# Equations involving arithmetic functions of primorials

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 $$1\leq z and $$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.

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.

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:

• 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:

• 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 $$(53,52,3)$$.
• 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 $a$. – user142929 Sep 26 '19 at 9:47
• I was inspired in Florian Luca, Equations Involving Arithmetic Functions of Factorials, Divulgaciones Matemáticas Vol 8. No. 1 (2000). – user142929 Sep 26 '19 at 9:59

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