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Here we denote the sum of divisors function as $\sigma(m)=\sum_{d\mid m}d$ and the Euler's totient function. We write two equations that satisfy odd perfect numbers (are easy to prove), and after we ask about two related conjecture. The computational evidence is showed.

Claim 1. If $n=q^{4\lambda+1}m^2$ is an odd perfect number written in its eulerian form (this is the MathWorld article dedicated to odd perfect numbers), then satisfies $$\sigma(m^2)\sigma\left(\frac{\sigma(n)}{m}\right)=3\sigma(n)\sigma(m).\tag{1}$$

Claim 2. If $n=q^{4\lambda+1}m^2$ is an odd perfect number written in its eulerian form, then satisfies $$m\varphi\left(\frac{\sigma(n)}{m}\right)=\varphi(\sigma(n)).\tag{2}$$

Remarks. To create previous simple equations we have not exploited all issues in Euler's theorem (or related theorems) for odd perfect numbers. The equation in the second claim is a simplification of $$\varphi(m^2)\varphi\left(\frac{\sigma(n)}{m}\right)=1\cdot\varphi(\sigma(n))\varphi(m).$$ We emphasize the factor $1$ in previous equation with the purpose to show the similarity of the equations $(1)$ and $(2)$.

Question. Prove or refute, finding a counterexample, the following conjectures/questions:

Conjecture 1. If two integers $1\leq x$ and $1\leq y$ satisfying that $x$ is odd, $y\mid x$ are solutions of $$\sigma(y^2)\sigma\left(\frac{\sigma(x)}{y}\right)=3\sigma(x)\sigma(y),$$ (thus also is required that the condition $y\mid \sigma(x)$ is satisfied). Then $x$ is deficient or perfect.

Conjecture 2. If two integers $1\leq x$ and $1\leq y$ satisfy that $x$ is odd, $y\mid x$ and are solutions of the symtem of equations $$\left. \begin{array}{l} \sigma(y^2)\sigma\left(\frac{\sigma(x)}{y}\right)=3\sigma(x)\sigma(y)\\ \varphi(y^2)\varphi\left(\frac{\sigma(x)}{y}\right)=\varphi(\sigma(x))\varphi(y) \end{array} \right\}$$ (thus also is required that the condition $y\mid \sigma(x)$ is satisfied). Then $x$ is deficient or perfect.

Many thanks.

This is a Wikipedia's article dedicated to deficient numbers. Thus in previous questions I am asking about counterexamples or well what work can be done about the conjectures (since this kind of questions are difficult, maybe I should accept partial answers showing remarkable effort in the study of such conjectures).

Computational fact 1. Let integers $1\leq x\leq 2\cdot 10^4$ and $1\leq y\leq 2\cdot 10^4$ satisfying that $x$ is odd, $y\mid x$ and $$\sigma(y^2)\sigma\left(\frac{\sigma(x)}{y}\right)=3\sigma(x)\sigma(y),$$ thus also is required in our algorithm $y\mid \sigma(x)$. Then $x$ is deficient.

Here we've the code written in Pari-GP (is a line, that you can see in Sage Cell Server: choice GP as Language and press Evaluate):

for (x = 1, 20000,for (y = 1, 20000, if (x%(y)==0&&x%2==1&&sigma(x)%y==0&&sigma(y^2)*sigma(sigma(x)/y)==3*sigma(x)*sigma(y), print(sigma(x)-2*x))))

Computational fact 2. Let integers $1\leq x\leq 15000$ and $1\leq y\leq 15000$ satisfying that $x$ is odd, $y\mid x$ and $$\left. \begin{array}{l} \sigma(y^2)\sigma\left(\frac{\sigma(x)}{y}\right)=3\sigma(x)\sigma(y)\\ \varphi(y^2)\varphi\left(\frac{\sigma(x)}{y}\right)=\varphi(\sigma(x))\varphi(y) \end{array} \right\},$$

thus also is required in our algorithm $y\mid \sigma(x)$. Then $x$ is deficient.

Here is the implementation (is a line) written in Pari-GP that you can see from Sage Cell Server.

for (x = 1, 15000,for (y = 1, 15000, if (x%y==0&&x%2==1&&sigma(x)%y==0&&sigma(x)%y==0&&sigma(y^2)*sigma(sigma(x)/y)==3*sigma(x)*sigma(y)&&eulerphi(y^2)*eulerphi(sigma(x)/y)==eulerphi(sigma(x))*eulerphi(y), print(sigma(x)-2*x))))

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    $\begingroup$ This is the link for Sage Cell Server. About the equations $(1)$ and the versions of $(2)$ I don't know if these are in the literature. Of course Conjecture 1 implies Conjecture 2. $\endgroup$ – user243301 May 8 '18 at 10:55
  • $\begingroup$ In the first paragraph I wanted to say that the Euler's totient function is denoted as $\varphi(m)$ in this post. $\endgroup$ – user243301 May 8 '18 at 12:23

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