# On Rassias' conjecture and $\varphi(x)(1+\varphi(y))=\sigma(z)$, with $\varphi(n)$ the Euler's totient function and $\sigma(n)$ the sum of divisors

Let for integers $n\geq 1$ the Euler's totient function $\varphi(n)$, and $\sigma(n)$ denotes the sum of divisors $\sum_{d\mid n}d$. I don't know if next question involving the equation $(1)$ was in the literature.

On assumption of the Rassias' conjecture, see this Wikipedia, it's easy to deduce that there exist infinitely many triples $(x,y,z)$ of different positive integers that satisfy $$\varphi(x)(1+\varphi(y))=\sigma(z).\tag{1}$$

Question. Is it possible to deduce, unconditionally, that do exist infinitely many triples $(x,y,z)$ of different integers satisfying $(1)$? If such question is yet known feel free to answer this question as a reference request (thus refer the literature and I try find and read those facts or propositions about this equation). Many thanks.

• I thought about this for a few minutes.I tried to decide if I should think that this is true. Pomerance and Erdos have studied the densities of the images of both $\phi$ and $\sigma$, but these thoughts got me nowhere. I suspect that if someone were to "just know" the answer, it would be Pomerance. – davidlowryduda Nov 30 '17 at 22:31
• Then we can dedicate this problem to Pomerance, he is a very good mathematician. Many thanks for your attention and help @mixedmath – user243301 Nov 30 '17 at 22:34
• It also follows from there being infinitely many Mersenne primes: let $x = 1$, $y = 2^p-1$, and $z = 2^{p-1}$, where $y$ is prime. – Dan Brumleve Dec 11 '17 at 1:07
• Many thanks @DanBrumleve for your nice contribution. The bounty ends in few minutes. I would have liked to pay the bounty, if you edit/had edited your comment as an answer. – user243301 Dec 11 '17 at 10:51
• Here is another way to get there: suppose for all $k$ there exists a $p$ such that both $p$ and $2 \cdot k \cdot p - 1$ are prime (or equivalently that $2 \cdot k \cdot p^2 - p$ is semiprime). Then take $2 \cdot k = \phi(x)$, $1 + \phi(y) = y = p$, and $\sigma(z) = 1 + z = 2 \cdot k \cdot p$. – Dan Brumleve Dec 11 '17 at 17:47