A293692 Numbers z such that x^2 + y^7 = z^2 for positive integers x and y.

12, 18, 33, 54, 126, 160, 272, 366, 375, 520, 531, 540, 594, 630, 756, 825, 945, 1028, 1044, 1094, 1350, 1372, 1506, 1536, 1575, 1980, 2050, 2219, 2304, 2619, 2940, 3250, 3500, 3645, 3906, 3925, 4097, 4224, 4390, 4625, 5500, 5844, 5988, 6048, 6192, 6283, 6422

Offset: 1

Author

XU Pingya, Oct 14 2017

Keywords

• nonn
• uned

Comments

Let i, j and k are nonnegative integers, n is positive integer. As [(n^)^(7i+1) * (2n+1)^(7j + 3) * (n + 1)^(7k)]^2 + [n)^(2i) * (2n + 1)^(2j + 1) * (n + 1)^(2k)]^7 = [n^(7i) * (2n + 1)^(7j + 3) * (n+1)^(7k+1)]^2, so that number of form n^(7i) * (2n + 1)^(7j + 3) * (n+1)^(7k+1) is a term in sequence.

When (x, y, z) is solution of x^2 + y^3 = z^2 (i.e., z = A070745(n)), (x^(7i+1) * y^(7j + 2) * z^(7k)]^2, x^(2i) * y^(2j + 1) * z^(2k), x^(7i) * y^(7j + 2) * z^(7k+1) is solution of x^2 + y^7 = z^2.

When (x, y, z) is solution of x^2 + y^5 = z^2, (i.e., z = A293284(n)), x^(7i+1) * y^(7j + 1) * z^(7k), x^(2i) * y^(2j + 1) * z^(2k), x^(7i) * y^(7j + 1) * z^(7k+1) is solution of x^2 + y^7 = z^2.

When (x, y, z) is solution of x^2 + y^7 = z^2, (x^(7i+1) * y^(7j + 2) * z^(7k), x^(7i) * y^(j + 1) * z^(7k), x^(7i) * y^(7j +2) * z^(7k)) is also.

If x^2 + y^7 = z^2 then y^7 = z^2 - x^2 = (z - x)(z + x) and so (z - x, z + x) is a divisor pair of z^7. - David A. Corneth, May 24 2025

Examples

			4^2 + 2^7 = 12^2, 12 is a term.
31^2 + 2^7 = 33^2, 33 is a term.
		

Crossrefs

Cf. A070745, A228946, A293284.

Programs

Mathematica

z[n_] := Count[n^2 - Range[(n^2 - 1)^(1/7)]^7, _?(IntegerQ[Sqrt[#]] &)] > 0; Select[Range[6550], z]