Your equation is of the form: (quadratic in n) = (quadratic in K) so you can complete the square on both sides and multiply both sides by a constant factor to obtain the Pell-like equation: x^2 - d y^2 = c where x is a linear function of n, and y is a linear function of K (both with integer coefficients). Each pair (x,y) in Z^2 can be identified with the algebraic integer: x + y sqrt(d) and it's a solution if and only if the (number-theoretic) norm x^2 - d y^2 is equal to c. But norms are multiplicative, so if you find a single solution (which you have done) together with a primitive unit (e.g. the smallest norm-1 element with x,y > 0) then you can generate all of the solutions to the Pell-like equation. Then the solutions to your original problem will be the subset of solutions which satisfy some congruence condition. Best wishes, Adam P. Goucher
Sent: Wednesday, April 24, 2019 at 4:10 AM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] Square hex numbers?
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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