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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.

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    $\begingroup$ 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. $\endgroup$ – davidlowryduda Nov 30 '17 at 22:31
  • $\begingroup$ Then we can dedicate this problem to Pomerance, he is a very good mathematician. Many thanks for your attention and help @mixedmath $\endgroup$ – user243301 Nov 30 '17 at 22:34
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    $\begingroup$ 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. $\endgroup$ – Dan Brumleve Dec 11 '17 at 1:07
  • $\begingroup$ 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. $\endgroup$ – user243301 Dec 11 '17 at 10:51
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    $\begingroup$ 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 $\sigma(z) = 1 + z = 2 \cdot k \cdot p$. $\endgroup$ – Dan Brumleve Dec 11 '17 at 17:47

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