Can you prove that there are infinite palindromic primes that when squared give a palindromic number?

closed as off-topic by R. van Dobben de Bruyn, abx, Gro-Tsen, Mark Sapir, Andrés E. Caicedo Aug 4 at 14:52

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  • 2,3,11,101 --- it is seems unknown if there are more. – Carlo Beenakker Aug 4 at 13:28
  • Crlo Beenakker yeah I know but can you prove that? – stavros panagiotidis Aug 4 at 13:56
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    I think Carlo might be suggesting it's well known as an open problem. If so, then please know that MO generally closes well known open problems. We generally want questions for which there is reasonable expectation that someone may know an answer. – Todd Trimble Aug 4 at 14:03
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    I think that it is bad form to change your question silently to a dramatically different one (are there only 4? → are there infinitely many?) after a counterexample is offered—especially since, as @AlexM politely observes, you could easily have found the counterexample by searching the OEIS. – LSpice Aug 4 at 16:36
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    @LSpice Carlo Beenakker suggested that he not ask that as a separate question, but incorporate it into this question. It's true that the former question should not have been erased but rather added to, but I think maybe the OP feels besieged and confused at this point, what he is supposed to do. – Todd Trimble Aug 4 at 16:57
up vote 8 down vote accepted

There are in fact more than those 4, and they have their own page on OEIS. Two conjectures, then, would be:

  • are they infinitely many? I suspect so.

  • are $2$ and $3$ the only ones formed with decimal digits other than $0$ and $1$? I suspect so.

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    I found that page by entering the sequence $2,3,11,101$ in the search field on OEIS and then browsing through the results. – Alex M. Aug 4 at 14:12
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    I'm pretty sure the second conjecture could be proven inductively, from the outermost digits in. First, both the prime and its square must have an odd number of digits (excluding $11$), so the outermost digits must be $1, 2,3$. Then the 2nd digit can't affect anything higher, so it's restricted. This continues to the middle digit. Further, some careful juggling could say that other than the primes $2,3$, all of the primes must be made of $0,1$. – user44191 Aug 4 at 20:58

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