# Solutions of the equation $\psi(\sigma(n))=2n$, where $\sigma(n)$ is the sum of divisors function and $\psi(n)$ the Dedekind psi function

For integers $$m\geq 1$$ let $$\sigma(m)$$ the sum of divisors function $$\sum_{1\leq d\mid m}d$$ and let $$\psi(m)$$ the Dedekind psi function (as reference I add the Wikipedia Dedekind psi function), then there exist integers $$n\geq 1$$ that satisfy $$\psi(\sigma(n))=2n.\tag{1}$$ I don't know it this equation is in the literature, compare this equation with the equation studied in the second page of [1] (or well from the last paragraph of the article Totient Function from the encyclopedia Wolfram MathWorld).

Up to $$10^4$$ these solutions are $$n=2,3,4,16,64$$ and $$4096$$. I believe that this sequence isn't in the OEIS, I've searched also the string psi(sigma(n)). It is easy to prove the following statement.

Claim. If $$2^{\alpha+1}-1$$ is a Mersenne prime, then $$n=2^{\alpha}$$ is a solution of the equation $$(1)$$.

Question. I would like to know if it is possible to do more work about the solutions of the equation $$\psi(\sigma(n))=2n.$$ What additional and reasonable* work can be done about it? Many thanks.

*I'm asking about if we can deduce more statements about the solutions (characterization of solutions and if there exist finitely/infinitely many solutions) of $$(1)$$.

Remarks. The functions $$\sigma(n)$$ and $$\psi(n)$$ are multuiplicative. It is unknown if there exist infinitely many Mersenne primes. As a side remark the integer $$n=3$$ also is a solution, for which $$2n-1=5\in$$A175611 from the OEIS.

## References:

[1] L. Alaoglu and P. Erdös, A conjecture in elementary number theory, Bull. Amer. Math. Soc. Volume 50, Number 12 (1944), 881-882.

• Why A175611 is relevant? – Max Alekseyev Nov 22 '19 at 15:33
• Many thanks for your attention @MaxAlekseyev It is a side comment because I don't know what is the relationship between the equation $\psi(\sigma(n))=2n$ and the sequences A175611 versus A000668 (Mersenne primes) from the OEIS. See the Claim. – user142929 Nov 22 '19 at 15:47
• All, I was inspired to state previous Claim for the equation $\psi(\sigma(n))=2n$ after I've known the comments for the sequence A001229 from the OEIS. – user142929 Nov 22 '19 at 16:40

There are no other solutions than $$n=3$$ and those from Claim: $$n=2^{p-1}$$ such that $$2^p-1$$ is prime.

Consider several cases.

1. $$n=2^ks$$ is even (here $$k\geqslant 1$$ and $$s$$ is odd). Then $$\frac{\psi(\sigma(n))}{n}=\frac{\psi((2^{k+1}-1)\sigma(s))}{(2^{k+1}-1)\sigma(s)}\cdot \frac{2^{k+1}-1}{2^k}\cdot \frac{\sigma(s)}s.\quad(1)$$ Note that $$\frac{\psi(ab)}{ab}=\prod_{p|ab}\left(1+\frac1p\right)\geqslant \prod_{p|a}\left(1+\frac1p\right)=\frac{\psi(a)}a$$ for positive integers $$a,b$$, therefore $$\frac{\psi((2^{k+1}-1)\sigma(s))}{(2^{k+1}-1)\sigma(s)}\geqslant \frac{\psi(2^{k+1}-1)}{2^{k+1}-1}\geqslant \frac{2^{k+1}}{2^{k+1}-1},\quad (2)$$ the last inequality holds due to $$k\geqslant 1$$ and turns into equality if and only if $$2^{k+1}-1$$ is a Mersenne prime. Substitute (2) to (1), we get $$\frac{\psi(\sigma(n))}{n}\geqslant \frac{2^{k+1}}{2^{k+1}-1}\cdot \frac{2^{k+1}-1}{2^k}\cdot \frac{\sigma(s)}s=2\frac{\sigma(s)}s\geqslant 2,$$ with equality if and only if $$s=1$$ and $$2^{k+1}-1$$ is a Mersenne prime. So for even $$n$$ all solutions come from Claim.

2. $$n$$ is odd. Denote $$\sigma(n)=p_1^{\alpha_1}\ldots p_m^{\alpha_m}$$. Then $$2n=\psi(\sigma(n))=(p_1^{\alpha_1}+p_1^{\alpha_1-1})\ldots (p_m^{\alpha_m}+p_m^{\alpha_m-1})$$. Note that $$p_i^{\alpha_i}+p_i^{\alpha_i-1}$$ is even unless $$p_i^{\alpha_i}=2$$. But $$2n$$ is not divisible by 4, therefore either

(i) $$m=1$$ and $$\sigma(n)=p^{\alpha}$$; or

(ii) $$m=2$$ and $$\sigma(n)=2p^{\alpha-1}$$ for odd $$p$$.

In the case (i) we get $$n=\frac{p+1}2 p^{\alpha-1}$$. If $$p=2$$, then since $$n$$ should be odd we get $$\alpha=2$$, $$n=3$$. It is a solution. If $$p$$ is odd and $$\alpha\geqslant 2$$, then $$p^\alpha=\sigma(n)=\sigma(\frac{p+1}2)\sigma(p^{\alpha-1})$$, but $$\sigma(p^{\alpha-1})$$ is greater than 1 and not dividible by $$p$$. If $$\alpha=1$$, we get $$\sigma(\frac{p+1}2)=p$$. Then $$n=\frac{p+1}2=q^s$$ for a prime $$q$$ (otherwise $$\sigma(\frac{p+1}2)$$ is composite), $$q$$ is odd (since $$q|n$$) and $$2q^s-1=p=\sigma(n)=q^s+q^{s-1}+\ldots+1$$ that fails modulo $$q$$.

In the case (ii) we get $$n=3\frac{p+1}2p^{\alpha-1}$$ and $$\sigma(n)\geqslant n+\frac{n}3=2(p+1)p^{\alpha-1}>2p^\alpha,$$ a contradiction.

• Many thanks I've understand all details. While I know that the theorems that great mathematicians, you and your colleagues of this MO, prove are more difficult and abstract than my question, let me tell you that I was impressed for your proof. I would like to dedicate thus your theorem to your excellence and the excellence of your colleagues. – user142929 Dec 1 '19 at 13:02
• How are you getting that $2n=\psi(\sigma(n))=(p_1^{\alpha_1}+p_1^{\alpha_1-1})\ldots (p_m^{\alpha_m}+p_m^{\alpha_m-1})$? – JoshuaZ Dec 1 '19 at 13:46
• @JoshuaZ $2n=\psi(\sigma(n))$ is given, then use the formula for $\psi(m)=m\prod_{p|m} (1+1/p)$ for $m=\sigma(n)$. – Fedor Petrov Dec 1 '19 at 14:55