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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 $\omega(m)>1$).

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  • $\begingroup$ 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. $\endgroup$ – user759001 Apr 14 at 20:09
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    $\begingroup$ Does applying the quadratic formula to Equation $(1)$ give Equation $(2)$? $\endgroup$ – Geoffrey Trang Apr 14 at 21:16
  • $\begingroup$ 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. $\endgroup$ – user759001 Apr 16 at 11:48

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