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Digit block repetition in prime squares

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  • woodhodgson@xtra.co.nz
    Considering squares with an odd number of digits, the cases where the squared integers are prime and the squares less than 10^9 appear to be only these (the
    Message 1 of 7 , Aug 28, 2012
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      Considering squares with an odd number of digits, the cases where the squared integers are prime and the squares less than 10^9 appear to be only these (the repetition meaning the square looks like AbA, where A is a string of digits and b any digit):

      121 = 11^2, 29929 = 173^2, 69169 = 263^2, 732947329 = 27073^2.

      Does anyone know how uncommon these are for larger squares?

      {Regarding the related question about squares with an even number of digits and a complete repetition AA - I couldn't find any solutions at all for "small" squares, whether or not the original numbers were prime, and suspect there may not be any at all}.
    • Jack Brennen
      I suspect that there are an infinite number of such p^2 of the form AbA, but that s strictly based on a simple heuristic argument. The next two after your
      Message 2 of 7 , Aug 28, 2012
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        I suspect that there are an infinite number of such p^2 of the form
        AbA, but that's strictly based on a simple heuristic argument.

        The next two after your listed group are:

        208139^2 = 43321843321

        252717253^2 = 63866009963866009


        Some experimentation for such numbers of the form n^2 seems to imply
        that n^2 is perhaps more likely to be of the form AbA when the number
        of digits in n is a multiple of 3. I can't really explain why, but I
        haven't spent a lot of time thinking about it.

        Statistics for the number of n with n^2 of form AbA:

        n = 2 digits, solutions = 3
        n = 3 digits, solutions = 4
        n = 4 digits, solutions = 0
        n = 5 digits, solutions = 2
        n = 6 digits, solutions = 3
        n = 7 digits, solutions = 1
        n = 8 digits, solutions = 0
        n = 9 digits, solutions = 21



        On 8/28/2012 12:23 PM, woodhodgson@... wrote:
        > Considering squares with an odd number of digits, the cases where the squared integers are prime and the squares less than 10^9 appear to be only these (the repetition meaning the square looks like AbA, where A is a string of digits and b any digit):
        >
        > 121 = 11^2, 29929 = 173^2, 69169 = 263^2, 732947329 = 27073^2.
        >
        > Does anyone know how uncommon these are for larger squares?
        >
        > {Regarding the related question about squares with an even number of digits and a complete repetition AA - I couldn't find any solutions at all for "small" squares, whether or not the original numbers were prime, and suspect there may not be any at all}.
        >
        >
        >
        >
        >
        > ------------------------------------
        >
        > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
        > The Prime Pages : http://primes.utm.edu/
        >
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        >
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      • woodhodgson@xtra.co.nz
        Interesting, thank you. I am still wondering about the even AA case - I think there may be no solutions.
        Message 3 of 7 , Aug 29, 2012
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          Interesting, thank you. I am still wondering about the even "AA" case - I think there may be no solutions.

          --- In primenumbers@yahoogroups.com, Jack Brennen <jfb@...> wrote:
          >
          > I suspect that there are an infinite number of such p^2 of the form
          > AbA, but that's strictly based on a simple heuristic argument.
          >
          > The next two after your listed group are:
          >
          > 208139^2 = 43321843321
          >
          > 252717253^2 = 63866009963866009
          >
          >
          > Some experimentation for such numbers of the form n^2 seems to imply
          > that n^2 is perhaps more likely to be of the form AbA when the number
          > of digits in n is a multiple of 3. I can't really explain why, but I
          > haven't spent a lot of time thinking about it.
          >
          > Statistics for the number of n with n^2 of form AbA:
          >
          > n = 2 digits, solutions = 3
          > n = 3 digits, solutions = 4
          > n = 4 digits, solutions = 0
          > n = 5 digits, solutions = 2
          > n = 6 digits, solutions = 3
          > n = 7 digits, solutions = 1
          > n = 8 digits, solutions = 0
          > n = 9 digits, solutions = 21
          >
          >
          >
          > On 8/28/2012 12:23 PM, woodhodgson@... wrote:
          > > Considering squares with an odd number of digits, the cases where the squared integers are prime and the squares less than 10^9 appear to be only these (the repetition meaning the square looks like AbA, where A is a string of digits and b any digit):
          > >
          > > 121 = 11^2, 29929 = 173^2, 69169 = 263^2, 732947329 = 27073^2.
          > >
          > > Does anyone know how uncommon these are for larger squares?
          > >
          > > {Regarding the related question about squares with an even number of digits and a complete repetition AA - I couldn't find any solutions at all for "small" squares, whether or not the original numbers were prime, and suspect there may not be any at all}.
          > >
          > >
          > >
          > >
          > >
          > > ------------------------------------
          > >
          > > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
          > > The Prime Pages : http://primes.utm.edu/
          > >
          > > Yahoo! Groups Links
          > >
          > >
          > >
          > >
          > >
          >
        • Maximilian Hasler
          ... obviously p^2 =concat(A,A) = A * (10^k+1) with 10^(k-1)
          Message 4 of 7 , Aug 29, 2012
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            > Interesting, thank you. I am still wondering about the even "AA" case - I
            > think there may be no solutions.
            >

            obviously p^2 =concat(A,A) = A * (10^k+1) with 10^(k-1) < A < 10^k
            so this is not possible.

            Maximilian


            [Non-text portions of this message have been removed]
          • Jack Brennen
            True, not solvable when p is a prime. There are solutions for the general case including non-primes. I think that the smallest are (and yes, there s a
            Message 5 of 7 , Aug 29, 2012
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              True, not solvable when p is a prime.

              There are solutions for the general case including non-primes. I think
              that the smallest are (and yes, there's a pattern):

              36363636364^2 = 1322314049613223140496
              45454545455^2 = 2066115702520661157025
              54545454546^2 = 2975206611629752066116
              63636363637^2 = 4049586776940495867769
              72727272728^2 = 5289256198452892561984
              81818181819^2 = 6694214876166942148761
              90909090910^2 = 8264462810082644628100

              The trick here, for these seven examples, is to find 22 digit squares
              which are divisible by 100000000001 (10^11+1). The reason these are
              the smallest is because 10^11+1 is the smallest number of that form
              which is not square-free.



              On 8/29/2012 12:31 PM, Maximilian Hasler wrote:
              >> Interesting, thank you. I am still wondering about the even "AA" case - I
              >> think there may be no solutions.
              >>
              >
              > obviously p^2 =concat(A,A) = A * (10^k+1) with 10^(k-1) < A < 10^k
              > so this is not possible.
              >
              > Maximilian
              >
              >
              > [Non-text portions of this message have been removed]
              >
              >
              >
              > ------------------------------------
              >
              > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
              > The Prime Pages : http://primes.utm.edu/
              >
              > Yahoo! Groups Links
              >
              >
              >
              >
              >
            • woodhodgson@xtra.co.nz
              Of course, trivial for prime p, should have seen that instantly.
              Message 6 of 7 , Aug 29, 2012
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                Of course, trivial for prime p, should have seen that instantly.

                --- In primenumbers@yahoogroups.com, Maximilian Hasler <maximilian.hasler@...> wrote:
                >
                > > Interesting, thank you. I am still wondering about the even "AA" case - I
                > > think there may be no solutions.
                > >
                >
                > obviously p^2 =concat(A,A) = A * (10^k+1) with 10^(k-1) < A < 10^k
                > so this is not possible.
                >
                > Maximilian
                >
                >
                > [Non-text portions of this message have been removed]
                >
              • woodhodgson@xtra.co.nz
                Thanks, all clear. There is a significant gap to later candidates, as the next case where 10^k + 1 is not square-free is for k=21. The first entry in that
                Message 7 of 7 , Aug 30, 2012
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                  Thanks, all clear. There is a significant gap to later candidates, as the next case where 10^k + 1 is not square-free is for k=21.

                  The first entry in that family appears to be

                  183673469387755102041183673469387755102041 = 428571428571428571429^2


                  --- In primenumbers@yahoogroups.com, Jack Brennen <jfb@...> wrote:
                  >
                  > True, not solvable when p is a prime.
                  >
                  > There are solutions for the general case including non-primes. I think
                  > that the smallest are (and yes, there's a pattern):
                  >
                  > 36363636364^2 = 1322314049613223140496
                  > 45454545455^2 = 2066115702520661157025
                  > 54545454546^2 = 2975206611629752066116
                  > 63636363637^2 = 4049586776940495867769
                  > 72727272728^2 = 5289256198452892561984
                  > 81818181819^2 = 6694214876166942148761
                  > 90909090910^2 = 8264462810082644628100
                  >
                  > The trick here, for these seven examples, is to find 22 digit squares
                  > which are divisible by 100000000001 (10^11+1). The reason these are
                  > the smallest is because 10^11+1 is the smallest number of that form
                  > which is not square-free.
                  >
                  >
                  >
                  > On 8/29/2012 12:31 PM, Maximilian Hasler wrote:
                  > >> Interesting, thank you. I am still wondering about the even "AA" case - I
                  > >> think there may be no solutions.
                  > >>
                  > >
                  > > obviously p^2 =concat(A,A) = A * (10^k+1) with 10^(k-1) < A < 10^k
                  > > so this is not possible.
                  > >
                  > > Maximilian
                  > >
                  > >
                  > > [Non-text portions of this message have been removed]
                  > >
                  > >
                  > >
                  > > ------------------------------------
                  > >
                  > > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
                  > > The Prime Pages : http://primes.utm.edu/
                  > >
                  > > Yahoo! Groups Links
                  > >
                  > >
                  > >
                  > >
                  > >
                  >
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