数理解析研究所講究録
第 1874 巻 2014 年 215-223

215

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

Tomohiro Yamada

Abstract
We shall solve the simultaneous equations$\sigma(2^{a})=p^{f}1q^{g_{1}},$ $\sigma(3^{b})=$$p^{f_{2}}q^{g_{2}}, \sigma(5^{C})=p^{f_{3}}q^{g_{3}}.$

1

Introduction

We denote by$\sigma(N)$the sum of divisors of $N.$

In the preprint [14], 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. In this preprint, we showedthat the simultaneous equation
for $2k+1$ prime powers$p_{i}^{e_{i}}(i=1,2, \ldots, 2k+1)$ cannot have small solutions$\cdots,p_{2k+1}.$$\sigma(p_{i}^{e}i)=q_{1}^{f_{1i}}\cdots q_{k}^{f_{ki}}$$p_{1},$

Here we use the method in the preprint to solve the simultaneous equations$\sigma(2^{a})=p^{f_{1}}q^{g_{1}},$$\sigma(3^{b})=p^{f_{2}}q^{g_{2}},$ $\sigma(5^{c})=p^{f_{3}}q^{g_{3}}.$

Wakulicz[12] has shown that all solutions of $2^{n}-5^{m}=3$ are $(n, m)=$and $(7, 3)$ , from which Makowski and Schinzel[6] derived thathave only the solution $(a, c)=(4,2)$ . We note that it is easy
to show thathas no nontrivial solution andalso has no nontrivial solution.
$(2,0),$ $(3,1)$$\sigma(2^{a})=\sigma(5^{c})$$\sigma(2^{a})=\sigma(3^{b})$$\sigma(3^{b})=\sigma(5^{c})$Bugeaud and Mignotte[3] has shown that neither of
can be perfect power except $\sigma(3^{4})=11^{2}$ . Moreover, they have shown thatthe only perfect powers
with $x=z^{t},$ $z\leq 10$ areand$\sigma(2^{a}),$$\frac{x^{n}-1}{x-1}$$\frac{7^{4}-1}{7-1}=20^{2}.$$*2000$Mathematics Subject Classification: llA05, llA25.

\dagger Key words and phrases: Odd perfect numbers.

$\sigma(3^{b}),$$\sigma(5^{c})$$\frac{3^{5}-1}{3-1}=11^{2}$

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2 PRELIMINARY LEMMAS
Now we shall state our result.
Theorem 1.1. The simultaneous equations$c>0,$
has only the following solutions:with$\sigma(2^{a})=p^{f_{1}}q^{g_{1}},$ $\sigma(3^{b})=p^{f_{2}}q^{g_{2}},$ $\sigma(5^{c})=$$p^{f_{3}}q^{g_{3}}$$a,$ $b,$$f_{1},$$f_{2},$$f_{3},$$g_{1},$ $g_{2},$$g_{3}\geq 0$

1. $(a, b, c)=(1,1,1)$ .

2. $(a, b, c)=(4,1,2)$ ,3. $(a, b, c)=(4,4,2)$ and

4. $(a, c)=(4,2)$ andis prime.

$\sigma(3^{b})$Our results are related to the Nagell-Ljunggren equation$\frac{x^{n}-1}{x-1}=y^{m},$ $x\geq 2,$ $y\geq 2,$ $n\geq 3,$$q\geq 2$(1),

which has been conjectured to have only finitely many solutions. Some ofrecent remarkable results are [2], [3], [8] and [9].
Now we are led to conjecture that there exists an integer$n_{0}$

such thatthe equation$\frac{x^{n}-1}{x-1}=y^{m}z^{l},$(2)$x\geq 2,$ $y\geq 2,$ $z\geq 2,$ $n,$ $m,$ $l\geq n_{0}$has only finitely many solutions. Theorem 1.1 can be seen to support thisconjecture.

2

Preliminary Lemmas

In this section, we introduce some preliminary lemmas. One is Matveev’s
lower bound for linear forms of logarithms [7].

Lemma 2.1. Let
be nonzero integers such that, let $A_{j} \geq\max\{O.16, \log a_{j}\}.$
are not all zero. For each $j=1,$$a_{1},$$a_{2},$$\ldots,$$\log a_{1},$$a_{n}$$\ldots,$$\ldots,$$\log a_{n}$

$n$

Put
$B= \max\{1, |b_{1}|A_{1}/A_{n}, |b_{2}|A_{2}/A_{n}, \ldots, |b_{n}|\},$$\Omega=A_{1}A_{2}\ldots A_{n},$$C_{0}=1+\log 3-\log 2$ ,$C_{1}(n)= \frac{16}{n!}e^{n}(2n+3)(n+2)(4(n+1))^{n+1}(\frac{1}{2}en)(4.4n+5.5\log n+7)$

(3)

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3 MAIN THEORY
and$\Lambda=b_{1}\log a_{1}+\ldots+b_{n}\log a_{n}$(4).

Then we have$\log|\Lambda|>-C_{1}(n)(C_{0}+\log B)\max\{1, \frac{n}{6}\}\Omega$(5).

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[15] and rediscovered by many
authors such as Dickson[4] and Kanold[5]. See also Theorem 6. $4A.1$ in [11].
Lemma 2.3 is proved in [3], as mentioned above.
$\cdot$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)=$$(2,1,6)$ or $a-b=n=1$ , or $n=2$ and $a+b$ is a power of 2.
Lemma 2.3. Let
be positive integers with $x,$ $f>1$ and $e>$
. The equation $(a^{e}-1)/(a-1)=x^{f}$ has no solution but $(a, e, x, f)=$$(3,5,11,2),$ $(7,4,20,2)$ in integers $2\leq a\leq 10,$ $e>2,$ $x>1,$ $f>1.$$a,$ $e,$ $x,$ $f$

$a,$

$2$

Using Lemmas 2.2 and 2.3, we can prove the following lemma.
Lemma 2.4. If $(a^{e}-1)/(a-1)=p^{f_{1}}q^{f_{2}}$ for some integers
andprime $p<q$ , then we have $(a, e,p, q, f_{1}, f_{2})=(2,6,3,7,2,1),$ $e=r$ $ore=r^{2}$for some prime . Moreover, in the case $e=r$ , then we have $p\geq r$ . In thecase $e=r^{2}$ , 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.$$a,$ $e,.f_{1},$$f_{2}$$r$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$e_{i}\log a_{i}<E_{i}=C_{i}\log p\log q(\log\log p+C_{i+3})$,(6)

where $C_{1}=1.5\cross 10^{10},$ $C_{2}=1.3\cross 10^{12},$ $C_{3}=1.9\cross 10^{12},$ $C_{4}=1.3\cross 10^{10},$ $C_{5}=$$1.1\cross 10^{12},$ $C_{6}=1.6\cross 10^{12}.$

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3 MAIN THEORY

Proof. We may assume that$e_{1},$$e_{2},$$e_{3}>10^{10}\log q$

and $q>10.$

Let $\Lambda_{i}=f_{1}\log a_{i}+g_{1}\log q+\log(a_{i}-1)-e_{i}\log 2=\log(1-a_{i}^{-e_{i}})$ for$i=1,2,3.$

Matveev’s theorem gives$-\log|\Lambda_{1}|<C(3)(C_{0}+\log(e_{1}\log 2/\log q))\log 2\log p\log q$

,

(7)

(8)

$-\log|\Lambda_{2}|<C(4)(C_{0}+\log(e_{2}\log 3/\log q))\log 2\log 3\log p\log q$

and$-\log|\Lambda_{3}|<C(4)(C_{0}+\log(e_{3}\log 5/\log q))\log 2\log 5\log p\log q$.

(9)

Now we shall show (6) in the case $i=1$ . Since $0<|\Lambda_{1}|=-\log(1-$, we have-log $|\Lambda_{1}|>\log(2^{e_{1}}-1)\geq(1-10^{-10})e_{1}\log 2.$$2^{-e_{1}})< \frac{1}{2^{e}1-1}$

Combining upper and lower bounds for$\Lambda_{1}$, we obtain$\frac{e_{1}\log 2}{\log q}<(1+10^{-10})\frac{C_{0}+\log(10^{10})}{C_{0}}C(3)\log 2\log(e_{1}\log 2/\log q)\log p$. (10)

This gives (6) in the case $i=1.$
Next we shall prove (6) in the case $i=2$ . Since $0<|\Lambda_{2}|=-\log(1-$, we have $\log|\Lambda_{2}|>\log(3^{e_{2}}-1)\geq(1-10^{-10})e_{2}\log 3.$$3^{-e}2)< \frac{1}{3^{e}2-1}$-

Combining upper and lower bounds for$\Lambda_{1}$, we obtain$\frac{e_{2}\log 3}{\log q}<(1+10^{-10})\frac{C_{0}+\log(10^{10})}{C_{0}}C(4)\cross\log 2\log 3\log(e_{2}\log 3/\log q)\log p.$(11)

Since $0<| \Lambda_{2}|=-\log(1-3^{-e_{2}})<\frac{1}{3^{e_{2}}-1}$ , we have $-\log|\Lambda_{2}|>\log(3^{e_{2}}-$
1 $\geq(1-10^{-10})e_{2}\log 3$ and therefore$)$$\frac{e_{2}\log 3}{\log q}<(1+10^{-10})\frac{C_{0}+\log(10^{10})}{C_{0}}C(4)\log 2\log 3\log(e_{2}\log 3/\log q)\log p.$(12)This gives (6) in the case $i=2.$

A similar argument yields (6) in the case $i=3$ . This completes theproof of the lemma.
$\square$Next, we shall show that we cannot have all of$a_{i}^{ei\prime}s$small.

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3 MAIN THEORY
Lemma 3.2. Let$f_{3}g_{1}-f_{1}g_{3}$andbe the smallest among. Let $h_{1}=f_{2}g_{3}-f_{3}g_{2},$ $h_{2}=$and $H= \max|h_{i}|$ . Then$x$$a_{i}^{e_{i}}s$$h_{3}=f_{1}g_{2}-f_{2}g_{1}$$\log x\leq\log(7H/4)+C(3)(C_{0}+\log((e_{1}+2)H))\log 2\log 3\log 5$.

(13)

Proof. We begin by observing that$(2^{e1}-1)^{h_{1}}( \frac{3^{e_{2}}-1}{2})^{h_{2}}(\frac{5^{e_{3}}-1}{4})^{h_{3}}=1$

(14)

.

Now we put$\Lambda=(e_{1}h_{1}-h_{2}-2h_{3})\log 2+e_{2}h_{2}\log 3+e_{3}h_{3}\log 5$$=h_{1} \log\frac{2^{e_{1}}}{2^{e_{1}}-1}+h_{2}\log\frac{3^{e}2}{3^{e_{2}}-1}+h_{3}\log\frac{5^{e}3}{5^{e_{3}}-1}$

(15)

.

Then we have$0<| \Lambda|\leq H(\frac{1}{2^{e_{1}}-1}+\frac{1}{3^{e_{2}}-1}+\frac{1}{5^{e_{3}}-1)}\leq\frac{7H}{4x}$(16)$\log|\Lambda|\leq-\log x+\log(7H/4)$ .

(17)

and therefore
It follows from the assumption $e_{i}>0$ thatlower bound gives$\Lambda\neq 0$. Hence Matveev’s$\log|\Lambda|\geq-C(3)(C_{0}+\log((e_{1}+2)H))\log 2\log 3\log 5$

.

(18)

Combining (17) and (18), we obtain (13).
The third step is to obtain upper bounds for each$\square$$e_{i}.$

Lemma 3.3. We have $e_{1}<1.1\cross 10^{59},$ $e_{2}<10^{63}$ and $e_{3}<1.5\cross 10^{63}.$

Proof. We begin by considering the case $q|x$ . In this case, we have $\log q<$$\log x<\log(7H/4)+C(3)(C_{0}+\log((e_{1}+2)H))\log 2\log 3\log 5$ . We notethat $H\leq C_{2}C_{3}\log p\log$ q(log log $p+C_{5}$ ) $(\log\log p+C_{6})$ . By Lemma 3.1,
we have $f_{i}\leq C_{i}\log q$ (log log $p+C_{i+3}$ ), $g_{i}\leq C_{i}\log p(\log\log p+C_{i+3})$ andtherefore $H<C_{2}C_{3}(\log q)^{2}$ (log log $q+C_{5}$ ) $(\log\log q+C_{6})$ . Hence we obtain$\log p<\log q<5.8\cross 10^{12}.$Now we consider the case
. Put$(a_{i}^{e_{i}}-1)/(a_{i}-1),$
be the others and$q\nmid x$

$j,$

$i$

to be the index such that $x=$$k$$\Lambda’=e_{j}h_{j}\log a_{j}+e_{k}h_{k}\log a_{k}-h_{j}\log(a_{j}-1)-h_{J}\prime\log(a_{j}-1)+h_{3}\log x$$=h_{j} \log\frac{a_{j}^{e}j}{a_{j}^{e_{j}}-1}+h_{k}\log\frac{a_{k}^{e_{k}}}{a_{k}^{e_{k}}-1}.$

(19)

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3 MAIN THEORY
Now Lemma 2.3 implies that $(a^{e}-1)/(a-1)=p^{f}$ with $a\in\{2,3,5\}$
implies that $f=1$ unless $(a, e,p, f)=(3,5,11,2)$ . Therefore we see that
$x=p_{1}$ or $(p_{1}, x)=(11,11^{2})$ , and $(a_{j}^{e}j-1)/(a_{j}-1)$ and $(a_{k}^{e_{k}}-1)/(a_{k}-1)$
must be divisible by$p_{2}.$

Then we have

.

(20)

$0< \Lambda’<H(\frac{1}{a_{1}^{e_{1}}-1}+\frac{1}{a_{2^{2}}^{e}-1})\leq\frac{3H}{2q}$

Similarly to the above, Matveev’s theorem now gives$\log|\Lambda’|\geq-C(4)(C_{0}+\log(E_{3}H/\log x))\log 2\log 3\log 5\log x$ .

(21)

Combining (20) and (21), we obtain$\log q\leq\log(3H/2)+C(4)(C_{0}+\log(E_{3}H/\log x))\log 2\log 3\log 5\log x$ .

(22)

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 (13) and (22), we
obtain $\log q<6.0\cross 10^{25}$ . Moreover, $\log p=\log x<\log(7H/4)+C(3)(C_{0}+$$\log((e_{1}+2)H))\log 2\log 3\log 5$ gives $\log p<7.1\cross 10^{12}.$Now we conclude that in both cases, we have $\log p<7.1\cross 10^{12}$ and. Observing that $(e_{1}-1)\log 2<f_{1}\log p+g_{1}\log q,$ $(e_{2}-$$1)\log 3<f_{2}\log p+g_{2}\log q$ and $(e_{3}-1)$ log5 $<f_{3}\log p+g_{3}\log q$ , we have$e_{1}<1.1\cross 10^{59},$ $e_{2}<10^{63}$ and $e_{3}<1.5\cross 10^{63}.$$\log q<6.0\cross 10^{25}$$\square$

The last step is to reduce our upper bounds into feasible ones.

Lemma 3.4. $x\leq 1550712.$

Since $x\geq 2^{H}-1$ , we have$| \Lambda|<\frac{7H}{4x}<\frac{7\cross 2^{H}}{4(2^{H}-1)}\exp(\log H-H\log 2)$

.

(23)

Let $M$ be the matrix defined by $m_{12}=m_{13}=m_{21}=m_{23}=0$ andbe the reduced matrix of $M.$
$m_{11}=m_{22}=\gamma$ and$m_{3i}=\lfloor C\gamma\log a_{i}\rfloor.$$L$

Now we know that $H<H_{0}=1.5\cross 10^{126}$ and Lemma 3.7 of de Weger’sbook[13] with $C=10^{380},$ $\gamma=2$ gives that $X_{1}>H_{0}$ and we see that (23)has no solutions with $X_{1}>H>H_{1}=854$ . So that $H\leq 854.$

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4 CONSEQUENCES FROM THE $ABC$ CONJECTURE
Iterating this argument with $C=10^{10},$ $\gamma=3$ gives that $X_{1}>H_{1}$and we see that $H\leq H_{2}=30$ . Finally, iterating this argument with
$C=150000,$ $\gamma=3$ gives that $X_{1}>H_{2}$ and we see that $H\leq 19.$

Now we have $|\Lambda|\geq-15\log 2+8\log 3+\log 5$ for $H\leq 19$ . Since$-15\log 2+8\log 3+\log 5=0.001128\cdots$ , we conclude that $x\leq 1550712.$$\frac{7H}{4x}>$

The final step is checking all possibilities of$x.$

If $x=2^{e_{1}}-1$ , then $e_{1}\in\{2,3,4,5,6,7,9,11,13,17,19\}$ . If $x=(3^{e_{2}}-$$1)/2$ , then $e_{2}\in\{2,3,4,5,7,9,11,13\}$ . Moreover, if $x=(5^{e_{3}}-1)/4$, then$e_{3}\in\{2,3,5,7\}.$

Here we exhibit only the proof of $x\neq 2^{9}-1$ . If $x=2^{9}-1=7\cross 73,$
then $(p, q)=(7,73)$ . So that must divide either $3^{e_{2}}-1$ or $5^{e_{3}}-1$ . If$p|(3^{e}2-1)$ , then 6, which is impossible by Lemma 2.4. If $P|5^{e_{3}}-1,$
then 6, which contradicts 2.4 again. Thus cannot be $2^{9}-1.$$p$$|e_{2}$$|e_{3}$4$x$Consequences from the ab conjecture$c$

In Aug. 31. 2012, Mochizuki[10] claims to prove the abc conjecture. If
Mochizuki’s proof is right, Mochizuki’s theorem gives that, if $(x^{n}-1)/(x-$1with $n\geq 3,$ $lm\geq 2$ and $y<z$ , then for any given, up toonly finitely many counterexamples, we have$)$$=y^{m}z^{l}$$\epsilon>0$

1. $(n, m, l)=(3,1,2)$ ,2. $(n, l)=(3,2),$ $m\geq 2$ and $\log y<\epsilon\log z,$
3. $(n, m, l)=(3,1,3),$ $(4,1,2)$ and $\log y<(1+\epsilon)\log z$ , or
4. $l=1,$ $m\geq 2$ and $\log y<\frac{1+\epsilon}{(n-2)m-(n-1)}\log z.$
Moreover, Mochizuki’s theorem implies that for any fixed$(x^{n}-$solutions. Another consequence ofMochizuki’s theorem is that $(x_{1}^{n_{1}}-1)/(x-1)=y^{m_{1}}z^{l_{1}}$ and $(x_{2}^{n_{2}}-1)/(x-1)=$
have only finitely many solutions inwith $y<z$ and$y,$ $z,$$1)/(x-1)=y^{m}z^{l}$ has at most two integer$y^{m2}z^{l_{2}}$$(x_{1}, x_{2}, n_{1}, n_{2}, y, z, m_{1}, m_{2}, l_{1}, l_{2})$$n_{1},$ $n_{2}\geq 3,$$l_{1},$$l_{2}\geq 1.$

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Tomohiro Yamada
Center for Japanese language and culture
Osaka University562-8558
8-1-1, Aomatanihigashi, Minoo, OsakaJapan-mail: tyamada1093@gmail.com$e$