=============================================== woodhodgson@xtra.co.nz Message 1 of 7 Aug 28, 2012 ----------------------------------------------- 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 Message 2 of 7 Aug 28, 2012 ----------------------------------------------- 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}. > > =============================================== woodhodgson@xtra.co.nz Message 3 of 7 Aug 29, 2012 ----------------------------------------------- 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 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}. > > > =============================================== Maximilian Hasler Message 4 of 7 Aug 29, 2012 ----------------------------------------------- > 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 Message 5 of 7 Aug 29, 2012 ----------------------------------------------- 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 > > > > =============================================== woodhodgson@xtra.co.nz Message 6 of 7 Aug 29, 2012 ----------------------------------------------- Of course, trivial for prime p, should have seen that instantly. --- In primenumbers@yahoogroups.com, 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 Message 7 of 7 Aug 30, 2012 ----------------------------------------------- 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 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