1 Introduction
Let
$\sigma (n)$
and
$\omega (n)$
denote the sum of the positive divisors of n and the number of distinct prime factors of n, respectively. A natural number n is perfect if
$\sigma (n)=2n$
. More generally, given a fixed integer k, the number n is called multiperfect or k-fold perfect if
$\sigma (n)=kn$
. The famous Euclid–Euler theorem asserts that an even number is perfect if and only if it has the form
$2^{p-1}(2^p-1)$
, where both p and
$2^p-1$
are primes. It is not known if there are odd perfect numbers.
In 2012, Pollack and Shevelev [Reference Pongsriiam10] introduced the concept of near-perfect numbers. A positive integer n is near-perfect with redundant divisor d if d is a proper divisor of n and
$\sigma (n)=2n+d$
. Note that when
$d=1$
, we get a special kind of near-perfect numbers called quasiperfect.
Pollack and Shevelev constructed the following three types of even near-perfect numbers.
-
Type A.
$n=2^{p-1}(2^p-1)^2$
where both p and
$2^p-1$
are primes and
$2^p-1$
is the redundant divisor. -
Type B.
$n=2^{2p-1}(2^p-1)$
where both p and
$2^p-1$
are primes and
$2^p(2^p-1)$
is the redundant divisor. -
Type C.
$n=2^{t-1}(2^t-2^k-1)$
,
$t\ge k+1$
where
$2^t-2^k-1$
is prime and
$2^k$
is the redundant divisor.
In 2013, Ren and Chen [Reference Ren and Chen12] proved that all near perfect numbers n with
$\omega (n)=2$
are of types A, B and C together with 40. It is an open problem to classify all even near-perfect numbers. However, from the definition, it is easy to see that all odd near-perfect numbers are squares. Tang et al. [Reference Tang, Ren and Li14] showed that there is no odd near-perfect number n with
$\omega (n)=3$
and Tang et al. [Reference Tang, Ma and Feng13] proved that the only odd near-perfect number n with
$\omega (n)=4$
is
$173369889=3^4\cdot 7^2\cdot 11^2\cdot 19^2$
. Thus, for any other odd near-perfect number n, if it exists, we have
$\omega (n)\ge 5$
.
There are several papers discussing perfect and multiperfect numbers of various forms. For example, Luca [Reference Luca7] proved that there are no perfect Fibonacci or Lucas numbers, while Broughan et al. [Reference Broughan, Gonzalez, Lewis, Luca, Huguet and Togbe2] showed that no Fibonacci number (larger than 1) is multiperfect. Assuming the
$ABC$
-conjecture, Klurman [Reference Klurman5] proved that any integer polynomial of degree
$\ge 3$
without repeated factors can take only finitely many perfect values. Pollack and Shevelev [Reference Pollack and Shevelev9] studied perfect numbers with identical digits in base g,
$g\ge 2$
. He found that in each base g, there are only finitely many examples and that when
$g=10$
, the only example is
$6$
. Later, Luca and Pollack [Reference Luca and Pollack8] established the same results for multiperfect numbers with identical digits.
In this short note, we study near-perfect numbers in the Fibonacci and Lucas sequences, near-perfect values taken by integer polynomials and near-perfect numbers with identical digits. Recall that the Fibonacci sequence
$(F_n)_{n\ge 0}$
is given by
$F_0=0$
,
$F_1=1$
and
$F_{n+2}=F_{n+1}+F_n$
for all
$n\ge 0$
and the Lucas sequence
$(L_n)_{n\ge 0}$
is given by
$L_0=2$
,
$L_1=1$
and
$L_{n+2}=L_{n+1}+L_n$
for all
$n\ge 0$
. A natural number is called a repdigit in base g if all of the digits in its base g expansion are equal.
Here we prove the following results.
Theorem 1.1
-
(a) There are no odd near-perfect Fibonacci or Lucas numbers.
-
(b) There are no near-perfect Fibonacci numbers
$F_n$
with
$\omega (F_n)\le 3$
. -
(c) The only near-perfect Lucas number
$L_n$
with two distinct prime factors is
$L_6=18$
.
Theorem 1.2. Suppose
$P(x)\in \mathbb {Z}[x]$
with
$\deg P(x)\ge 3$
has no repeated factors. Then there are only finitely many n such that
$P(n)$
is an odd near-perfect number. Furthermore, if we assume that the
$ABC$
-conjecture holds, then
$P(n)$
takes only finitely many near-perfect values with two distinct prime factors.
Theorem 1.3. Let
$2\le g\le 10$
.
-
(a) There are only finitely many repdigits in base g which are near-perfect and have two distinct prime factors. All such numbers are strictly less than
${(g^3-1)}/{(g-1)}$
. In particular, when
$g=10$
, the only repdigit near-perfect number with two distinct prime divisors is
$88$
. -
(b) There are no odd near-perfect repdigits in base g.
2 Preliminary results
In this section, we collect several auxiliary results. We begin with the famous and remarkable theorem of Bugeaud et al. [Reference Bugeaud, Mignotte and Siksek4] about perfect powers in the Fibonacci and Lucas sequences.
Theorem 2.1 (Bugeaud–Mignotte–Siksek).
The only perfect powers among the Fibonacci numbers are
$F_0=0$
,
$F_1=F_2=1$
,
$F_6=8$
and
$F_{12}=144$
. For the Lucas numbers, the only perfect powers are
$L_1=1$
and
$L_3=4$
.
In [Reference Pollack11], Pongsriiam gave the description of the Fibonacci numbers satisfying
$\omega (F_n)=3$
. We state his results in the following theorems.
Theorem 2.2. The only solutions to the equation
$\omega (F_n)=3$
are given by
-
(a)
$n=16, 18 \ \text {or} \ 2p \ \text {for some prime} \ p\ge 19,$
-
(b)
$n=p, p^2\ or\ p^3 \ \text {for some prime} \ p\ge 5,$
-
(c)
$n=pq \ \text {for some distinct primes} \ p, q\ge 3.$
Theorem 2.3. Assume that
$\omega (F_n)=3$
and
$n=p_1p_2$
, where
$p_1<p_2$
are odd primes. Then
$F_{p_1}=q_1$
,
$F_{p_2}=q_2$
and
$F_n=q^{a_1}_1q_2q^{a_3}_3$
, where
$q_1, q_2, q_3$
are distinct primes,
$q_3$
is a primitive divisor of
$F_n$
(that is, a prime divisor which does not divide any
$F_m$
for
$0<m<n$
),
$a_3\ge 1$
and
$a_1\in \{1,2\}$
. Furthermore,
$a_1=2$
if and only if
$q_1=p_2$
.
Let us also recall the
$ABC$
-conjecture. For
$n\in \mathbb {Z}\setminus \{0\}$
, the radical of n is defined by
$\text {rad}(n)=\prod _{p|n}p$
.
Conjecture 2.4 (
$ABC$
-conjecture).
For each
$\epsilon>0$
, there exists
$M_{\epsilon }>0$
such that whenever
$a,b,c\in \mathbb {Z}\setminus \{0\}$
satisfy the conditions
$$ \begin{align*} \gcd (a,b,c)=1 \quad \text{and} \quad a+b=c, \end{align*} $$
then
$$ \begin{align*} \max \{|a|, |b|, |c|\}\leq M_{\epsilon}\,{\mathrm{rad}}(abc)^{1+\epsilon}. \end{align*} $$
The next lemma is important for the proof of Theorem 1.2.
Lemma 2.5 [Reference Klurman5, Corollary 2.4].
Assume that the
$ABC$
-conjecture is true. Suppose that
$f(x)\in \mathbb {Z}[x]$
is nonconstant and has no repeated roots. Fix
$\epsilon>0$
. Then,
$$ \begin{align} \prod_{p|f(m)}p\gg |m|^{\deg f-1-\epsilon}. \end{align} $$
We also need the finiteness result for the solutions of the hyperelliptic equation.
Theorem 2.6 (Baker [Reference Baker1]).
All solutions in integers
$x,y$
of the diophantine equation
$$ \begin{align*} y^2=a_0x^n+a_1x^{n-1}+\cdots+a_n, \end{align*} $$
where
$n\ge 3$
,
$a_0\ne 0$
,
$a_1,\ldots ,a_n$
are integers and where the polynomial on the right-hand side possesses at least three simple zeros, satisfy
$$ \begin{align*} \max(|x|,|y|)<\exp\exp\exp\{(n^{10n}\mathcal{H})^{n^2}\}, \end{align*} $$
where
$\mathcal {H}=\max _{0\le j\le n}|a_j|$
.
The next two theorems characterise those perfect powers which are also repdigits.
Theorem 2.7 (Bugeaud–Mignotte [Reference Bugeaud and Mignotte3]).
Let a and b be integers with
$2\le b\le 10$
and
$1\le a\le b-1$
. The integer N with all digits equal to a in base b is not a pure power, except for
$N=1, 4, 8$
or
$9$
, for
$N=11111$
written in base
$b=3$
, for
$N=1111$
written in base
$b=7$
and for
$N=4444$
written in base
$b=7$
.
Theorem 2.8 (Ljunggren [Reference Ljunggren6]).
The only integer solutions
$(x,n,y)$
with
$|x|>1$
,
$n>2$
and
$y>0$
to the exponential equation
$$ \begin{align*} \frac{x^n-1}{x-1}=y^2 \end{align*} $$
are
$(x,n,y)=(7,4,20)$
and
$(x,n,y)=(3,5,11)$
.
3 Proofs
Proof of Theorem 1.1.
(a) Since any odd near-perfect number is square, the result follows from Theorem 2.1.
(b) It is easy to show that there are no near-perfect numbers of the form
$p^k$
,
$k\ge 0$
, where p is prime. Suppose that there exists an even near-perfect number of type A belonging to the Fibonacci sequence. For
$p=2$
or
$p=3$
, one gets the numbers
$18$
and
$196$
which do not belong to the Fibonacci sequence.
Assume now that
$p\ge 5$
. The equation
$F_n=2^{p-1}(2^p-1)^2$
implies that
$16\mid F_n$
. From this, it follows that
$12\mid n$
. Hence,
$3=F_4\mid F_n=2^{p-1}(2^p-1)^2$
, which is impossible because
$p\ge 5$
and
$2^p-1$
is prime. A similar argument can be used to show that there are no type B or type C near-perfect Fibonacci numbers.
Suppose now that
$F_n$
is a near-perfect Fibonacci number with
$\omega (F_n)=3$
. Since
$F_n$
is even, by Theorems 2.2 and 2.3,
$n=3p$
,
$p>3$
and
$F_n=2q_1q^{\alpha }_2$
, where
$F_p=q_1$
and
$q_2$
is a primitive divisor of
$F_n$
and
$\alpha \ge 1$
. If
$q_1\ge 7$
, then
$$ \begin{align*} 2=\frac{\sigma(F_n)}{F_n}-\frac{d}{F_n}<\frac{3}{2}\cdot\frac{q_1+1}{q_1}\cdot\frac{q_2}{q_2-1}<\frac{3}{2}\cdot\frac{8}{7}\cdot\frac{11}{10}<2, \end{align*} $$
which is impossible. Thus,
$q_1=5$
. Then
$F_n=F_{15}=2\cdot 5\cdot 61$
, which is not a near-perfect number.
(c) Clearly
$L_6=18$
is a near-perfect number of type A. Using the fact that no member of the Lucas sequence is divisible by
$8$
, it is easy to verify that there are no other near-perfect Lucas numbers with two distinct prime divisors.
Proof of Theorem 1.2.
For odd near-perfect numbers, the result follows unconditionally from Baker’s Theorem 2.6. Note that if m is a sufficiently large near-perfect number with
$\omega (m)=2$
, then
$\text {rad}(m)\ll \sqrt {m}$
. Assume
$P(n)$
is a near-perfect number with a large value of n,
$\text {deg}\ P=d\ge 3$
and
$\omega (P(n))=2$
. Fix
$\epsilon>0$
. Applying (2.1),
$$ \begin{align*} n^{d-1-\epsilon}\ll\text{rad}(P(n))\ll n^{d/2}, \end{align*} $$
which gives
$$ \begin{align*} \tfrac{1}{2}d\ge d-1-\epsilon \end{align*} $$
or
$d\le 2+\epsilon <3$
. This contradiction implies the result.
Proof of Theorem 1.3.
Fix
$g\ge 2$
. Let
$U_n={(g^n-1)}/{(g-1)}$
.
(a) First we consider the near-perfect numbers of type A. We may assume that
$g>2$
(since every binary repdigit is odd). Thus, to find repdigit near-perfect numbers, we need to solve the equation
$$ \begin{align*} N=aU_n=2^{p-1}(2^p-1)^2, \quad \text{where} \ a\in\{1,\ldots,g-1\} \ \text{ and } \ 2^p-1 \ \text{is prime}. \end{align*} $$
For the sake of contradiction, assume that
$n\ge 3$
. It is clear that
$2^p-1\mid U_n$
for otherwise
$(2^p-1)^2\mid a$
and then
$$ \begin{align*} g>a\ge(2^p-1)^2>\sqrt{N}\ge\bigg(\frac{g^n-1}{g-1}\bigg)^{1/2}=\sqrt{g^{n-1}+\cdots+1}>g^{{(n-1)}/{2}}\ge g, \end{align*} $$
which is impossible. Thus,
$U_n=2^b(2^p-1)^2$
or
$U_n=2^b(2^p-1)$
for some nonnegative integer b. Consider the first case. If g is even, then
$U_n$
is odd, therefore
$b=0$
. Hence,
$U_n=(2^p-1)^2$
which has no solutions for
$n\ge 3$
by Theorem 2.8. Thus, g must be odd and n must be even. Write
$n=2m$
. We then get
$$ \begin{align*} 2^b(2^p-1)^2=\frac{g^{2m}-1}{g-1}=(g^m+1)\bigg(\frac{g^m-1}{g-1}\bigg). \end{align*} $$
Note that
$g^m+1>{(g^m-1)}/{(g-1)}$
and
$2^p-1>2^b$
. Moreover,
$$ \begin{align*}\text{gcd}\bigg(g^m+1,\frac{g^m-1}{g-1}\bigg)\le2.\end{align*} $$
Therefore,
$g^m+1=2(2^p-1)^2$
and
${(g^m-1)}/{(g-1)}=2^{b-1}$
. The latter equation has no solutions in view of our assumption
$2\le g\le 10$
and Theorem 2.7.
Now suppose that
$U_n=2^b(2^p-1)$
. If g is even, then
$U_n$
is odd, therefore
$b=0$
. Hence,
$$ \begin{align*} a=2^{p-1}(2^p-1)>2^p-1=\frac{g^n-1}{g-1}=g^{n-1}+\cdots+1>g^{n-1}>g, \end{align*} $$
which contradicts the assumption
$1\le a\le g-1$
. Thus, g must be odd and n must be even. Put
$n=2m$
. We then obtain
$$ \begin{align*} 2^b(2^p-1)=\frac{g^{2m}-1}{g-1}=(g^m+1)\bigg(\frac{g^m-1}{g-1}\bigg). \end{align*} $$
Since
$g^m+1>{(g^m-1)}/{(g-1)}$
and
$2^p-1>2^b$
, it follows that
$2^p-1\mid g^m+1$
, and we get
$g^m+1=2(2^p-1)$
and
${(g^m-1)}/{(g-1)}=2^{b-1}$
. Since
${(g^m-1)}/{(g-1)}$
is even and g is odd, we see that m is even. Hence,
$m=2m_1$
and so
$2(2^p-1)=g^m+1=g^{2m_1}+1\equiv 2\ (\text {mod}\ 8)$
. Then
$2^p-1\equiv 1\ (\text {mod} \ 4)$
, but this is impossible for any prime
$p\ge 2$
. Observe that for this case, we did not use the assumption
$2\le g\le 10$
.
Suppose now
$aU_n$
is near-perfect of type B, where
$1\le a<g$
and
$n\ge 3$
. We may write
$$ \begin{align*} aU_n=2^{2p-1}(2^p-1). \end{align*} $$
Suppose first that
$U_n$
is odd. Since
$1<U_n\mid 2^{2p-1}(2^p-1)$
, it follows that
$U_n=2^p-1$
. Thus,
$a=2^{2p-1}$
. However, since
$n\ge 3$
,
$$ \begin{align*} g^2<U_3\le U_n=2^p-1<2^p, \quad \text{whence}\quad g<2^{p/2}<2^{2p-1}=a, \end{align*} $$
which contradicts
$a<g$
. If
$U_n$
is even, then since
$U_n=1+g+\cdots +g^{n-1}$
, it follows that g is odd and n is even. Write
$n=2m$
. We have
$$ \begin{align} (g^m+1)\bigg(\frac{g^m-1}{g-1}\bigg)=U_n\mid2^{2p-1}(2^p-1). \end{align} $$
If
$2\mid m$
, then
$g^m+1$
has a prime divisor
$q\equiv 1\pmod {4}$
contradicting (3.1). Hence,
${2\nmid m}$
. Thus,
$U_m$
is odd. Since
$m>1$
and
$2^p-1$
is prime, (3.1) implies that
$U_m=2^p-1$
. Hence,
${g^m+1\mid 2^{2p-1}}$
. So
$g^m+1$
is a power of 2. However,
$$ \begin{align*} g^m+1=(g+1)(g^{m-1}-g^{m-2}+\cdots+1). \end{align*} $$
The second factor here is odd, so must equal 1. Thus,
$m=1$
, which is a contradiction.
In a similar manner, one can show finiteness of repdigits in base g among near-perfect numbers of type C.
(b) The result is an immediate consequence of Theorem 2.7.
Theorem 1.3 asserts that repdigit near-perfect numbers of types A, B and C have at most two digits in base g,
$2\le g\le 10$
. For
$g\in \{2,3,4,6\}$
, there are no repdigit near-perfect numbers with two distinct prime factors. For
$g=5$
, the only repdigit near-perfect numbers with two distinct prime factors are 12, 18 and 24. For
$g=7$
, the only repdigit near-perfect numbers with two distinct prime factors are 24 and 40. For
$g=8$
, the only repdigit near-perfect number with two distinct prime factors is 18. For
$g=9$
, the only repdigit near-perfect numbers with two distinct prime factors are 20 and 40. Finally, in base
$g=10$
, the only repdigit near-perfect number with two distinct prime factors is 88.
Acknowledgement
The author would like to thank the anonymous referee for the helpful comments.
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