1
$\begingroup$

Let $x$ be an even number. Suppose that we concatenate it with its successor to form $x (x+1)$ (not multiplication, but concatenation). For example, if we start with $x = 2$, we would get $23$. If we start with $x = 4$, we would get $45$. If we start with $x = 6$ we would get $67$. If we start with $8$ we would get $89$. If we start with $10$ we would get $1011$, and so forth.

Are there infinitely many prime numbers of the form $x (x+1)$?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Probably yes, but I don't think we can prove this with current knowledge. (I did not downvote.) $\endgroup$
    – GH from MO
    Commented May 4, 2022 at 12:25
  • 4
    $\begingroup$ This is definitely beyond the reach of current methods. This is in spirit (and in asymptotic density) similar to the question of existence of palindromic primes. $\endgroup$
    – Wojowu
    Commented May 4, 2022 at 12:39
  • $\begingroup$ Hi, Wojowu! What do you mean with similarity in terms of asymptotic density? $\endgroup$
    – Seven9
    Commented May 4, 2022 at 19:09

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .