
COMMENTS

2*(x^2)*((x^2)1) = 3*((y^2)1) has only these five positive solutions.
x*(x1)/2 = (2^z)1 has only these five positive solutions.
Richard K. Guy notes, as Example 29: "True, but why the coincidence?"
Algebraically, y solutions = {1, 3, 7, 29, 6761} can be derived from x solutions as follows: y = sqrt(((2*x^2  1)^2 + 5)/6). From this relationship it becomes clear that the form (((2*x^2  1)^2 + 5)/6) can only be an integer square for x is in {1, 2, 3, 6, 91}. Thus, x and y solutions are also unique integer solutions to the following equivalency: (2x^2  1)^2 = 6y^2  5. From this relationship the following statement naturally follows: ((sqrt(6*y^2  5) + 1)/2  sqrt((sqrt(6*(y^2)  5) + 1)/2))/2 = (2^z  1) = {0, 1, 3, 15, 4095} = A076046(n), the RamanujanNagell triangular numbers; z = {0, 1, 2, 4, 12} = (A060728(n)  3).  Raphie Frank, Jun 26 2013


FORMULA

x = sqrt((sqrt(6*(y^2)  5) + 1)/2) = (sqrt(2^(z + 3)  7) + 1)/2; y = {1, 3, 7, 29, 6761} and z = (A060728(n)  3) = A215795(n) = {0, 1, 2, 4, 12}.  Raphie Frank, Jun 23 2013
