# Attempt to get a characterization of even perfect numbers from an equation involving the Dedekind psi function

In this post we denote the Dedekind psi function as $$\psi(n)$$ for integers $$n\geq 1$$. This is an important arithmetic fuction in several subjects of mathematics. As reference I add the Wikipedia Dedekind psi function.

An even perfect number is an even integer $$n\geq 1$$ satisfying that $$\sigma(n)=2n$$, where $$\sigma(n)=\sum_{1\leq d\mid n}d$$ is the sum of divisors function.

Claim. Invoking Euclid–Euler theorem it is easy to prove that each even perfect number satisfies $$4\psi(n)^2-(12n+3)\psi(n)=-9n^2,\tag{1}$$ alternatively $$\psi(n)=\frac{12n+3+3\sqrt{1+8n}}{8}.\tag{2}$$

Question. I would like to know if it is possible to get a characterization for even perfect numbers from $$(1)$$: what I evoke is if we can to prove that if an integer $$n\geq 1$$ satisfies $$(1)$$ then $$n$$ is an even perfect number. Many thanks.

The case $$n=1$$ is easy. By contradiction also is easy to rule out/discard that $$n$$ is a power of two and that $$n$$ is odd. Thus the only discussion is what happens with numbers of the form $$n=2^{\alpha}m\tag{3}$$ where $$\alpha\geq 1$$ is integer and $$(2,m)=1$$ for integers $$m>1$$.

Our case $$(3)$$ leads to

$$2^{\alpha+1}m\prod_{\substack{\text{primes }p\\p\mid m}}(p+1)^2-(2^{\alpha+2}m+1)\prod_{\substack{\text{primes }p\\p\mid m}}p(p+1)=2^{\alpha+1}m \prod_{\substack{\text{primes }p\\p\mid m}}p^2,\tag{4}$$ and if I was right from $$(4)$$ on assumption that $$m=p$$ is a prime (that is we assume that the number of distinct primes dividing $$m$$ is $$\omega(m)=1$$) I get that $$p+1=\lambda 2^{\alpha+1}$$ with $$\lambda=1$$ using the congruence $$\prod_{\substack{\text{primes }p\\p\mid m}}(p+1)\equiv 0\text{ mod }2^{\alpha+1}$$.

Thus if my argument works I need to rule out/discard the case of integers $$m$$ with more than $$1$$ distinct primes factors in their prime factorization (I want to remove the case $$\omega(m)>1$$).

• If it is impossible to remove the case in which $m$ has at least two distinct prime factors in its factorization, or you find a mistake in the cases that I've studied, feel free to add a comment or your feedback. Apr 14, 2020 at 20:09
• Does applying the quadratic formula to Equation $(1)$ give Equation $(2)$? Apr 14, 2020 at 21:16
• I removed my previous comment since I understand that you was asking just about clarification about $(2)$: yes (I think that there aren't typos), the quadratic formula gives $(2)$ from $(1)$. Many thanks for your attention @GeoffreyTrang , feel free to provide me your feedback about any of my unanswered questions in Mathematics Stack Exchange. Apr 16, 2020 at 11:48
• Stated another way, I believe you're asking for a proof of $\sigma_1(2\,n)=\sum\limits_{d|2\,n} d=4\,n\iff\psi(2\,n)=2\,n \prod\limits_{p|2\,n}\left(1+\frac{1}{p}\right)=\frac{24\,n+3+3\sqrt{1+16\,n}}{8}$ where $(n\land d)\in \mathbb{N}\land p\in \mathbb{P}$? Jul 18, 2021 at 15:04
• Now I don't know, I'm in a library, and I did the calculations a year ago. Many thanks @StevenClark Jul 19, 2021 at 8:56

If a positive integer $$n$$ satisfies $$4\psi(n)^2-(12n+3)\psi(n)+9n^2=0\tag1$$ then $$n$$ is an even perfect number.

Proof :

Solving $$(1)$$ for $$\psi(n)$$ gives $$\psi(n)=\frac{12n+3\pm 3\sqrt{1+8n}}{8}$$ There has to be a positive integer $$k$$ such that $$1+8n=k^2$$, so $$\psi\bigg(\frac{k^2-1}{8}\bigg)=\frac{12\cdot\frac{k^2-1}{8}+3\pm 3k}{8}=\frac 3{16}(k\pm 1)^2$$ Since $$k$$ has to be odd, there has to be a positive integer $$m$$ such that $$k=2m+1$$ to have $$\psi\bigg(\frac{m(m+1)}{2}\bigg)=\frac 3{4}\bigg(m+\frac{1\pm 1}{2}\bigg)^2$$ There has to be a positive integer $$\ell$$ such that $$m+\dfrac{1\pm 1}{2}=2\ell$$, so $$\psi\bigg(\ell(2\ell\mp 1)\bigg)=3\ell^2$$ Since $$\psi$$ is the multiplicative function with $$\gcd(\ell,2\ell\mp 1)=1$$, $$\psi(\ell)\psi(2\ell\mp 1)=3\ell^2$$ For $$\ell=1$$, this does not hold. Suppose here that $$\ell\gt 1$$ is odd. Then, LHS is even while RHS is odd, which is impossible. So, $$\ell$$ has to be even, and there has to be a positive integer $$a,s$$ ($$s$$ is odd) such that $$\ell=2^as$$, so $$\psi(2^a)\psi(s)\psi(2^{a+1}s\mp 1))=3\cdot 2^{2a}s^2,$$ i.e. $$3\cdot 2^{a-1}\psi(s)\psi(2^{a+1}s\mp 1)=3\cdot 2^{2a}s^2,$$ i.e. $$\psi(s)\psi(2^{a+1}s\mp 1)=2^{a+1}s^2$$

Suppose here that there are $$a,s$$ such that $$\psi(s)\psi(2^{a+1}s+1)=2^{a+1}s^2$$. Then, LHS is larger than RHS, which is impossible.

So, we have to have $$\psi(s)\psi(2^{a+1}s-1)=2^{a+1}s^2$$

Let $$\displaystyle 2^{a+1}s-1=\prod_{i=1}^{N} p_i^{b_i}$$ where $$p_i$$ are distinct primes and $$b_i$$ are positive integers. Then, $$\psi(s)\prod_{i=1}^{N} p_i^{b_i-1}(p_i+1)=\bigg(1+\prod_{i=1}^{N} p_i^{b_i}\bigg)s$$Since RHS is not divisible by any $$p_i$$, we have to have $$b_i=1$$ for which we have $$\psi(s)\prod_{i=1}^{N} (p_i+1)=\bigg(1+\prod_{i=1}^{N} p_i\bigg)s$$Suppose here that $$N\geqslant 2$$. Then, LHS is larger than RHS, which is impossible. Therefore, $$N=1$$ implies $$\psi(s)=s$$, i.e. $$s=1$$. So, $$n$$ has to be of the form $$n=2^a(2^{a+1}-1)$$ where $$2^{a+1}-1$$ is prime. This means that $$n$$ has to be an even perfect number. $$\quad\blacksquare$$

• Many thanks, I'm trying to read the proof in next days before expires the bounty. I would like to dedicate the post, and this characterization (now yours) of even perfect numbers, to your person due your excellence in mathematics. Jul 19, 2021 at 15:10