) $l=4,n=1$ , iii) $(l,m,n)=(4,1,2)$ . iv) $l=3,n\leq 2$ or v) $(l,m,n)=(3,1,3)$ .

2 Preliminary Lemmas

In this section, we introduce some preliminary lemmas. One is Matveev’s lower bound for linear forms of logarithms [10].

Lemma 2.1.

Let $a_{1},a_{2},\ldots,a_{n}$ be positive integers such that $\log a_{1},\ldots,\log a_{n}$ are not all zero and $A_{j}\geq\max\{0.16,\log a_{j}\}$ for each $j=1,2,\ldots,n$ . Moreover, we put

[TABLE]

and

[TABLE]

Then we have

[TABLE]

The others concern to some arithmetical properties of values of cyclotomic polynomials. Lemma 2.2 is a basic and well-known result of this area. Lemma 2.2 has been proved by Zsigmondy [19] and rediscovered by many authors such as Dickson [6] and Kanold [7]. We need only the special case $b=1$ , where this lemma had already been proved by Bang [1]. See also Theorem 6.4A.1 in [14].

Lemma 2.2.

If $a>b\geq 1$ 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 $(a,b,n)=(2,1,6)$ , $a-b=n=1$ , or $n=2$ and $a+b$ is a power of $2$ .

Let $o_{p}(a)$ denote the residual order of $a\pmod{p}$ . Lemma 2.2 immediately gives the following result.

Lemma 2.3.

If $(a^{e}-1)/(a-1)=p^{f_{1}}q^{f_{2}}$ for some integers $a,e,f_{1},f_{2}$ and primes $p<q$ , then we have $(a,e,p,q,f_{1},f_{2})=(2,6,3,7,2,1)$ , $e=r$ or $e=r^{2}$ for some prime $r$ . Moreover, in the case $e=r$ , then we have $p=r,o_{q}(a)=r$ or $o_{p}(a)=o_{q}(a)=r$ . In the case $e=r^{2}$ , we have $(p,q,f_{1},f_{2})=((a^{r}-1)/(a-1),(a^{r^{2}}-1)/(a^{r}-1),1,1)$ or $(a,e,p,f_{1})=(2^{m}-1,4,2,m+1)$ for some integer $m$ .

The following lemma is proved in [3], as mentioned in the introduction.

Lemma 2.4.

Let $a,e,x,f$ be positive integers with $a,x,f>1$ and $e>2$ . The equation $(a^{e}-1)/(a-1)=x^{f}$ has no solution but $(a,e,x,f)=(3,5,11,2),(7,4,20,2)$ in integers $2\leq a\leq 10,e>2,x>1,f>1$ .

Using results mentioned in the introduction, we can immediately solve some special case of our main theorem.

Lemma 2.5.

Choose $a<b$ from the first three primes $2,3,5$ . If $(a^{e}-1)/(a-1)=p^{k}$ and $(b^{f}-1)/(b-1)=p^{l}$ for some integers $e,f,k,l$ and some prime $p$ , then $(a^{e},b^{f})=(2^{5},5^{3})$ and $p=31,k=l=1$ .

Proof.

In the case $k=l=1$ and $(a^{e}-1)/(a-1)=(b^{f}-1)/(b-1)$ , as observed in the introduction, we have $(a^{e},b^{f})=(2^{5},5^{3})$ .

Lemma 2.4 yields that the perfect power case must arise from $(3^{5}-1)/2=11^{2}$ or $(3^{2}-1)/2=2^{2}$ . In this case, we must have $2^{e}-1=2$ or $11$ or $(5^{f}-1)/4=2$ or $11$ , which is clearly impossible. ∎

3 Main Theory

For convenience, we put $a_{1}=2,a_{2}=3,a_{3}=5$ and $e_{1}=a+1,e_{2}=b+1,e_{3}=c+1$ .

Lemma 3.1.

For each $i=1,2,3$ , we have

[TABLE]

where $C_{1}=1.422\times 10^{10},C_{2}=1.226\times 10^{12},C_{3}=1.795\times 10^{12},C_{4}=23.3,C_{5}=27.8,C_{6}=28.1$ .

Proof.

Let ${\Lambda}_{i}=f_{i}\log a_{i}+g_{i}\log q+\log(a_{i}-1)-e_{i}\log 2=\log(1-a_{i}^{-e_{i}})$ for $i=1,2,3$ . It immediately follows from Matveev’s theorem that

[TABLE]

and

[TABLE]

for $j=2,3$ .

Now we shall show (7) in the case $i=1$ . We may assume that $e_{1}>10^{10}\log q/\log 2$ . Since $0<\left\lvert{\Lambda}_{1}\right\rvert=-\log(1-2^{-e_{1}})<\frac{1}{2^{e_{1}}-1}$ , we have $-\log\left\lvert{\Lambda}_{1}\right\rvert>\log(2^{e_{1}}-1)>(1-10^{-10})e_{1}\log 2$ . Combining upper and lower bounds for ${\Lambda}_{1}$ , we obtain

[TABLE]

Hence, observing that $1.244\times 10^{10}\log p\geq 1.244\times 10^{10}\log 2$ , we obtain

[TABLE]

giving (7) in the case $i=1$ .

Next we shall prove (7) in the case $i=2$ . We may assume that $e_{2}>10^{10}\log q/\log 3$ as with the previous case. Since $0<\left\lvert{\Lambda}_{2}\right\rvert=-\log(1-3^{-e_{2}})<\frac{1}{3^{e_{2}}-1}$ , we have $-\log\left\lvert{\Lambda}_{2}\right\rvert>\log(3^{e_{2}}-1)>(1-10^{-10})e_{2}\log 3$ . From $0<\left\lvert{\Lambda}_{2}\right\rvert=-\log(1-3^{-e_{2}})<\frac{1}{3^{e_{2}}-1}$ , we see that $-\log\left\lvert{\Lambda}_{2}\right\rvert>\log(3^{e_{2}}-1)\geq(1-10^{-10})e_{2}\log 3$ and therefore

[TABLE]

This gives (7) in the case $i=2$ .

Similarly, (7) in the case $i=3$ follows from

[TABLE]

A similar argument yields (7) in the case $i=3$ . This completes the proof of the lemma. ∎

Next, we shall show that we cannot have all of $a_{i}^{e_{i}}$ ’s large.

Lemma 3.2.

Let $x$ be the smallest among $a_{i}^{e_{i}}$ ’s. Let $h_{1}=f_{2}g_{3}-f_{3}g_{2},h_{2}=f_{3}g_{1}-f_{1}g_{3}$ and $h_{3}=f_{1}g_{2}-f_{2}g_{1}$ and $H=\max\left\lvert h_{i}\right\rvert$ . Then

[TABLE]

Proof.

We begin by observing that

[TABLE]

Now we put

[TABLE]

Then we have

[TABLE]

and therefore

[TABLE]

It follows from the assumption $e_{i}>0$ that ${\Lambda}\neq 0$ . Hence Matveev’s lower bound gives

[TABLE]

Combining (18) and (19), we obtain (14). ∎

The third step is to obtain upper bounds for each $e_{i}$ .

Lemma 3.3.

Unless $x=p=31$ , we have $e_{1}<4.44\times 10^{52},e_{2}<2.54\times 10^{54}$ and $e_{3}<2.55\times 10^{54}$ and $H<2.89\times 10^{68}$ .

Proof.

We may assume without the loss of generality that $p<q$ . We begin by considering $q\mid x$ . In this case, we have

[TABLE]

We note that $H\leq C_{2}C_{3}\log p\log q(\log\log p+C_{5})(\log\log p+C_{6})$ since it follows from Lemma 3.1 that $f_{i}<e_{i}\log a_{i}/\log p_{i}<C_{i}\log q(\log\log p+C_{i+3})$ and $g_{i}<e_{i}\log a_{i}/\log q_{i}<C_{i}\log p(\log\log p+C_{i+3})$ . Hence we obtain $\log p<\log q<4.35\times 10^{12}$ .

Now we consider the case $p<q$ and $q\nmid x$ . Put $i$ to be the index such that $x=(a_{i}^{e_{i}}-1)/(a_{i}-1)$ , $j,k$ be the others and

[TABLE]

It follows from Lemma 2.5 that if $(a_{j}^{e_{j}}-1)/(a_{j}-1)=p^{f_{j}}$ or $(a_{k}^{e_{k}}-1)/(a_{k}-1)=p^{f_{k}}$ , then $a_{i}^{e_{i}}=2^{5}$ or $5^{3}$ and $x=p=31$ . Hence we see that both numbers $(a_{j}^{e_{j}}-1)/(a_{j}-1),(a_{k}^{e_{k}}-1)/(a_{k}-1)$ must be divisible by $q$ unless $x=p=31$ .

Thus we obtain

[TABLE]

As in the previous case, Matveev’s theorem now gives

[TABLE]

Combining (22) and (23), we obtain

[TABLE]

Since $E_{3}=C_{3}\log p\log q(\log\log p+C_{6})\leq C_{3}\log x\log q(\log\log x+C_{6})$ and $H<C_{2}C_{3}(\log q)^{2}(\log\log q+C_{5})(\log\log q+C_{6})$ , combining (14) and (24), we obtain $\log q<3.45\times 10^{27}$ . Moreover, we have

[TABLE]

So that, we conclude that in both cases, we have $\log p<7.22\times 10^{12}$ and $\log q<3.45\times 10^{27}$ . Now Lemma 3.1 immediately gives that $e_{1}<4.44\times 10^{52},e_{2}<2.54\times 10^{54}$ and $e_{3}<2.55\times 10^{54}$ . Finally, the upper bound $H<2.89\times 10^{68}$ follows from $H<C_{2}C_{3}(\log p)(\log q)(\log\log p+C_{6})(\log\log q+C_{5})$ . ∎

Now, using the lattice reduction algorithm, we shall obtain feasible upper bounds.

Lemma 3.4.

We have $\log x<354.8$ . Moreover, if $p<q$ and $q$ divides $x$ , then $\log x<249.5$ .

Proof.

We begin by noting that we can assume $x\neq 31$ without the loss of generality.

In order to reduce our upper bounds, we use the LLL lattice reduction algorithm introduced in [8]. Let $M$ be the matrix defined by $m_{12}=m_{13}=m_{21}=m_{23}=0$ and $m_{11}=m_{22}=\gamma$ and $m_{3i}=\left\lfloor C\gamma\log a_{i}\right\rfloor$ for $i=1,2,3$ , where $C$ and $\gamma$ are constants chosen later. Let $L$ denote the LLL-reduced matrix of $M$ and $l(L)$ the shortest length of vectors in the lattice generated by the column vectors of $L$ .

From the previous lemma, we know that ${\Lambda}$ has coefficients with absolute values at most $H\max\{e_{1}+3,e_{2},e_{3}\}<7.37\times 10^{122}$ . It is implicit in the proof of Lemma 3.7 of de Weger’s book [17] that if $l(\Gamma)>X_{1}\sqrt{16+4\gamma}$ and $X_{1}\geq 7.37\times 10^{122}$ , then $\left\lvert{\Lambda}\right\rvert>X_{1}/(C\gamma)$ .

Taking $C=10^{370},\gamma=2$ , we can confirm that $l(\Gamma)>X_{1}\sqrt{16+4\gamma}$ and therefore we obtain that $\left\lvert{\Lambda}\right\rvert>3.685\times 10^{-248}$ . Hence we have

[TABLE]

We choose the index $i$ such that $x=(a_{i}^{e_{i}}-1)/(a_{i}-1)$ and let $j,k$ be the others. From the above estimate for $x$ , we derive that

[TABLE]

We consider the case $p<q$ and $q$ does not divide $x$ . From (24) we obtain $\log q<3.337\times 10^{17}$ . Lemma 3.1 gives that

[TABLE]

and similar upper bounds hold for $\left\lvert h_{k}\right\rvert$ and $\left\lvert e_{k}\right\rvert$ , respectively. Hence ${\Lambda}$ has coefficients with absolute values at most $3.852\times 10^{51}$ . Using the LLL-reduction again with $C=10^{157}$ and $\gamma=2$ , we obtain $\left\lvert{\Lambda}\right\rvert>1.926\times 10^{-106}$ and therefore $\log x<\log(7H/4)-\log\left\lvert{\Lambda}\right\rvert<354.8$ .

Next, we consider the case $p<q$ and $q$ divides $x$ . In this case, we have $\log p<\log q\leq\log x<727.94$ . We choose the index $i$ such that $x=(a_{i}^{e_{i}}-1)/(a_{i}-1)$ and let $j,k$ be the others. Lemma 3.1 gives that

[TABLE]

and similar upper bounds hold for $\left\lvert h_{k}\right\rvert$ and $\left\lvert e_{k}\right\rvert$ , respectively. Combining these upper bounds with (27), we see that ${\Lambda}$ has coefficients with absolute values at most $2.159\times 10^{36}$ . We use the LLL-reduction again with $C=10^{111}$ and $\gamma=2$ , we obtain $\left\lvert{\Lambda}\right\rvert>1.079\times 10^{-75}$ and therefore $\log x<\log(7H/4)-\log\left\lvert{\Lambda}\right\rvert<249.5$ . This proves the lemma. ∎

4 The final step

The final step is checking all possibilities of $x$ . We note that from The Cunningham Project (see [15] or [5]), we know all prime factors of $x$ ’s below our upper bounds.

For $x=(a_{i}^{e_{i}}-1)/(a_{i}-1)$ , we should check the residual orders of the other prime $a_{j}$ modulo $x$ . A summary is given in Tables 1-6, where $\textrm{P}n$ denotes a prime with $n$ digits and $(n)$ indicates that the residual order is a multiple of $n$ . For example, putting $x=2^{347}-1=pq$ with $p<q$ , $o_{q}(3)$ is divisible by $6$ since $q-1$ is divisible by $2^{3}\times 3^{2}$ and $3^{(q-1)/8},3^{(q-1)/3}\not\equiv 1\pmod{q}$ , although $3^{(q-1)/4}\equiv 1\pmod{q}$ , which yields that $(3^{e_{2}}-1)/2=p^{f_{2}}q^{g_{2}}$ with $g_{2}>0$ is impossible.

If $p=x=2^{e_{1}}-1$ is prime, then $e_{1}\leq 511$ and therefore $e_{1}$ must belong to the set

[TABLE]

Among them, there exists no $e_{1}$ such that the $o_{x}(3)=1$ , a prime or the square of a prime, as we can see from Table 1. Hence, by Lemma 2.3, we must have $(3^{e_{2}}-1)/2=q^{g_{2}}$ . By Lemma 2.5, $(5^{e_{3}}-1)/4$ must be divisible by $p=x$ . Hence, by Lemma 2.3, $o_{x}(5)=1$ or $o_{x}(5)$ must be a prime or a prime square and therefore, from Table 1, $e_{1}=e_{3}=2$ or $e_{1}=5,e_{3}=3$ . If $e_{1}=e_{3}=2$ , then $(5^{e_{3}}-1)/4=6=2\times 3$ and therefore $(3^{e_{2}}-1)/2$ must be a power of $2$ , yielding that $e_{2}=2$ . If $e_{1}=5$ and $e_{3}=3$ , then $p=31$ and $(3^{e_{2}}-1)/2=q^{g_{2}}$ , yielding that $e_{2}=5$ or $(3^{e_{2}}-1)/2=q$ .

If $x=2^{e_{1}}-1$ is not a prime power, then $e_{1}\leq 359$ and therefore $e_{1}$ must belong to the set

[TABLE]

Hence we can write $x=2^{e_{1}}-1=pq$ for distinct primes $p<q$ . By Lemma 2.5, $(3^{e_{2}}-1)/2=p^{g_{2}}$ and $(5^{e_{3}}-1)/4=p^{g_{3}}$ cannot simultaneously hold. In other words, at least one of these two integers must be divisible by $q$ . But, for no $e_{1}$ in the above set, $o_{q}(5)$ is $1$ , a prime or prime-square, as can be seen from Table 2. Hence $(3^{e_{2}}-1)/2$ must be divisible by $q$ . The only $e_{1}$ for which $o_{q}(3)$ is $1$ , a prime or prime-square is $e_{1}=4$ . Then we must have $x=2^{4}-1=3\times 5$ and $(p,q)=(3,5)$ . But this implies that $e_{2}$ is divisible by $4$ and $(3^{e_{2}}-1)/2$ must be divisible by $2$ . Hence $(3^{e_{2}}-1)/2$ cannot be of the form $p^{f_{2}}q^{g_{2}}$ . Hence it cannot occur that $x=2^{e_{1}}-1$ is not a prime power.

If $x=(3^{e_{2}}-1)/2=p^{f_{2}}$ is prime or prime power, then

[TABLE]

For none of them, $o_{p}(2)=1,6$ or a prime power. Hence, as above cases, $(5^{e_{3}}-1)/4$ must be divisible by $p$ . Since $o_{p}(5)$ must be $1$ or a prime power, we must have $e_{2}\in\{2,3,5\}$ . If $e_{2}=2$ , then $p=2$ and $e_{3}=2$ , which yields that $q=3$ and $e_{1}=2$ . If $e_{2}=3$ , then $p=13$ and $e_{3}=4$ , which is impossible since $(5^{e_{3}}-1)/4=156=2^{2}\times 3\times 13$ has three distinct prime factors. If $e_{2}=5$ , then $p=11$ and $e_{3}=5$ . Hence $(5^{e_{3}}-1)/4=781=11\times 71$ . This implies that $2^{e_{1}}-1=11^{f_{1}}71^{g_{1}}$ , which is impossible since $2^{10}-1=3\times 31$ and $2^{35}-1=31\times 71\times 127\times 122921$ .

If $x=(3^{e_{2}}-1)/2$ is not a prime power, then

[TABLE]

Hence we can write $x=(3^{e_{1}}-1)/2=pq$ for distinct primes $p<q$ with $p,q\neq 31$ . However, $o_{q}(2)$ or $o_{q}(5)$ can never be $1,6$ or a prime power among the above $e_{2}$ ’s. Hence both $2^{e_{1}}-1$ and $(5^{e_{3}}-1)/4$ must be a power of $p$ . By Lemma 2.5, we must have $p=31$ , which is impossible as mentioned above.

If $x=(5^{e_{3}}-1)/4$ is a prime power, then

[TABLE]

Among them, no $e_{3}$ gives a prime power (or one) residual order $3\pmod{x}$ and only $e_{3}=3$ makes the residual order $2\pmod{x}$ acceptable in view of Lemma 2.3. Hence $p=31,e_{3}=3,e_{1}=5$ and $(3^{e_{2}}-1)/2=q^{f_{2}}$ , which implies that $e_{2}=5$ or $(3^{e_{2}}-1)/2=q$ .

If $x=(5^{e_{3}}-1)/4$ is not a prime power, then

[TABLE]

Hence we can write $x=(5^{e_{3}}-1)/4=pq$ for distinct primes $p<q$ . None of such $e_{3}>2$ gives an acceptable residual order $2\pmod{q}$ or $3\pmod{q}$ in view of Lemma 2.3. Hence we see that neither $2^{e_{1}}-1$ nor $(3^{e_{2}}-1)/2$ can be divisible by $q$ and both must be a power of $p$ , contrary to Lemma 2.5. Hence we must have $e_{3}=2,(p,q)=(2,3)$ . This yields that $e_{1}=e_{2}=2$ .

This completes the proof of Theorem 1.1.

Bibliography19

The reference list from the paper itself. Each links out to its DOI / PubMed record.

1[1] A. S. Bang, Taltheoretiske Undersøgelser , Tidsskrift Math. 5 IV (1886), 70–80 and 130–137.
2[2] Yann Bugeaud, Guillaume Hanrot and Maurice Mignotte, Sur l’équation diophantienne x n − 1 x − 1 = y q superscript 𝑥 𝑛 1 𝑥 1 superscript 𝑦 𝑞 \frac{x^{n}-1}{x-1}=y^{q} , III , Proc. London Math. Soc. 84 (2002), 59–78
3[3] Y. Bugeaud and M. Mignotte, On integers with identical digits , Mathematika 46 (1999), 411–417.
4[4] Yann Bugeaud and Preda Mihăilescu, it On the Nagell-Ljunggren equation x n − 1 x − 1 = y q superscript 𝑥 𝑛 1 𝑥 1 superscript 𝑦 𝑞 \frac{x^{n}-1}{x-1}=y^{q} , Math. Scand. 101 (2007), 177–183.
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8[8] A. K. Lenstra, H. W. Lenstra Jr. and L. Lovász, Factoring polynomials with rational coefficients , Math. Ann. 261 (1982), 515–534.

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On the simultaneous equations σ(2a)=pf1qg1,σ(3b)=pf2qg2,σ(5c)=pf3qg3\sigma(2^{a})=p^{f_{1}}q^{g_{1}},\sigma(3^{b})=p^{f_{2}}q^{g_{2}},\sigma(5^{c})=p^{f_{3}}q^{g_{3}}σ(2a)=pf1​qg1​,σ(3b)=pf2​qg2​,σ(5c)=pf3​qg3​1112010 Mathematics

Abstract

1 Introduction

Theorem 1.1**.**

2 Preliminary Lemmas

Lemma 2.1**.**

Lemma 2.2**.**

Lemma 2.3**.**

Lemma 2.4**.**

Lemma 2.5**.**

Proof.

3 Main Theory

Lemma 3.1**.**

Proof.

Lemma 3.2**.**

Proof.

Lemma 3.3**.**

Proof.

Lemma 3.4**.**

Proof.

4 The final step

On the simultaneous equations $\sigma(2^{a})=p^{f_{1}}q^{g_{1}},\sigma(3^{b})=p^{f_{2}}q^{g_{2}},\sigma(5^{c})=p^{f_{3}}q^{g_{3}}$ 1112010 Mathematics

Theorem 1.1.

Lemma 2.1.

Lemma 2.2.

Lemma 2.3.

Lemma 2.4.

Lemma 2.5.

Lemma 3.1.

Lemma 3.2.

Lemma 3.3.

Lemma 3.4.