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
KEYWORD
nonn
AUTHOR
XU Pingya, Oct 16 2017
STATUS
approved