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

1,1

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

From Robert Israel, Nov 09 2025: (Start)

The first term that satisfies x^2 + y^7 = z^2 for two or more (x,y) pairs is 67620, as 67620^2 = 66836^2 + 14^7 = 57380^2 + 20^7.

If z is a term, then so is t^7 * z for every t > 1. (End)

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Example

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

Maple

N:= 10000: # for terms <= N
Res:= {}:
for y from 1 while y^7 < 4 * N^2 do
  for d in select(t -> t^2 < y^7, numtheory:-divisors(y^7)) do
     dp:= y^7/d;
     if dp - d mod 2 = 0 and d + dp <= 2*N then
        Res:= Res union {(d + dp)/2};
fi od od:
sort(convert(Res, list)); # Robert Israel, Nov 09 2025

Mathematica

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

Crossrefs

Cf. A070745, A228946, A293284.

Sequence in context: A277725 A319746 A079479 * A349027 A348965 A162694

Adjacent sequences: A293689 A293690 A293691 * A293693 A293694 A293695

Keyword

nonn,uned

Author

XU Pingya, Oct 14 2017

Status

approved