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Let $\sigma(n)=\sum_{1\leq d\mid n}d$ the sum of divisors, $\varphi(n)$ the Euler's totient function and we denote the primorial $\prod_{k=1}^n p_k$ as $N_n$, where $p_k$ denotes the $k$-th prime number. For integers $1\leq z<n$ and $m\geq 1$, and $a\geq 0$ I've consider if it is possible to find the solutions of some cases of the following equations of next problems.

Problem 1. Solve for positive integers $1\leq z,n, m$ and $a\geq 0$ $$\sigma\left(\frac{N_n}{N_z}\right)=2^aN_m\tag{1}$$ where also is required the condition $z<n$.

Problem 2. Solve for positive integers $1\leq z,n, m$ and $a\geq 0$ $$\varphi\left(\frac{N_n}{N_z}\right)=2^aN_m\tag{2}$$ where also is required the condition $z<n$.

Question. Is it possible to make a remarkable progress in studying the solutions of Problem 1? Is it possible to make a remarkable progress in studying the solutions of Problem 2? Many thanks.

I denote the solutions as $(n,z,m;a)$ or $(n,z,m)$ for a given integer $a$, and from th context we know if it are solutions of the corresponding problem. Below I add the cases for wich I've calculated some solutions using a Pari/GP program, over the segments of integers $1\leq z,n,m\leq 100$.

Here I add a claim for which I did an easy draft for its proof.

Claim. The equation $\varphi\left(\frac{N_n}{N_z}\right)=2^aN_m$ implies $n-z\leq a+1$ (the identity holds for the case $a=0$). Similarly we've a claim for the other Problem 1. Further we know that the Euler's totient function and the sum of divisors function are multiplicative functions, and each (odd) prime number $p$ of the form $4\lambda+1$ contributes as $\sigma(p)\equiv2\text{ mod }4$ and $\varphi(p)\equiv 0\text{ mod }4$.

Some solutions for Problem 1. This is a summary of the solutions that I know for Problem 1, when $1\leq z,n,m\leq 100$ and $z<n$:

  • For the case $a=0$ the solutions $(n,z,m)=(3,2,2)$ and $(10,9,3)$.
  • For the case $a=1$ the solutions $(n,z,m)=(2,1,1)$,$(5,4,2)$,$(17,16,3)$ and $(81,80,4)$.
  • For the case $a=2$ the solutions $(n,z,m)=(3,1,2)$, $(4,3,1)$ and $(9,8,2)$.
  • For the case $a=3$ the solutions $(n,z,m)=(4,2,2)$, $(15,14,2)$ and $(52,51,3)$.

Some solutions for Problem 2. This is a summary of the solutions that I know for Problem 2, when $1\leq z,n,m\leq 100$ and $z<n$:

  • For the case $a=0$ the solutions $(n,z,m)=(2,1,1)$,$(4,3,2)$,$(11,10,3)$ and $(47,46,4)$.
  • For the case $a=1$ the solutions $(n,z,m)=(3,2,1)$, $(5,3,3)$,$(6,5,2)$,$(18,17,3)$,$(20,18,5)$ and $(82,81,4)$.
  • For the case $a=2$ the solutions $(n,z,m)=(3,1,1)$, $(4,2,2)$,$(6,4,3)$ and $(11,9,4)$.
  • For the case $a=3$ the solutions are $(n,z,m)=(4,1,2)$,$(5,2,3)$,$(7,6,1)$,$(11,8,5)$, $(14,12,4)$ and $(53,52,3)$.
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  • $\begingroup$ For large values of our parameters, our integers in $(n,z,m;a)$, I evoke that maybe my Claim can be improved by counting how many primes there are of the form $4\lambda+1$ in comparison with the size of $a$. $\endgroup$ – user142929 Sep 26 '19 at 9:47
  • $\begingroup$ I was inspired in Florian Luca, Equations Involving Arithmetic Functions of Factorials, Divulgaciones Matemáticas Vol 8. No. 1 (2000). $\endgroup$ – user142929 Sep 26 '19 at 9:59
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First we should compute the left side of the first equation: $$\begin{array}{rcl} \displaystyle\sigma\left(\frac{N_n}{N_z}\right) & = & \displaystyle\sigma\left( \prod_{z < k \leq n}p_k \right) \\ \\ & = & \displaystyle\prod_{z < k \leq n}\dfrac{p_k^2-1}{p_k-1} \\ \\ & = & \displaystyle\prod_{z < k \leq n}(p_k+1) \end{array}$$ Then problem 1 is solving: $$\displaystyle\prod_{z < k \leq n}(p_k+1) = 2^a N_m$$ The reason why this equation is hard is the case $p_k+1=2^b$, then $p_k$ is Mersenne prime, and we don't know if the set of Mersenne's primes has infinite cardinality or not.

Your problem2 is solving: $$\varphi\left(\frac{N_n}{N_z}\right)=\displaystyle\prod_{z < k \leq n}(p_k-1)=2^aN_m$$

Same of problem 1:

If $p_k-1=2^b$ , then $p_k$ is Fermat prime number and we don't know if the set of Fermat's prime numbers has infinite cardinality or not.

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    $\begingroup$ Many thanks for your great contribution. $\endgroup$ – user142929 Sep 26 '19 at 11:26
  • $\begingroup$ And your discussion for Problem 1 also works for the equation $$\psi\left(\frac{N_n}{N_z}\right)=2^aN_m,$$ that is a different problem, a problem that involves the Dedekind psi function $\psi(l)$. Many thanks for you great answer again. $\endgroup$ – user142929 Apr 30 at 19:05

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