On Sinha's note on perfect numbers

TL;DR

This paper proves that odd perfect numbers cannot be of the forms 2^n+1 or n^n+1, contributing to the understanding of the structure of perfect numbers.

Contribution

It establishes the non-existence of odd perfect numbers of specific algebraic forms, advancing the classification of perfect numbers.

Findings

No odd perfect number of the form 2^n+1 exists.

No odd perfect number of the form n^n+1 exists.

Abstract

We shall show that there is no odd perfect number of the form $2^n+1$ or $n^n+1$.

Equations

$$C=\sum _{2^{U+1}(2m+1)<16,}\frac{1-\log \log (2^{U+1}m)}{2^{U+1}m}.$$

$$\log a>\frac{((4m+2)/e^C{)}^{2^{U+1}}}{2^U}.$$

$$\begin{array}{rl}& \log (4m+2)-C\\ <& \frac{\exp \left(\frac{1+\log t_0}{2^{U+1}}\right)}{2^{U+1}}\left(\log (2^U\log a)+(U+1)(1+\log t_0)\log 2+\frac{{\log }^2{t}_0}{2}+c\right).\end{array}$$

$$M_i=\frac{a^{2^U{P}_1{P}_2\cdots P_i}+1}{a^{2^U{P}_1{P}_2\cdots P_{i-1}}+1}$$

$$L_i=M_0{M}_1\dots M_i=a^{2^U{P}_1{P}_2\cdots P_i}+1$$

$$a^{n/P_i}+1=X^2,\frac{a^n+1}{a^{n/P_i}+1}=p{Y}^2$$

$$a^{n/P_i}+1=p{X}^2,\frac{a^n+1}{a^{n/P_i}+1}=Y^2$$

$$\frac{a^n+1}{a^{n/P_i}+1}=\sum _{j=0}^{P_i-1}(-1{)}^j{a}^{j(n/P_i)}\equiv P_i(\mathrm{mod}{a}^{n/P_i}+1),$$

$$M_i=\sum _{j=0}^{P_i-1}(-1{)}^j{2}^{2^U{P}_1{P}_2\dots P_{i-1}j}\equiv P_i(\mathrm{mod}{L}_{i-1}),$$

$$\begin{array}{rl}\prod _{o_p(a)=2^{U+1}d}\frac{p}{p-1}<& \exp \sum _{o_p(a)=2^{U+1}d}\frac{1}{p-1}\le \sum _{k=1}^{k_0(d)}\frac{1}{2^{U+1}kd}\\ \le & \exp \frac{1+\log (2^{U}d\log a/\log (2^{U+1}d))}{2^{U+1}d},\end{array}$$

$$\frac{\sigma (a^n+1)}{a^n+1}=\prod _{\begin{array}{c}{o}_p(a)=2^{U+1}d,\\ d\mid P_1{P}_2\dots P_r\end{array}}\frac{p}{p-1}<\exp \left(C+\sum _{d\mid P_1{P}_2\dots P_r}\frac{\log (2^{U}d\log a)}{2^{U+1}d}\right).$$

$$\sum _{d\mid P_1{P}_2\dots P_r}\frac{\log (2^{U}d\log a)}{2^{U+1}d}=\frac{U\log 2+\log \log a}{2^{U+1}}.$$

$$\sum _{i=0}^{\infty }\frac{i}{q^i}=\sum _{j=0}^{\infty }\sum _{i=j+1}^{\infty }\frac{1}{q^i}=\sum _{j=0}^{\infty }\frac{1}{q^j(q-1)}=\frac{q}{(q-1{)}^2},$$

$$\begin{array}{rl}& \sum _{d\mid P_1{P}_2\dots P_r}\frac{\log (2^{U}d\log a)}{2^{U+1}d}\\ <& \sum _{f_1,f_2,\dots ,f_r\ge 0}\frac{\log (2^U\log a)+f_1\log p_1+f_2\log p_2+\cdots +f_r\log p_r}{2^{U+1}{p}_1^{f_1}{p}_2^{f_2}\cdots p_r^{f_r}}\\ =& \prod _{i=1}^t\frac{p_i}{p_i-1}\left(\frac{\log (2^U\log a)}{2^{U+1}}+\sum _{k=1}^t\frac{\log p_k}{2^{U+1}(p_k-1)}\right)\\ =& \left(\frac{1}{2^{U+1}}\prod _{i=1}^r\frac{p_i}{p_i-1}\right)\left(\frac{\log (2^U\log a)}{2^{U+1}}+\sum _{k=1}^r\frac{\log p_k}{p_k-1}\right).\end{array}$$

$$\prod _{i=1}^r\frac{p_i}{p_i-1}<\prod _{k=1}^r\frac{2^{U+1}k+1}{2^{U+1}k}<\exp \frac{1+\log r}{2^{U+1}}$$

$$\begin{array}{rl}\sum _{k=1}^r\frac{\log p_k}{p_k-1}<& \sum _{k=1}^r\frac{\log k+(U+1)\log 2}{2^{U+1}k}\\ <& \frac{1}{2^{U+1}}\left((U+1)(1+\log r)\log 2+\frac{{\log }^{2}r}{2}+c\right).\end{array}$$

$$\begin{array}{rl}& \sum _{d\mid P_1{P}_2\dots P_r}\frac{\log (2^{U}d\log a)}{2^{U+1}d}\\ <& \frac{\exp \left(\frac{1+\log r}{2^{U+1}}\right)}{2^{U+1}}\left(\log (2^U\log a)+(U+1)(1+\log r)\log 2+\frac{{\log }^{2}r}{2}+c\right).\end{array}$$

$$\log (4m+2)=\log \frac{\sigma (a^n+1)}{a^n+1}<C+\frac{U\log 2+\log \log a}{2^{U+1}}$$

$$\begin{array}{rl}& \log (4m+2)-C\\ <& \frac{\exp \left(\frac{1+\log t_0}{2^{U+1}}\right)}{2^{U+1}}\left(\log (2^U\log a)+(U+1)(1+\log t_0)\log 2+\frac{{\log }^2{t}_0}{2}+c\right)\end{array}$$

$$2^{10}+1=5^2\times 41,\frac{2^n+1}{2^{10}+1}=41p{y}^2$$

$$\prod _{o_p(a)=2d}\frac{p}{p-1}<\exp \frac{1+\log (d\log 2/\log (2d))}{2d}<\exp \frac{\log d}{2d}$$

$$\sum _{\begin{array}{c}d=3^{f_2}87211^{f_1},\\ f_1>0,f_2\ge 0\end{array}}\frac{\log d}{2d}<\frac{87211}{116280}\left(\frac{\log 3}{174422}+\frac{\log 87211}{87210}\right)<\frac{1}{9000}.$$

$$\sum _{\begin{array}{c}d=3^{f_2}{p}_1^{f_1},\\ f_1>0,f_2\ge 0\end{array}}\frac{\log d}{2d}<\frac{3p_1}{2(p_1-1)}\left(\frac{\log 3}{2p_1}+\frac{\log p_1}{p_1-1}\right)<0.24$$

$$n^n+1=(n^{2^u}+1)\times \frac{n^{2^{u}s}+1}{n^{2^u}+1}=N_1{N}_2,$$

$$\frac{n^{2^{u}s}+1}{n^{2^u}+1}=Z^2.$$

Full text

1 Introduction

A positive integer $N$ is called perfect if $\sigma(N)=2N$, where $\sigma(N)$ denotes the sum of divisors of $N$. As is well known, an even integer $N$ is perfect if and only if $N=2^{k-1}(2^{k}-1)$ with $2^{k}-1$ prime. In contrast, one of the oldest unsolved problems is whether there exists an odd perfect number or not. Moreover, it is also unknown whether there exists an odd $m$-perfect number for an integer $m\geq 2$, i.e., an integer $N$ with $\sigma(N)=mN$ or not.

Sinha [5] showed that $28$ is the only even perfect number of the form $x^{n}+y^{n}$ with $\gcd(x,y)=1$ and $n\geq 2$ and also the only even perfect number of the form $a^{n}+1$ with $n\geq 2$. On the other hand, it is not even proved or disproved that there exists no odd perfect number of the form $x^{2}+1$ with $x$ an integer. Klurman [1] proved that if $P(x)$ is a polynomial of degree $\geq 3$ without repeated factors, then there exist only finitely many odd perfect numbers of the form $P(x)$ with $x$ an integer. Luca [4] (cited in Theorem 9.8 of [2]) showed that no Fermat number can be perfect.

In this article, we would like to prove that there exists no odd perfect number of the form $2^{n}+1$ or $n^{n}+1$.

Indeed, we prove a more general result.

Theorem 1.1**.**

*Let $m$ and $U$ be nonnegative integers. We put

$s_{0}=\left\lfloor 2^{U}\log a/(U+1)\log 2\right\rfloor$ and $t_{0}=2s_{0}+1$ if $U=0$ and $a+1$ is square and $t_{0}=2s_{0}$ otherwise. Let $c=1.093\cdots=(\log 2)/2+(\log 3)/3-(\log^{2}3)/2$ and $C=C(U)$ be the constant defined by*

[TABLE]

If $a^{n}+1$ is an odd $(4m+2)$-perfect number and $n=2^{U}$, then

[TABLE]

If $a^{n}+1$ is an odd $(4m+2)$-perfect number and $n=2^{U}v$ with $v>1$ odd, then

[TABLE]

Moreover, no integer of the form $2^{n}+1$ can be $(4m+2)$-perfect.

For example, if $a^{128s}+1$ is odd $(4m+2)$-perfect, then $a\geq 10$ and, if $a^{256s}+1$ is odd $(4m+2)$-perfect, then $a\geq 18$. Furthermore, if $a^{16}+1$ is odd $(4m+2)$-perfect, then $a>\exp\exp 19.4$ and, if $a^{32}+1$ is odd $(4m+2)$-perfect, then $a>\exp\exp 40.8$. We note that $C(0)=0.9807\cdots$, $C(1)=0.1758\cdots$, $C(2)=0.03348\cdots$ and $C(U)=0$ for $U\geq 3$.

We shall prove that an odd perfect number of the form $n^{n}+1$ must be of the form $2^{m}+1$ and deduce the following result from the above result.

Theorem 1.2**.**

$28$* is the only $(4m+2)$-perfect number of the form $n^{n}+1$ with $m,n\geq 0$ an integer.*

Thus, we conclude that $28$ is the only perfect number of the form $n^{n}+1$.

2 Proof of Theorem 1.1

Assume that $a^{n}+1$ is an odd $(4m+2)$-perfect number. By Euler’s result, we must have $a^{n}+1=px^{2}$ for a prime $p$ and an integer $x$.

Write $n=2^{U}p_{1}^{e_{1}}p_{2}^{e_{2}}\ldots p_{r}^{e_{r}}$ with $p_{1}>p_{2}>\cdots>p_{r}$ odd primes and let $P_{i}=p_{i}^{e_{i}}$ for $i=1,2,\ldots,r$ and $s=\omega(a^{2^{U}}+1)$. We put $o_{p}(x)$ to be the multiplicative order of $x$ modulo $p$.

We can factor $a^{n}+1=M_{0}M_{1}\cdots M_{r}$, where $M_{0}=a^{2^{U}}+1$ and

[TABLE]

for $i=1,2,\ldots,r$. Moreover, let

[TABLE]

and $M_{i}=E_{i}Y_{i}^{2},L_{i}=D_{i}X_{i}^{2}$ with $D_{i}$ and $E_{i}$ squarefree. Clearly, we have $a^{n}+1=L_{r}=px^{2}$ and therefore $D_{r}=p$.

We begin by showing that $p_{i}\equiv 1\pmod{2^{U+1}}$ for every $i$. If $\gcd((a^{n}+1)/(a^{n/P_{i}}+1),a^{n/P_{i}}+1)=1$, then

[TABLE]

or

[TABLE]

for some integers $X$ and $Y$. If $U=0$, then we clearly have $p_{i}\equiv 1\pmod{2^{U+1}}$. If $U>0$, then $n/p_{i}^{e_{i}}$ is even and (3) is clearly impossible. The impossibility of (4) follows from Ljunggren’s result [3] that $(a^{f}+1)/(a+1)$ with $a\geq 2,f\geq 3$ cannot be square.

Hence, we must have $\gcd((a^{n}+1)/(a^{n/P_{i}}+1),a^{n/P_{i}}+1)>1$. Observing that

[TABLE]

$p_{i}$ must divide $a^{n/P_{i}}+1$. Thus, proceeding as in the proof of Theorem 4.12 of [2], we see that $2^{U+1}$ divides $o_{p_{i}}(a)$ and $o_{p_{i}}(a)$ divides $2n/P_{i}$. In particular, $p_{i}\equiv 1\pmod{2^{U+1}}$ for every $i$.

Nextly, we show that for each $i=1,2,\ldots,r$, we have either (i) $\gcd(L_{i-1},M_{i})=1$ and $\omega(D_{i-1})<\omega(D_{i})$ or (ii) $p_{i}$ is the only prime dividing $\gcd(L_{i-1},M_{i})$ and $p_{i}$ divides $a^{2^{U}}+1$.

If $\gcd(L_{i-1},M_{i})=1$, then we must have $D_{i}=D_{i-1}E_{i-1}$ and $X_{i}=X_{i-1}Y_{i-1}$. It follows from Ljunggren’s result mentioned above that $E_{i-1}\neq 1$. Since $D_{i}$ is squarefree, we have $\omega(D_{i-1})<\omega(D_{i})$.

Assume that $\gcd(L_{i-1},M_{i})>1$. Since

[TABLE]

we see that $p_{i}$ is the only prime dividing both $L_{i-1}$ and $M_{i}$.

Now $p_{i}$ must divide $L_{i-1}$ and therefore, proceeding as above, we see that $2^{U+1}$ divides $o_{p_{i}}(a)$ and $o_{p_{i}}(a)$ divides $2^{U+1}P_{1}P_{2}\cdots P_{i-1}$. Hence, $o_{p_{i}}(a)=2^{U+1}d$ and therefore $p_{i}\equiv 1\pmod{2^{U+1}d}$ for some $d$ dividing $P_{1}P_{2}\cdots P_{i-1}$. But, since $p_{1}>\cdots>p_{i-1}>p_{i}$, we must have $o_{p_{i}}(a)=2^{U+1}$ and therefore $p_{i}$ must divide $a^{2^{U}}+1$.

It is clear that (ii) occurs at most $s$ times. Moreover, we observe that in the case (ii), $p_{i}$ is the only possible prime which divides $D_{i-1}$ but not $D_{i}$. Hence, we must have $\omega(D_{i-1})\leq\omega(D_{i})+1$ for each $i$. Now we see that (i) also occurs at most $s$ times.

We can easily see that $\omega(D_{0})=0$ if and only if $U=0$ and $a+1$ is a square. Thus we conclude that $r\leq 2s+1$ if $D_{0}=a+1$ with $U=0$ is square and $r\leq 2s$ otherwise.

If a prime $p$ divides $a^{2^{U}d}+1$ but $a^{2^{U}e}+1$ for any $e<d$, then the multiplicative order of $2\pmod{p}$ is equal to $2^{U+1}d$ and therefore $p=2^{U+1}kd+1$ for some integer $k$. Moreover, the number of such primes is at most $k_{0}(d)=\left\lfloor 2^{U}d\log a/\log(2^{U+1}d)\right\rfloor$ and therefore $s\leq s_{0}$.

Hence, for each $d$,

[TABLE]

so that

[TABLE]

If $r=0$, then we immediately see that

[TABLE]

If $r>0$, then, observing that

[TABLE]

we have

[TABLE]

Since each $p_{i}\equiv 1\pmod{2^{U+1}}$, we have

[TABLE]

and observing that $\sum_{k=1}^{t}\log k/k\leq(\log t)^{2}/2+c$ for $t\geq 1$,

[TABLE]

Thus, we obtain

[TABLE]

We see that $r\leq t_{0}$, where we recall that $s\leq s_{0}=\left\lfloor 2^{U}\log a/(U+1)\log 2\right\rfloor$. Hence, we conclude that

[TABLE]

if $r=0$ and

[TABLE]

otherwise. Thus (1) and (2) follows.

Now we consider the case $a=2$. If $U\geq 4$, then the right-hand side of (1) and (2) is $<0.53<\log 2$ and therefore $a^{n}+1$ cannot be $(4m+2)$-perfect.

If $U\leq 3$, then $2^{2^{U}}+1$ is prime and therefore $s=1$. Clearly, for $n=2^{U}$ with $U\leq 3$, $2^{n}+1=2^{2^{U}}+1$ is not $(4m+2)$-perfect. Hence, we must have $r\leq 2$ and $n=2^{U}p_{1}^{e_{1}}$ or $2^{U}p_{1}^{e_{1}}p_{2}^{e_{2}}$.

If $n=2^{U}p_{1}^{e_{1}}$, then, iterating the argument given before, we must have $p_{1}=2^{2^{U}}+1$. Thus, $n=3^{e_{1}}$, $2\times 5^{e_{1}}$, $2^{2}\times 17^{e_{1}}$ or $2^{3}\times 257^{e_{1}}$.

However, for $n=3^{e_{1}}$ with $e_{1}\geq 3$, we see that both primes $19$ and $87211$ divide $2^{n}+1$ exactly once since $19$ and $87211$ divide $2^{27}+1$ exactly once and the only prime dividing both $(2^{n}+1)/(2^{27}+1)$ and $2^{27}+1$ is $3$. This implies that $2^{n}+1$ cannot be of the form $px^{2}$ and therefore $2^{n}+1$ cannot be $(4m+2)$-perfect if $n=3^{e_{1}}$ with $e_{1}\geq 3$. Similarly, $41$ and $101$ divide $2^{n}+1$ exactly once if $n=2\times 5^{e_{1}}$ and $e_{1}\geq 2$. Clearly, none of $2^{3}+1,2^{9}+1,2^{10}+1$ is $(4m+2)$-perfect. Thus $2^{n}+1$ cannot be $(4m+2)$-perfect if $n=3^{e_{1}}$ or $2\times 5^{e_{1}}$. Similarly, $2^{n}+1$ cannot be $(4m+2)$-perfect if $n=2^{2}\times 17^{e_{1}}$ or $2^{3}\times 257^{e_{1}}$.

If $n=2^{U}p_{1}^{e_{1}}p_{2}^{e_{2}}$, then, iterating the argument given before, $p_{1}>p_{2}=2^{2^{U}}+1$.

If $U=1$ and $n=10p_{1}^{e_{1}}$, then we must have

[TABLE]

since $(2^{n}+1)/(2^{10}+1)$ cannot be square by Ljunggren’s result. Thus, we must have $p_{1}=41$. However, this implies that $2^{n}+1$ must be divisible by $821$ and $10169$ exactly once, which contradicts to the fact that $2^{n}+1=px^{2}$. If $U=1$ and $n=2\times 5^{e_{2}}p_{1}^{e_{1}}$ with $e_{2}\geq 2$, then, since three primes $41,101,8101$ divide $2^{50}+1$ exactly once, at least two of these primes divide $2^{n}+1$. Thus $2^{n}+1$ cannot be $(4m+2)$-perfect if $n=2p_{1}^{e_{1}}p_{2}^{e_{2}}$. Similarly, $2^{n}+1$ cannot be $(4m+2)$-perfect for $n=2^{U}p_{1}^{e_{1}}p_{2}^{e_{2}}$ with $U=2,3$.

Now we assume that $n=3^{e_{2}}p_{1}^{e_{1}}$.

If $n=3^{e_{2}}p_{1}^{e_{1}}$ with $e_{2}\geq 4$, then, at least two of three primes $19,163,87211$ divide $2^{n}+1$ exactly once and therefore $2^{n}+1$ cannot be $(4m+2)$-perfect for such $n$. If $n=27p_{1}^{e_{1}}$, then we must have $p_{1}=19$ or $87211$. We cannot have $p_{1}=19$ since $571$ and $87211$ divide $2^{n}+1$ exactly once for $n=27\times 19^{e_{1}}$. Assume that $p_{1}=87211$. We observe that, for $d=3^{f_{2}}87211^{f_{1}}$ with $f_{1}>0$, we have

[TABLE]

and, proceeding as in (9),

[TABLE]

Thus, $\sigma(2^{n}+1)/(2^{n}+1)<e^{1/9000}\sigma(2^{27}+1)/(2^{27}+1)<2$ and therefore $2^{n}+1$ cannot be $(4m+2)$-perfect.

If $n=9p_{1}^{e_{1}}$, then we must have $p_{1}=19$ and therefore two primes $571$ and $174763$ divide $2^{n}+1$ exactly once, which is a contradiction.

Finally, assume that $n=3p_{1}^{e_{1}}$. If $p_{1}\geq 11$, then, like (16),

[TABLE]

and $\sigma(2^{n}+1)/(2^{n}+1)<(13/9)e^{0.24}<2$, which is a contradiction.

The only remaining case is $n=3p_{1}^{e_{1}}$ with $p_{1}=5$ or $7$. We observe that $2^{15}+1=3^{2}\times 11\times 331$ and $2^{21}+1=3^{2}\times 43\times 5419$. Thus $2^{n}+1$ must be divisible by at least two distinct primes exactly once, which is a contradiction again. Now we conclude that $2^{n}+1$ can never be $(4m+2)$-perfect.

3 Proof of Theorem 1.2

Sinha’s result clearly implies that $28$ is the only even perfect number of the form $n^{n}+1$. Thus, we may assume that $n^{n}+1$ is an odd $(4m+2)$-perfect number. Clearly $n$ must be even and we can write $n=2^{u}s$ with $u>0$ and $s$ odd.

As before, we must have $n^{n}+1=px^{2}$ for some prime $p$ and integer $x$.

Assume that $s>1$. Then we must have

[TABLE]

say.

If $N_{1}$ and $N_{2}$ have a common prime factor $p$, then $p$ divides $d_{2}$ and therefore $p$ divides $2^{u}s=n$. This is impossible since $\gcd(n^{n}+1,n)=1$. Thus, we see that $\gcd(N_{1},N_{2})=1$ and therefore $N_{1}=X^{2},N_{2}=pY^{2}$ or $N_{1}=pX^{2},N_{2}=Y^{2}$.

We can easily see that $n^{2^{u}}+1$ cannot be square since $u>0$ and therefore

[TABLE]

However, this is also impossible from Ljunggren’s result.

Now we must have $s=1$ and $n^{n}+1=2^{u2^{u}}+1$, which we have just proved not to be $(4m+2)$-perfect in Theorem 1.1. This proves Theorem 1.2.