# 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. – user759001 Apr 14 at 20:09
• Does applying the quadratic formula to Equation $(1)$ give Equation $(2)$? – Geoffrey Trang Apr 14 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. – user759001 Apr 16 at 11:48