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Question: Do there exist infinitely many primes of the form $p*q,$ where $*$ denotes concatenation and $p$ and $q$ are both prime?

On my way into the grocery store the other day, I had the above curiosity pop into my head. Originally, I thought about the number of primes of the form $p * p,$ but I quickly realized that there are none. Explicitly, the number $n$ of digits of such an integer is twice the number of digits of $p,$ hence we have that $p * p = p(10^n + 1).$ For instance, we have that $55 = 5 \cdot 11 = 5(10^1 + 1),$ $1111 = 11 \cdot 101 = 11(10^2 + 1),$ $101101 = 101 \cdot 1001 = 101(10^3 + 1),$ etc. Revising my question a bit, I was immediately able to find several such pairs of primes $p * q$ for distinct primes $p$ and $q.$ For instance, the pairs $(2, 3),$ $(5, 9),$ and $(13, 7)$ give the primes $23,$ $59,$ and $137,$ respectively.

Unfortunately, I am not very familiar with techniques used to show that there are infinitely many primes of a certain form. Even worse, it is not always true that $p * q$ is prime whenever $p$ and $q$ are distinct primes (e.g., $57 = 3 \cdot 19$ is not prime); otherwise, we could invoke Dirichlet's Theorem. I would appreciate any observations or insights into the matter. Thank you for your consideration.

Dylan C. Beck
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  • One annoying feature is that the answer could depend on the base used. I like this question but I have no idea how to approach it. – Integrand Feb 13 '21 at 21:34
  • Could you elaborate on what you mean by "depend[s] on the base used?" – Dylan C. Beck Feb 13 '21 at 21:39
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    Related: How many of the primes can be subdivided into concatenations of smaller primes?. (OEIS A105184.) – Vepir Feb 13 '21 at 21:50
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    For instance, $101_{10}3_{10}=1013_{10}$, a prime. OTOH, in base $2$ we have $1100101_211_2=110010111_2 = 407_{10}$ and $407=11\cdot 37$. – Integrand Feb 13 '21 at 21:54
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    I doubt we can prove that there are infinite many such primes, but with the $1/\ln(n)$-approach, we can probably establish a heuristic predecting infinit many such primes for every "starting" prime $p$ no matter in which base we concatenate. Maybe, someone works it out. – Peter Feb 14 '21 at 11:11
  • An interesting problem is to find $p$ such that we need large $q$ to get a solution. If you are interested, I can do some calculations. – Peter Feb 14 '21 at 11:18
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    A partial result I got with PARI/GP : If "$$" denotes conatenation in base $10$, then if $p\le 10^9$ is a prime number, then there is a prime number $q\le 1\ 567$ such that $pq$ is prime. The (only) "hardest" case is $p=974\ 739\ 163$ upto this limit. – Peter Feb 14 '21 at 11:45
  • @Peter, you mean that if we are given a prime $p,$ we can show that there are infinitely many primes $q$ such that $p*q$ is prime? – Dylan C. Beck Feb 14 '21 at 20:34

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