ON NEAR-PERFECT NUMBERS OF SPECIAL FORMS

Abstract

We discuss near-perfect numbers of various forms. In particular, we study the existence of near-perfect numbers in the Fibonacci and Lucas sequences, near-perfect values taken by integer polynomials and repdigit near-perfect numbers.

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 [10] 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.

1. 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.

2. 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.

3. 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 [12] 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. [14] showed that there is no odd near-perfect number n with $\omega (n)=3$ and Tang et al. [13] 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 [7] proved that there are no perfect Fibonacci or Lucas numbers, while Broughan et al. [2] showed that no Fibonacci number (larger than 1) is multiperfect. Assuming the $ABC$ -conjecture, Klurman [5] proved that any integer polynomial of degree $\ge 3$ without repeated factors can take only finitely many perfect values. Pollack and Shevelev [9] 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 [8] 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

1. (a) There are no odd near-perfect Fibonacci or Lucas numbers.

2. (b) There are no near-perfect Fibonacci numbers $F_n$ with $\omega (F_n)\le 3$ .

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 $2\le g\le 10$ .

1. (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 $88$ .

2. (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. [4] 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 [11], 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

1. (a) $n=16, 18 \ \text {or} \ 2p \ \text {for some prime} \ p\ge 19,$

2. (b) $n=p, p^2\ or\ p^3 \ \text {for some prime} \ p\ge 5,$

3. (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 [5, 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,

(2.1) $$ \begin{align} \prod_{p|f(m)}p\gg |m|^{\deg f-1-\epsilon}. \end{align} $$

We also need the finiteness result for the solutions of the hyperelliptic equation.

Theorem 2.6 (Baker [1]).

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 [3]).

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 [6]).

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

(3.1) $$ \begin{align} (g^m+1)\bigg(\frac{g^m-1}{g-1}\bigg)=U_n\mid2^{2p-1}(2^p-1). \end{align} $$

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.