21x^6 + 7y^6 - 28z^6  = w^2


We show diophantine equation 21x^6 + 7y^6 - 28z^6  = w^2 has infinitely many integer solutions.

21x^6 + 7y^6 - 28z^6  = w^2.................................................(1)

For details,ax^6 + by^6 + cz^6 = w^2 Ⅴ

Let (a,b,c)=(21, 7, -28),(p,q)=(2, -2) the we obtain

u^2 = 207532836t^4+296475480t^3+741188700t^2+197650320t+148237740...........(2)

This quartic equation is birationally equivalent to an elliptic curve below.

Y^2 = X^3-1117935X -230135850.

Rank is 3 and generator is (X,Y)=[6825/4,425565/8],[-266,6958],[11074,1159928].

Hence we can obtain infinitely many integer solutions for equation (1).

Numerical example:

[X,Y,Z,W] = [4, 8, 5, 1218].