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 $(53,52,3)$.

Equations Involving Arithmetic Functions of Factorials, Divulgaciones Matemáticas Vol 8. No. 1 (2000). $\endgroup$ – user142929 Sep 26 '19 at 9:59