A293694 Numbers z such that x^2 + y^8 = z^2 for positive integers x and y.

20, 34, 65, 135, 320, 369, 544, 1040, 1095, 1305, 1350, 1404, 1620, 1625, 1746, 1971, 2056, 2160, 2379, 2754, 3060, 3281, 3996, 4100, 4470, 5120, 5265, 5904, 6625, 7825, 7830, 8194, 8575, 8704, 8796, 10250, 10935, 11125, 11700, 12500, 13154, 14500, 15579

Offset

1,1

Comments

Let i, j and k are nonegtive integers, m > n are positive integers. As [(m^2 - n^2)^(4i+1) * (2mn)^(4j + 2) * (m^2 + n^2)^(4k)]^2 + [(m^2 - n^2)^i * (2mn)^(j + 1) * (m^2 + n^2)^k]^8 = [(m^2 - n^2)^(4i) * (2mn)^(4j + 2) * (m^2 + n^2)^(4k+1)]^2, so that number of form (m^2 - n^2)^(4i) * (2mn)^(4j + 2) * (m^2 + n^2)^(4k+1) is a term in sequence.

When (x, y, z) is solution of x^2 + y^4 = z^2 (i.e., z = A271576(n)), (x^(4i+1) * y^(4j + 2) * z^(4k), x^i * y^(j + 1) * z^k, x^(4i) * y^(4j + 1) * z^(4k+1)) is solution of x^2 + y^8 = z^2.

When (x, y, z) is solution of x^2 + y^6 = z^2 (i.e., z = A293690(n)), (x^(4i+1) * y^(4j + 1) * z^(4k), x^i * y^(j + 1) * z^k, x^(4i) * y^(4j + 1) * z^(4k+1) is solution of x^2 + y^8 = z^2.

When (x, y, z) is solution of x^2 + y^8 = z^2, (x^(4i+1) * y^(4j) * z^(4k), x^(4i) * y^(j + 1) * z^k, x^(4i) * y^(4j) * z^(4k+1)) is also

Links

Karl-Heinz Hofmann, Table of n, a(n) for n = 1..13695

Example

12^2 + 2^8 = 20^2, 20 is a term.
63^2 + 2^8 = 65^2, 65 is a term.

Mathematica

z[n_] := Count[n^2 - Range[(n^2 - 1)^(1/8)]^8, _?(IntegerQ[Sqrt[#]] &)] > 0; Select[Range[16000], z]

Crossrefs

Cf. A009003, A228946, A271576, A293283, A293690, A293692.

Sequence in context: A127906 A108667 A108669 * A039343 A043166 A043946

Adjacent sequences: A293691 A293692 A293693 * A293695 A293696 A293697

Keyword

nonn

Author

XU Pingya, Oct 16 2017

Status

approved