Hi Dan There is a general way to get all solutions to your problem. You are looking for 3n^2 + 3n + 1 = K^2. First transform this to 3(2n + 1)^2 + 1 = 4K^2, which further simplifies to x^2 - 3y^2 = 1, where x = 2K and y = 2n + 1. This is a form of Pell's equation in x and y with solutions x = [(2 + sqrt(3))^m + (2 - sqrt(3))^m]/2 and y = [(2 + sqrt(3))^m - (2 - sqrt(3))^m]/sqrt(12) for m = 0, 1, 2,.... This leads to K = [(2 + sqrt(3))^m + (2 - sqrt(3))^m]/4 and n = [(2 + sqrt(3))^m - (2 - sqrt(3))^m]/sqrt(12), where m = 1, 3, 5,... to produce integer solutions. The first few solutions are (n, K) = (0, 1), (7, 13), (104, 181), (1455, 2521), (20272, 35113), (282359, 489061), (3932760, 6811741). Note that K_(i+1) = 14K_(i) + K_(i-1) and n_(i+1) = 14n_(i) + n_(i-1) + 6, so the solution string can be easily extended. This method also works for the more general Fermat equation, x^2 - Cy^2 = +H or -H. Multiple families of solutions are possible depending on C and H. -Dick Hess On 4/23/2019 8:10 PM, Dan Asimov wrote:
Just happened to notice that in the centered hexagonal numbers
H_n = 1 + 6 T_n = 3n^2 + 3n + 1
we have H_7 = 13^2. Also H_0 = 1^2.
So I tried to find all such cases of
H_n = K^2,
but am not sure of the best way to proceed. Suggestions?
—Dan
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