Let $\sigma (n)$ be the sum-of-divisors function. For example, $\sigma(7)=1+7=2^3$.

I know some results about triplets of positive integers $(n,a,b)$ where $a,b\ge 2$ such that $\sigma (n)=a^b$, but I'd like to ask the following question :

What is known and unknown about the triplets?

I'd like to know any relevant references as well.

Added : It is my understanding that so-called friend Don gave me an excellent answer for each fixed $b$. By the way, it'd be nice if somebody could show me some references for each fixed $a$.

Added : I've known the following results for a fixed $a$.

  • There exist pairs $(n,b)$ such that $\sigma(n)=2^b$ if and only if $n$ is a product of distinct Mersenne primes.

  • There exist no pairs $(n,b)$ such that $\sigma(n)=3^b$.

I would like to know if more general results for each fixed $a$ are known.

Added : When I was trying to expand the case for $a=3$, I found that there exist no pairs $(n,b)$ such that $\sigma(n)=p^b$ where $p$ is a Fermat prime.

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    $\begingroup$ Numbers $n$ such that $\sigma(n)$ is a square are tabulated at oeis.org/A006532 where it is noted that under standard conjectures there are infinitely many of them. $\endgroup$ Mar 24 '15 at 12:08
  • 1
    $\begingroup$ And numbers such that $\sigma(n)$ is a non-trivial power are tabulated at oeis.org/A065496 $\endgroup$ Mar 24 '15 at 12:26
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    $\begingroup$ Here is a comment on the fixed $a$ problem: One can show that there are only finitely many proper prime powers $p^e$ for which $\sigma(p^e)$ is supported on a given finite set of primes. See, e.g., the book of Shorey and Tijdeman on Exponential Diophantine Equations. This means that the problem in some sense reduces to understanding the shifted primes $\sigma(p)=p+1$. For instance, $\sigma(n) = 2^b$ has infinitely many solutions $(n,b)$ if and only if there are infinitely many Mersenne primes. $\endgroup$ Mar 25 '15 at 3:49
  • $\begingroup$ Let me supplement my previous comment by saying that it's not realistic at this time to expect definitive results for fixed $a$. For example, consider the equation $\sigma(n) = 6^b$. Probably there are infinitely many solutions $n,b$, since if $p=2\cdot 6^m-1$ is prime, then $n=2p$ gives rise to a solution. On the other hand, it seems hopeless to prove at this time that there are infinitely many solutions. For then there are infinitely many $p$ with $p+1$ a $3$-smooth number. We don't know anything nearly this strong! The ``smoothest'' we can get $p+1$ is $p^{0.2931}$ (Baker-Harman)! $\endgroup$ Mar 26 '15 at 14:34
  • $\begingroup$ @so-calledfriendDon : Thank you for your comments. Expecting definitive results may not be realistic, but expecting small results should be realistic. For example, one can prove that there exist no pairs $(n,b)$ such that $\sigma(n)=p^b$ where $p$ is a Fermat prime. Anyway, I'd like to know any relevant references for fixed $a$. $\endgroup$
    – mathlove
    Mar 26 '15 at 16:49

For each fixed $k$, there are infinitely many squarefree numbers $n$ for which $\sigma(n)$ is a $k$th power. See:

W.D. Banks, J.B. Friedlander, C. Pomerance and I.E. Shparlinski, Multiplicative structure of values of the Euler function, High Primes and Misdemeanours: Lectures in Honour of the Sixtieth Birthday of Hugh Cowie Williams (A. Van der Poorten, ed.), Fields Inst. Comm. 41 (2004), pp. 29-47.

(The main object of study there is the Euler function $\phi$. But since $\phi(n) = \prod_{p\mid n}(p-1)$ on squarefree $n$, while $\sigma(n) = \prod_{p\mid n}(p+1)$, the methods carry over essentially unchanged.)

There is also more recent work of Banks and Luca (google "power totients"), Freiberg ("products of shifted primes"), and Freiberg and Pomerance (google "square totients").

As an example of an unsolved problem, I think we don't know if there are infinitely many prime powers $p^e$ for which $\sigma(p^e)$ is a power. For related work, look up the Nagell--Ljunggren equation.

  • $\begingroup$ I'm very glad to read this answer! Thank you very much. $\endgroup$
    – mathlove
    Mar 24 '15 at 22:05

I apologize for answering my own question, but I think that the answer might be of some interest to others.

When I was trying to expand the cases where $a$ is a Fermat prime, I got the following theorem.

Theorem. For each odd prime $q$, there exist only finitely many pairs $(n,b)$ such that $\sigma(n)=q^b$.

Lemma 1. For every pairs of primes $(p,r)$, $\frac{p^{r^2}-1}{p-1}$ is not of the form $q^b$.

Proof. For $r=2$, supposing that $\frac{p^{2^2}-1}{p-1}=(p+1)(p^2+1)$ is of the form $q^b$ gives us that both $p+1$ and $p^2+1$ are of the form $q^b$. Since $\gcd(p+1,p^2+1)\le 2$, one has $q=2$. Since $p^2+1\ge 2^2+1=5$, one has $p^2+1\equiv 0\pmod 4$, which is impossible.

For $r\ge 3$, supposing that $\frac{p^{r^2}-1}{p-1}$ is of the form $q^b$ gives us $\frac{p^r-1}{p-1}$ is of the form $q^b$. Since $p^r\equiv 1\pmod q$, it follows that $\gcd\left(p^r-1,\frac{p^{r^2}-1}{p^r-1}\right)$ is a divisor of $r$. So, one has $r=q$. Setting $p^r=p^q=cq+1$ gives us $$\frac{p^{r^2}-1}{p^r-1}=\frac{p^{q^2}-1}{p^q-1}\equiv \sum_{i=0}^{q-1}(cq+1)^i\equiv 1+\sum_{i=1}^{q-1}(1+icq)\equiv q+q^2\cdot\frac{c(q-1)}{2}\equiv q\pmod{q^2}.$$ Thus, one knows that $\frac{p^{r^2}-1}{p^r-1}$ can be divided by $q$ at most once. It follows that $\frac{p^{r^2}-1}{p^r-1}=q$. So, one has $q=\frac{p^{r^2}-1}{p^r-1}\gt \frac{p^r-1}{p-1}\ge q$, which is a contradiction. QED

Lemma 2. If $\frac{p^r-1}{p-1}$ is of the form $q^b$ for primes $p,r$ where $p\not=q$, then $r$ is a divisor of $q-1$.

Proof. Supposing that $r$ is not a divisor of $q-1$ gives us that there exist positive integers $m,n$ such that $rm-(q-1)n=1$. Since one has $p^r\equiv 1\pmod q$ and $p^{q-1}\equiv 1\pmod q$, one has $p^1\cdot p^{(q-1)n}\equiv p^{rm}\Rightarrow p\cdot (p^{q-1})^{n}\equiv (p^r)^m\Rightarrow p\equiv 1\pmod q$. Since $\gcd\left(p-1,\frac{p^r-1}{p-1}\right)$ is a divisor of $r$, one has $r=q$. Since $\frac{p^r-1}{p-1}\equiv q\pmod{q^2}$, one has $\frac{p^r-1}{p-1}=q$. Thus, it follows that $q=\frac{p^r-1}{p-1}\ge \frac{2^r-1}{2-1}\gt r=q$, which is a contradiction. QED

Theorem. For each odd prime $q$, there exist only finitely many pairs $(n,b)$ such that $\sigma(n)=q^b$.

Proof. By lemma 1, it is sufficient to prove that there exist only finitely many pairs of primes $(p,r)$ such that $\frac{p^r-1}{p-1}=q^b$.

Suppose that there exist infinitely many pairs of primes $(p,r)$ where $r\ge 3$ and $p\not=q$ such that $\frac{p^r-1}{p-1}=q^b$. Since $r$ is a divisor of $q-1$ by lemma 2, there exist only finitely many such $r$. So, for an $r$, there exist infinitely many primes $p$ such that $\frac{p^r-1}{p-1}=q^b$. Let $n=r-1\ge 2$. Let us separate $(p,r)$ into $n$ groups by the values of $m\pmod n$. For an $l\ (0\le l\le n)$, there exsit infinitely many pairs $(p,r)$ such that $\frac{p^r-1}{p-1}=q^m$ and $m\equiv l\pmod n$. Since $p^n\lt \frac{p^r-1}{p-1}\lt (p+1)^n$, one has $l\not=0$. Then, $$\frac{p^r-1}{p-1}=1+p+p^2+\cdots+p^n=q^{l+nk}=C_1x^n\ \ (C_1=q^l\ge q,x=q^c)$$ has infinitely many solutions $(p,x)$.

Setting $np=y-1,C_1n^n=C_2$ gives us $$n^n+n^{n-1} (y-1)+⋯+n(y-1)^{n-1}+(y-1)^n=C_2 x^n.$$ So, there exist integers $a_0,a_1,\cdots,a_{n-1}$ such that $$a_0+a_1y+a_2y^2+\cdots+a_{n-2}y^{n-2}=C_2x^n-y^n$$ where $a_{n-1}=0$.

Setting $C_3=\max\{|a_0|,|a_1|,\cdots,|a_{n-2}|\}$ gives us $$C_2x^n-y^n\lt C_3(1+y+y^2+\cdots+y^{n-2})\lt C_3(y+1)^{n-2}\lt C_3(2y)^{n-2}.$$ So, setting $\frac xy=t,\frac{C_3\cdot 2^{n-2}}{C_2}$ gives us $$t^n-\frac{1}{C_2}\lt\frac{C_4}{y^2}.$$

So, $$\left(\frac xy-\alpha_1\right)\left(\frac xy-\alpha_2\right)\cdots\left(\frac xy-\alpha_n\right)\lt\frac{C_4}{y^2}$$ has infinitely many solutions $(x_1,y_1),(x_2,y_2),\cdots,(x_k,y_k),\cdots$ where $|y_1|\lt|y_2|\lt\cdots$ and $\alpha_1,\alpha_2,\cdots,\alpha_n$ are distinct irrational solutions of $t^n-\frac{1}{C_2}=0$. Since $|y_k|\to\infty$ as $k\to\infty$, $$\left(\frac{x_k}{y_k}-\alpha_1\right)\left(\frac{x_k}{y_k}-\alpha_2\right)\cdots\left(\frac{x_k}{y_k}-\alpha_n\right)\to 0.$$ So, one may suppose $\left|\frac{x_k}{y_k}-\alpha_1\right|\to 0$. Let $d=\min|\alpha_{i_1}-\alpha_{i_2}|\ \ (i_1\not=i_2)$. Here, if one takes a sufficiently large $N$, one has $\left|\frac{x_k}{y_k}-\alpha_1\right|\lt\frac d2$ for $k\gt N$. One also has $\left|\frac{x_k}{y_k}-\alpha_i\right|\gt\frac d2\ \ (i\not=1)$. Thus, it follows that $$\left|\frac{x_k}{y_k}-\alpha_1\right|\lt\frac{C_4}{{y_k}^2}\times\left(\frac 2d\right)^{n-1}=\frac{C_5}{{y_k}^2}$$where $C_5=C_4\left(\frac 2d\right)^{n-1}$.

Here, by the Roth-Ridout theorem, one has $$\left|\frac{x_k}{y_k}-\alpha_1\right|\gt\frac{C_6}{{y_k}^{1.5}}$$ where $C_6$ is a constant independent of $x_k,y_k$. Thus, since one has $$\frac{C_6}{{y_k}^{1.5}}\lt \left|\frac{x_k}{y_k}-\alpha_1\right|\lt\frac{C_5}{{y_k}^2}\Rightarrow \frac{C_6}{{y_k}^{1.5}}\lt \frac{C_5}{{y_k}^2},$$ one has $y_k\lt C_7$ where $C_7$ is a constant independent of $x_k,y_k$. This is a contradiction. QED


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