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\theoremstyle{plain}
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\begin{center}
\vskip 1cm{\LARGE\bf The 4-Nicol Numbers Having \\
\vskip .1in
 Five Different Prime Divisors }
\vskip 1cm \large {Qiao-Xiao Jin and Min Tang\footnote{Corresponding author.
This work was supported
by the National Natural Science Foundation of China, Grant No.\
10901002 and the SF of the Education Department of Anhui Province,
Grant No.\ KJ2010A126.}}\\
Department of Mathematics \\
Anhui Normal University\\
Wuhu 241000 \\
P. R. China\\
\href{mailto:tangmin@mail.ahnu.edu.cn}{\tt tangmin@mail.ahnu.edu.cn}\\
\end{center}

\vskip .2in

\begin{abstract}
A positive integer $n$ is called a Nicol number if $n\mid
\varphi(n)+\sigma(n)$, and a $t$-Nicol number if
$\varphi(n)+\sigma(n)=tn$. In this paper, we show that if $n$ is a
4-Nicol number that has five different prime divisors, then $n=2^{\alpha_{1}}\cdot 3\cdot 5^{\alpha_{3}}\cdot p^{\alpha_{4}}\cdot q^{\alpha_{5}}$, or
$n=2^{\alpha_{1}}\cdot 3\cdot 7^{\alpha_{3}}\cdot p^{\alpha_{4}}\cdot q^{\alpha_{5}}$ with $p\leq 29$.
\end{abstract}

\section{Introduction}

For any positive integer $n$, let $\phi(n)$, $\omega(n)$ and
$\sigma(n)$ be the Euler function of $n$, the number of prime
divisors of $n$ and the sum of divisors of $n$, respectively. We
call $n$ is a Nicol number if $n\mid\varphi(n)+\sigma(n)$, and a
t-Nicol number if $\varphi(n)+\sigma(n)=tn$. It is well-known that
$t\geq 2$, and $n$ is prime if and only if
$\varphi(n)+\sigma(n)=2n$. In 1966, Nicol \cite{Nicol}
conjectured that Nicol numbers are all even, and proved that if
$\alpha$ is such that $p=2^{\alpha-2}\cdot7-1$ is prime, then
$n=2^{\alpha}\cdot3\cdot p$ is 3-Nicol number. In 1995, Ming-Zhi
Zhang \cite{Zhang} showed that if $n=p^{\alpha}q$ then $n$ cannot be
a Nicol number, where $p$ and $q$ are distinct primes and $\alpha$
is a positive integer. In 1997, Lin and Zhang
\cite{Lin} showed that if $\omega(n)=2$, then $n$ cannot be a Nicol
number. In 2008, Luca and Sandor \cite{Luca} showed that if
$n$ is a Nicol number and $\omega(n)=3$, then either $n\in
\{560,588,1400\}$ or $n=2^{\alpha}\cdot 3\cdot p$ with
$p=2^{\alpha-2}\cdot 7-1$ prime. In 2008, Wang \cite{Wang} studied the Nicol numbers
that
have four different prime divisors. In 2009,
Harris \cite{Harris} showed that the Nicol numbers that
have four different prime divisors must be one of the following
forms:

$1.\quad n=2^{3}\cdot 3^{3}\cdot5^{2}\cdot11$, $2^{4}\cdot
3^{3}\cdot5\cdot11$, $2^{7}\cdot 5\cdot 11\cdot 79$, $2^{3}\cdot
3^{3}\cdot5^{3}\cdot13^{2}$, $2^{2}\cdot 3^{2}\cdot 17\cdot 241$,
$2^{2}\cdot 3^{2}\cdot 17^{2}\cdot 2243$;

$2.\quad n\in \big\{2^{a}\cdot 3\cdot p_{3}\cdot p_{4} \big|
p_{4}=\displaystyle\frac{(7\cdot 2^{a-2}-1)p_{3}+9\cdot
2^{a-2}-1}{p_{3}-(7\cdot 2^{a-2}-1)}$, where $p_{3}$, $p_{4}$ are
distinct primes.$\big\}$

Moreover, Harris \cite{Harris} proved that all but finitely many Nicol
numbers that have 5 different prime divisors are divisible by 6 and
not 9.

In this paper, we study the 4-Nicol
numbers that have five different prime divisors and obtain the
following result:

\begin{theorem}\label{thm1} If $n$ is a 4-Nicol number with $\omega(n)=5$, then either $n=2^{\alpha_{1}}3^{\alpha_{2}}5^{\alpha_{3}} p^{\alpha_{4}}q^{\alpha_{5}}$, or $n=2^{\alpha_{1}}3^{\alpha_{2}}7^{\alpha_{3}} p^{\alpha_{4}}q^{\alpha_{5}}$ with $p\leq 29$, where $p$, $q$ are distinct primes, and
$\alpha_{i}(i=1,2,\cdots,5)$ are positive integers. \end{theorem}

By the Harris result and Theorem \ref{thm1}, we have the following result:

\begin{corollary} All but finitely many 4-Nicol numbers with 5 different prime
divisors have the form $n=2^{\alpha_{1}}\cdot 3\cdot 5^{\alpha_{3}}\cdot p^{\alpha_{4}}\cdot q^{\alpha_{5}}$, or
$n=2^{\alpha_{1}}\cdot 3\cdot 7^{\alpha_{3}}\cdot p^{\alpha_{4}}\cdot q^{\alpha_{5}}$ with $p\leq 29$. \end{corollary}

\bigbreak

Throughout this paper, let $a$ and $m$ be relatively prime positive
integers, the least positive integer $x$ such that $a^{x} \equiv 1
\pmod m$ is called the order of $a$ modulo $m$. We denote the order
of $a$ modulo $m$ by $\textnormal {ord}_{m}(a)$. Let $V_{p}(m)$ be the exponent of
the highest power of $p$ that divides $m$.
 \bigbreak


\section{Lemmas}
The following three lemmas are motivated by the work of
Luca and S\'{a}ndor \cite{Luca}. Here we make some minor revisions.

\begin{lemma}\label{lem1} Let $a$, $b$ be two natural numbers and $p$ be an odd prime. If $V_{p}(a-1)\geq 1$, then
$$V_{p}(a^{b}-1)=V_{p}(b)+V_{p}(a-1).$$\end{lemma}

\begin{proof} Let $V_{p}(b)=m$ and $V_{p}(a-1)=n$. We may assume that $b=p^{m}t$ with $p\nmid t$ and $a=1+p^{n}a_{0}$ with $p\nmid
a_{0}$.

Since $n\geq 1$, we have
$$a^{t}=(1+p^{n}a_{0})^{t}=1+C_{t}^{1}p^{n}a_{0}+\cdots+C_{t}^{t}(p^{n}a_{0})^{t}=1+p^{n}c, \quad p\nmid c.$$
 Thus
$$a^{tp}=(1+p^{n}c)^{p}=1+C_{p}^{1}p^{n}c+\cdots+C_{p}^{p}(p^{n}c)^{p}=1+p^{n+1}a_{1}, \quad p\nmid a_{1}.$$
By induction on $m$, for all $m\geq 0$ we have $a^{b}=a^{tp^{m}}=1+p^{m+n}a_{m}$ with $p\nmid a_{m}$.
Hence $$V_{p}(a^{b}-1)=m+n=V_{p}(b)+V_{p}(a-1).$$

 This completes the proof of Lemma \ref{lem1}.
\end{proof}\vskip 3mm

\begin{lemma}\label{lem2} Let $t$ be a natural number and $p$, $q$ be two primes.
We have $$V_{p}(q^{t}-1)\leq V_{p}(q^{f}-1)+V_{p}(t),$$ where
$f=\textnormal{ord}_{p}(q)$, if $p\neq 2$; and $f=2$, if
$p=2$.\end{lemma}

\begin{proof} (i) $p=2$. By \cite[Lemma 1]{Luca}, we have
$$V_{2}(q^{t}-1)\leq V_{2}(q^{2}-1)+V_{2}(t).$$

(ii) $p>2$. Now consider the following two cases:

 {\bf Case 1.} $q^{t}\not\equiv 1 \pmod p$. The above inequality is obvious.

 {\bf Case 2.} $q^{t}\equiv 1 \pmod p$. Then $\textnormal{ord}_{p}(q)\mid
t$ and $\textnormal{ord}_{p}(q)\mid p-1$.
 Let $V_{p}(t)=m$. We may assume that $t=\textnormal{ord}_{p}(q)\cdot p^{m}\cdot k$ with $p\nmid k$. Thus $$\begin{array}{ll}
V_{p}(q^{t}-1)& =V_{p}\big((q^{\textnormal{ord}_{p}(q)})^{p^{m}\cdot
k}-1\big)\\
&=V_{p}(q^{\textnormal{ord}_{p}(q)}-1)+V_{p}({p^{m}\cdot
k})\\
&=V_{p}(q^{\textnormal{ord}_{p}(q)}-1)+m\\
&=V_{p}(q^{\textnormal{ord}_{p}(q)}-1)+V_{p}(t).\end{array}$$

 This completes the proof of Lemma \ref{lem2}.
 \end{proof}

 \begin{lemma}\label{lem3} Let
$n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{k}^{\alpha_{k}}$
be the standard factorization of $n$ and $X=\max\{\alpha_{j}\mid j=1,2,\cdots,k\}$. We fix $i\in\{1,\cdots,k\}$ such that $X=\alpha_i$. If $n$ is a Nicol number, then we have
$$X-1\leq \sum_{j=1,j\neq i}^k V_{p_{i}}(p_{j}^{f_{j}}-1)+\displaystyle\frac{k-1}{\log{p_{i}}}\log(X+1),$$
where $f_{j}=ord_{p_{i}}(p_{j})$, if $p_{i}\neq 2$; and $f_{j}=2$,
if $p_{i}=2$.\end{lemma}

\begin{proof} Since $n\mid
\phi(n)+\sigma(n)$ and $p_{i}^{X-1}\mid \phi(n)$, we have $$p_{i}^{X-1}\mid \sigma(n)=\prod_{j=1}^k
\Big(\displaystyle\frac{p_{j}^{\alpha_{j}+1}-1}{p_{j}-1}\Big).$$ Hence $$p_{i}^{X-1}\Big|\prod_{j=1}^k (p_{j}^{\alpha_{j}+1}-1).$$
The above relation implies that $$X-1\leq \sum_{j=1}^k
V_{p_{i}}(p_{j}^{\alpha_{j}+1}-1)=\sum_{j=1,j\neq i}^k
V_{p_{i}}(p_{j}^{\alpha_{j}+1}-1).$$

 By Lemma \ref{lem2}
$$\begin{array}{ll} X-1
&\leq \displaystyle\sum\limits_{j=1,j\neq i}^k
V_{p_{i}}(p_{j}^{f_{j}}-1)+\sum\limits_{j=1,j\neq i}^k
V_{p_{i}}(\alpha_{j}+1)\\
&\leq \displaystyle\sum\limits_{j=1,j\neq i}^k
V_{p_{i}}(p_{j}^{f_{j}}-1)+\sum\limits_{j=1,j\neq i}^k
\displaystyle\frac{\log(\alpha_{j}+1)}{\log{p_{i}}}
\\
&\leq \displaystyle\sum\limits_{j=1,j\neq i}^k
V_{p_{i}}(p_{j}^{f_{j}}-1)+\displaystyle\frac{k-1}{\log{p_{i}}}\log(X+1).\end{array}$$


 This completes the proof of Lemma \ref{lem3}.
 \end{proof}


\begin{lemma}\label{lem4} If $n$ is  a 4-Nicol number and $\omega(n)=5$, then
$n$ must be one of the following three forms:

$1.\quad n=2^{\alpha_{1}}\cdot 3^{\alpha_{2}}\cdot
5^{\alpha_{3}}\cdot p^{\alpha_{4}}\cdot q^{\alpha_{5}}$, $p,q$ are
distinct primes and $7\leq p<q$.

$2.\quad n=2^{\alpha_{1}}\cdot 3^{\alpha_{2}}\cdot
7^{\alpha_{3}}\cdot p^{\alpha_{4}}\cdot q^{\alpha_{5}}$, $p<q$ are
distinct primes and $p\leq 29$.

$3.\quad n=2^{\alpha_{1}}\cdot 3^{\alpha_{2}}\cdot
11^{\alpha_{3}}\cdot 13^{\alpha_{4}}\cdot p^{\alpha_{5}}$, $p\leq
23$ is prime.\end{lemma}

\begin{proof} Let
$n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}p_{3}^{\alpha_{3}}p_{4}^{\alpha_{4}}p_{5}^{\alpha_{5}}$
be the standard factorization of $n$.
Put $l=\displaystyle\frac{\sigma(n)}{n}$.

Noting that $n$ is a
4-Nicol number and $\varphi(n)\sigma(n)<n^{2}$, we have
$4=\displaystyle\frac{\varphi(n)}{n}+\displaystyle\frac{\sigma(n)}{n}<l+l^{-1}$,
hence $l>2+\sqrt{3}$. By
$\displaystyle\frac{n}{\varphi(n)}>\displaystyle\frac{\sigma(n)}{n}=l$,
we have
$$\displaystyle\frac{n}{\varphi(n)}=\displaystyle\frac{p_{1}}{p_{1}-1}\frac{p_{2}}{p_{2}-1}\frac{p_{3}}{p_{3}-1}\frac{p_{4}}{p_{4}-1}
\frac{p_{5}}{p_{5}-1}>l>2+\sqrt{3}.$$

If $p_{2}\geq 5$ then
$$\displaystyle\frac{n}{\varphi(n)}=\displaystyle\frac{p_{1}}{p_{1}-1}\frac{p_{2}}{p_{2}-1}\frac{p_{3}}{p_{3}-1}
\frac{p_{4}}{p_{4}-1}\frac{p_{5}}{p_{5}-1}\leq \displaystyle2\cdot
\frac{5}{4}\cdot \frac{7}{6}\cdot \frac{11}{10}\cdot
\frac{13}{12}<2+\sqrt{3},$$ a contradiction. Thus $p_{2}=3$ and
$p_{1}=2$.

If $p_{3}\geq 13$ then
$$\displaystyle\frac{n}{\varphi(n)}=\displaystyle\frac{p_{1}}{p_{1}-1}\frac{p_{2}}{p_{2}-1}\frac{p_{3}}{p_{3}-1}
\frac{p_{4}}{p_{4}-1}\frac{p_{5}}{p_{5}-1}\leq \displaystyle2\cdot
\frac{3}{2}\cdot \frac{13}{12}\cdot \frac{17}{16}\cdot
\frac{19}{18}<2+\sqrt{3},$$ a contradiction, thus $p_{3}\leq 11$.

 {\bf Case 1.} $p_{3}=7$. Then $p_{4}\leq 29$. In fact, if
$p_{4}\geq 31$ then
$$\displaystyle\frac{n}{\varphi(n)}=\displaystyle\frac{p_{1}}{p_{1}-1}\frac{p_{2}}{p_{2}-1}\frac{p_{3}}{p_{3}-1}
\frac{p_{4}}{p_{4}-1}\frac{p_{5}}{p_{5}-1}\leq \displaystyle2\cdot
\frac{3}{2}\cdot \frac{7}{6}\cdot  \frac{31}{30}\cdot
\frac{37}{36}<2+\sqrt{3},$$ a contradiction.

 {\bf Case 2.} $p_{3}=11$. Then $p_{4}=13$ and $p_{5}\leq
23$. In fact, if $p_{4}\geq 17$ then
$$\displaystyle\frac{n}{\varphi(n)}=\displaystyle\frac{p_{1}}{p_{1}-1}\frac{p_{2}}{p_{2}-1}\frac{p_{3}}{p_{3}-1}
\frac{p_{4}}{p_{4}-1}\frac{p_{5}}{p_{5}-1}\leq \displaystyle2\cdot
\frac{3}{2}\cdot \frac{11}{10}\cdot \frac{17}{16}\cdot
\frac{19}{18}<2+\sqrt{3},$$ a contradiction.

If $p_{5}\geq 29$ then
$$\displaystyle\frac{n}{\varphi(n)}=\displaystyle\frac{p_{1}}{p_{1}-1}\frac{p_{2}}{p_{2}-1}\frac{p_{3}}{p_{3}-1}
\frac{p_{4}}{p_{4}-1}\frac{p_{5}}{p_{5}-1}\leq \displaystyle2\cdot
\frac{3}{2}\cdot \frac{11}{10}\cdot \frac{13}{12}\cdot
\frac{29}{28}<2+\sqrt{3},$$ a contradiction.

This completes the proof of Lemma \ref{lem4}.\end{proof}


\section{Proof of Theorem \ref{thm1}}
By Lemma \ref{lem4}, it is enough to show that there is no 4-Nicol
numbers $n=2^{\alpha_{1}}\cdot 3^{\alpha_{2}}\cdot
11^{\alpha_{3}}\cdot 13^{\alpha_{4}}\cdot p^{\alpha_{5}}$ with $p\leq
23$.

Assume that $n=2^{\alpha_{1}}\cdot 3^{\alpha_{2}}\cdot
11^{\alpha_{3}}\cdot 13^{\alpha_{4}}\cdot p^{\alpha_{5}}$ with $p\leq
23$ be a 4-Nicol number, then by
$\displaystyle\frac{\varphi(n)}{n}+\displaystyle\frac{\sigma(n)}{n}=4$
we have:
$$\begin{array}{ll}2^{\alpha_{1}+6}\cdot 3^{\alpha_{2}+1}\cdot 5\cdot 11^{\alpha_{3}-1}\cdot 13^{\alpha_{4}-1}\cdot p^{\alpha_{5}-1}\cdot
(133p+10)\cdot (p-1)
\\
=(2^{\alpha_{1}+1}-1)\cdot (3^{\alpha_{2}+1}-1)\cdot
(11^{\alpha_{3}+1}-1)\cdot (13^{\alpha_{4}+1}-1)\cdot
(p^{\alpha_{5}+1}-1).\end{array}$$

 {\bf Case 1.} $p=17$, $n=2^{\alpha_{1}}\cdot
3^{\alpha_{2}}\cdot 11^{\alpha_{3}}\cdot 13^{\alpha_{4}}\cdot
17^{\alpha_{5}}$. Then

\begin{align}&2^{\alpha_{1}+10}\cdot 3^{\alpha_{2}+2}\cdot
5\cdot 11^{\alpha_{3}-1}\cdot 13^{\alpha_{4}-1}\cdot
17^{\alpha_{5}-1}\cdot
757\nonumber\\
&=(2^{\alpha_{1}+1}-1)(3^{\alpha_{2}+1}-1)(11^{\alpha_{3}+1}-1)(13^{\alpha_{4}+1}-1)(17^{\alpha_{5}+1}-1).\label{e:1}\end{align}


 By Lemma \ref{lem3} we have $X\leq 35$. Noting that
$$\textnormal{ord}_{757}(2)=756, \textnormal{ord}_{757}(3)=9,
\textnormal{ord}_{757}(11)=\textnormal{ord}_{757}(13)=\textnormal{ord}_{757}(17)=189,$$ thus
$$757\nmid 2^{\alpha_{1}+1}-1, 757\nmid11^{\alpha_{3}+1}-1,
757\nmid 13^{\alpha_{4}+1}-1, 757\nmid17^{\alpha_{5}+1}-1 .$$ By
(\ref{e:1}) we have $757\mid 3^{\alpha_{2}+1}-1$, thus
$\alpha_{2}+1=9k, k\in \mathbb{Z}$. By $X\leq 35$, we have
$k=1,2,3$.

 {\bf Subcase 1}:   $k=1$, $\alpha_{2}+1=9$. Then
$$\begin{array}{ll}2^{\alpha_{1}+9}\cdot 3^{10}\cdot
5\cdot 11^{\alpha_{3}-1}\cdot 13^{\alpha_{4}-2}\cdot
17^{\alpha_{5}-1}\\
=(2^{\alpha_{1}+1}-1)(11^{\alpha_{3}+1}-1)(13^{\alpha_{4}+1}-1)(17^{\alpha_{5}+1}-1).\end{array}$$

$(i)$ $\alpha_{4}=2$. By $\textnormal{ord}_{61}(13)=3$ we have $61\mid
13^{3}-1$, this is impossible.

$(ii)$ $\alpha_{4}>2$. Then $13\mid (2^{\alpha_{1}+1}-1)(11^{\alpha_{3}+1}-1)(17^{\alpha_{5}+1}-1)$.
On the other hand, we have the following facts: If $13\mid 2^{\alpha_{1}+1}-1$, by $\textnormal{ord}_{13}(2)=12$, thus $12\mid
\alpha_{1}+1$, and noting that $\textnormal{ord}_{7}(2)=3$ we have $7\mid
2^{\alpha_{1}+1}-1$, which is impossible.
If $13\mid 11^{\alpha_{3}+1}-1$, by $\textnormal{ord}_{13}(11)=12$, thus $12\mid
\alpha_{3}+1$, and noting that $\textnormal{ord}_{7}(11)=3$ we have $7\mid
11^{\alpha_{3}+1}-1$, which is impossible.
If $13\mid 17^{\alpha_{5}+1}-1$, by $\textnormal{ord}_{13}(17)=6$, thus $6\mid
\alpha_{5}+1$, and noting that $\textnormal{ord}_{7}(17)=6$ we have $7\mid
17^{\alpha_{5}+1}-1$, which is impossible.
Thus $13\nmid (2^{\alpha_{1}+1}-1)(11^{\alpha_{3}+1}-1)(17^{\alpha_{5}+1}-1)$, a contradiction.

 {\bf Subcase 2}: $k=2$, $\alpha_{2}+1=18$. By
$\textnormal{ord}_{7}(3)=6$, we have $7\mid 3^{18}-1$, thus $7\mid
3^{\alpha_{2}+1}-1$, which contradicts (\ref{e:1}).

 {\bf Subcase 3}: $k=3$, $\alpha_{2}+1=27$. By
$\textnormal{ord}_{757}(3)=9$, we have $757\mid 3^{27}-1$, thus $757\mid
3^{\alpha_{2}+1}-1$, which contradicts (\ref{e:1}).

 {\bf Case 2.} $p=19$, $n=2^{\alpha_{1}}\cdot
3^{\alpha_{2}}\cdot 11^{\alpha_{3}}\cdot 13^{\alpha_{4}}\cdot
19^{\alpha_{5}}$. Then
\begin{align}&2^{\alpha_{1}+7}\cdot 3^{\alpha_{2}+3}\cdot
5\cdot 11^{\alpha_{3}-1}\cdot 13^{\alpha_{4}-1}\cdot
19^{\alpha_{5}-1}\cdot
43\cdot 59\nonumber\\
&=(2^{\alpha_{1}+1}-1)(3^{\alpha_{2}+1}-1)(11^{\alpha_{3}+1}-1)(13^{\alpha_{4}+1}-1)(19^{\alpha_{5}+1}-1).\label{e:2}\end{align}
 By Lemma \ref{lem3} we have $X\leq 33$. Noting that
$$\textnormal{ord}_{59}(2)=\textnormal{ord}_{59}(11)= \textnormal{ord}_{59}(13)=58,
\textnormal{ord}_{59}(3)=\textnormal{ord}_{59}(19)=29,$$ we have
$$59\nmid2^{\alpha_{1}+1}-1, 59\nmid11^{\alpha_{3}+1}-1,
59\nmid13^{\alpha_{4}+1}-1.$$ By (\ref{e:2}) we have $59\mid
3^{\alpha_{2}+1}-1$ or $59\mid19^{\alpha_{5}+1}-1$. If $59\mid
3^{\alpha_{2}+1}-1$, then $29\mid \alpha_{2}+1$. Since
$\textnormal{ord}_{28537}(3)=29$, we have $28537\mid
3^{\alpha_{2}+1}-1$, which contradicts with (\ref{e:2}), thus
$59\nmid 3^{\alpha_{2}+1}-1$ . If $59\mid 19^{\alpha_{5}+1}-1$, then
$29\mid \alpha_{5}+1$. Since $\textnormal{ord}_{233}(19)=29$, we
have $233\mid 19^{\alpha_{5}+1}-1$, which contradicts with
(\ref{e:2}), thus $59\nmid 19^{\alpha_{5}+1}-1$.

 {\bf Case 3.} $p=23$, $n=2^{\alpha_{1}}\cdot
3^{\alpha_{2}}\cdot 11^{\alpha_{3}}\cdot 13^{\alpha_{4}}\cdot
23^{\alpha_{5}}$. Then
\begin{align}&2^{\alpha_{1}+7}\cdot 3^{\alpha_{2}+3}\cdot
5\cdot 11^{\alpha_{3}+1}\cdot 13^{\alpha_{4}-1}\cdot
23^{\alpha_{5}-1}\cdot
31\nonumber\\
&=(2^{\alpha_{1}+1}-1)(3^{\alpha_{2}+1}-1)(11^{\alpha_{3}+1}-1)(13^{\alpha_{4}+1}-1)(23^{\alpha_{5}+1}-1).\label{e:3}\end{align}

 By Lemma \ref{lem3} we have $X\leq 34$. Noting that
$$\textnormal{ord}_{31}(2)=5, \textnormal{ord}_{31}(3)= \textnormal{ord}_{31}(11)=\textnormal{ord}_{31}(13)=30,
\textnormal{ord}_{31}(23)=10,$$ by $\textnormal{ord}_{31}(3)=
\textnormal{ord}_{31}(11)=\textnormal{ord}_{31}(13)=30$, we know
that $30\mid \alpha_{i}+1, i=2,3,4$. Noting that
$\textnormal{ord}_{61}(3)=10,
\textnormal{ord}_{19}(11)=\textnormal{ord}_{61}(13)=3,$ we have
$61\mid 3^{\alpha_{2}+1}-1, 19\mid 11^{\alpha_{3}+1}-1, 61\mid
13^{\alpha_{4}+1}-1$. Which contradicts with (\ref{e:3}), then we
have $$31\nmid3^{\alpha_{2}+1}-1, 31\nmid11^{\alpha_{3}+1}-1,
31\nmid13^{\alpha_{4}+1}-1.$$ By (\ref{e:3}) we know that $31\mid
23^{\alpha_{5}+1}-1$ or $31\mid 2^{\alpha_{1}+1}-1$.

If $31\mid 23^{\alpha_{5}+1}-1$, then by $\textnormal{ord}_{31}(23)=10$, we know that $10\mid
\alpha_{5}+1$. Noting that $\textnormal{ord}_{41}(23)=5$ we have $41\mid
23^{\alpha_{5}+1}-1$, which contradicts (\ref{e:3}).


If $31\mid 2^{\alpha_{1}+1}-1$, then $\alpha_{1}+1=5k, k\in
\mathbb{Z}$. By $X\leq 34$, we have $k=1,2,3,4,5,6$.

 {\bf Subcase 1}: $k=1$, $\alpha_{1}+1=5$, $n=2^{4}\cdot
3^{\alpha_{2}}\cdot 11^{\alpha_{3}}\cdot 13^{\alpha_{4}}\cdot
23^{\alpha_{5}}$. Put $m=3^{\alpha_{2}}\cdot 11^{\alpha_{3}}\cdot
13^{\alpha_{4}}\cdot 23^{\alpha_{5}}$. Then
$$\displaystyle\frac{\sigma(m)}{m}<\frac{m}{\varphi(m)}=\frac{3}{2}\cdot
\frac{11}{10}\cdot \frac{13}{12}\cdot
\frac{23}{22}=\displaystyle\frac{299}{160}=1.86875.$$
On the other hand, noting that $\varphi(n)+\sigma(n)=4n$, then $8\varphi(m)+31\sigma(m)=64m$,
thus $$\displaystyle\frac{\sigma(m)}{m}> 1.9264>1.86875,$$ a
contradiction.

 {\bf Subcase 2:} $k=2$, $\alpha_{1}+1=10$. Thus
$$\begin{array}{ll}2^{16}\cdot 3^{\alpha_{2}+2}\cdot
5\cdot 11^{\alpha_{3}}\cdot 13^{\alpha_{4}-1}\cdot
23^{\alpha_{5}-1}\\
=(3^{\alpha_{2}+1}-1)(11^{\alpha_{3}+1}-1)(13^{\alpha_{4}+1}-1)(23^{\alpha_{5}+1}-1).\end{array}$$
 Noting that the following facts:

{\bf $(i)$} If $2^{\alpha}\mid 3^{\alpha_{2}+1}-1$, then $\alpha\leq
4$. In fact, if $\alpha\geq 5$, then by $\textnormal{ord}_{32}(3)=8$, we have
$\alpha_{2}+1=8s,s\in \mathbb{Z}$. Noting that $\textnormal{ord}_{41}(3)=8$, thus
$41\mid 3^{\alpha_{2}+1}-1$, this is impossible.

{\bf $(ii)$} If $2^{\alpha}\mid 11^{\alpha_{3}+1}-1$, then
$\alpha\leq 3$. In fact, if $\alpha\geq 4$, then by $\textnormal{ord}_{16}(11)=4$,
we have $\alpha_{3}+1=4s,s\in \mathbb{Z}$. Noting that $\textnormal{ord}_{61}(11)=4$, thus
$61\mid 11^{\alpha_{3}+1}-1$, this is impossible.

{\bf $(iii)$} If $2^{\alpha}\mid 13^{\alpha_{4}+1}-1$, then
$\alpha\leq 3$. In fact, if $\alpha\geq 4$, then by $\textnormal{ord}_{16}(13)=4$,
thus $\alpha_{4}+1=4s,s\in \mathbb{Z}$. Noting that $\textnormal{ord}_{7}(13)=2$, thus
$7\mid 13^{\alpha_{4}+1}-1$, this is impossible.

{\bf $(iv)$} If $2^{\alpha}\mid 23^{\alpha_{5}+1}-1$, then
$\alpha\leq 4$. In fact, if $\alpha\geq 5$, then by $\textnormal{ord}_{32}(23)=4$,
we have $\alpha_{5}+1=4s,s\in \mathbb{Z}$. Noting that $\textnormal{ord}_{53}(23)=4$, thus
$53\mid 23^{\alpha_{5}+1}-1$, this is impossible.

Let
$$\begin{array}{ll}A&=(3^{\alpha_{2}+1}-1)(11^{\alpha_{3}+1}-1)(13^{\alpha_{4}+1}-1)(23^{\alpha_{5}+1}-1),\\
B&=2^{16}\cdot 3^{\alpha_{2}+2}\cdot 5\cdot 11^{\alpha_{3}}\cdot
13^{\alpha_{4}-1}\cdot 23^{\alpha_{5}-1}.\end{array}$$ We have $V_{2}(A)\leq
14$ and $V_{2}(B)=16$, this is impossible.

 {\bf Subcase 3}: $k=3$, $\alpha_{1}+1=15$. By
$\textnormal{ord}_{7}(2)=3$, we have $7\mid 2^{\alpha_{1}+1}-1$, which contradicts
(\ref{e:3}).

 {\bf Subcase 4}: $k=4$, $\alpha_{1}+1=20$. By
$\textnormal{ord}_{41}(2)=20$, we have $41\mid 2^{\alpha_{1}+1}-1$, which
contradicts (\ref{e:3}).

{\bf Subcase 5}: $k=5, \alpha_{1}+1=25$. By
$\textnormal{ord}_{601}(2)=25$, we have $601\mid
2^{\alpha_{1}+1}-1$, which contradicts (\ref{e:3}).

{\bf Subcase 6}: $k=6, \alpha_{1}+1=30$. By
$\textnormal{ord}_{151}(2)=15$, we have $151\mid
2^{\alpha_{1}+1}-1$, which contradicts (\ref{e:3}).

This completes the proof of Theorem \ref{thm1}.


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\bibitem{Lin}  Da-Zheng Lin and Ming-Zhi Zhang,  On the divisibility
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\bibitem{Luca}  F. Luca and J. S\'{a}ndor, On a problem of
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\bibitem{Nicol}  C. A. Nicol, Some Diophantine
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\bibitem{Wang}  Wei Wang, The research on Nicol problems,
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\bibitem{Zhang}  Ming-Zhi Zhang, A divisibility problem, {\it J.
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\noindent 2010 {\it Mathematics Subject Classification}: Primary 11A25.

\noindent \emph{Keywords: } Nicol number; Euler's totient function.

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\vspace*{+.1in}
\noindent
Received April 12 2011;
revised version received  July 7 2011.
Published in {\it Journal of Integer Sequences}, September 4 2011.
Revised, April 11 2012.

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