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Consider triples $(p,q,r)$ of prime numbers $p$, $q$ and $r$ such that $(p+1)(q+1)=(r+1)$. Here are some examples : $(2,3,11), (3,7,31)$.

How to prove there are infinitely many such triples?!

I define two integer numbers $n$ and $m$ or $(n,m)$ to be Isomorph iff $F(n)=F(m)$, where $F(n)$ is sum of divisor of $n$ (for example $(6,11), (10,17), (14, 23), (21, 31)$). If for prime numbers $p,q,r$ have: $(p+1)(q+1)=(r+1)$ so $(pq,r)$ are Isomorph and so it maybe a conjecture: there are infinitely many pairs of Isomorph numbers!

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3 Answers 3

With regard to your broader question, a much stronger result was proved by Ford, Luca and Pomerance. As a special case their result implies:

Let $\sigma(n)$ denote the sum of divisors function and $\phi(n)$ denote the Euler totient function. Then for any $k\ge1$, there exist $2k$ distinct integers $m_1, m_2, \ldots, m_k$ and $n_1, n_2, \ldots, n_k$ such that

$$\phi(m_1) = \phi(m_2) = \cdots = \phi(m_k) = \sigma(n_1) = \sigma(n_2) = \cdots = \sigma(n_k).$$

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Even if you fix one prime, say, $p=2$, the standard conjectures say there are infinitely many $q$ and $r$. For $p=2$, you want $q$ and $3q+2$ to be prime, and while no one can prove that there are infinitely many such $q$, no one has any doubt that there are infinitely many such $q$.

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I do not know how to prove there are infinetely many such triples. Consider now chains of those triples of length L: (p(n), p(n+1)=nextprime(p(n)), r); (p(n+1) , p(n+2)=nextprime(p(n+1)), s)); ... For example a chain of length L =5:

(11,13,167);(13,17,251);(17,19,359);(19,23,479);(23,29,719)

The question over your question is if we could consider there could be chains of arbitrary long length L; which is a much stronger statement, or else prove there exists some n, such that there is no L >n.

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