38x^6 + 20y^6 - 58z^6 = w^2


We show diophantine equation 38x^6 + 20y^6 - 58z^6 = w^2 has infinitely many integer solutions.

38x^6 + 20y^6 - 58z^6 = w^2...............................................................(1)

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

Let (a,b,c)=(38, 20, -58),(p,q)=(2, -9/10) the we obtain

u^2 = 152616244480t^4+440137036800t^3+540597120000t^2+296217600000t+105792000000..........(2)

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

Y^2 = X^3-48249902775375X -70387870133370376250.

Rank is 3 and generator is (X,Y)=[-2102099146/1369 , 14282355678942/50653],
                                 [-34952994449255603321866288302247100426/5941837236393758094026836422969 , -45535254581366611989383338050219337078420067281236562082/14483753556953643565709341333275626089455373203],
                                 [-17611734671430037532120524562935066559/6441482325295707611903533698624 , 104801957479909789936888696932889621007849502711300785761/16348530776617784120026851398644178974468397568].

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

Numerical example:

[X,Y,Z,W] = [13, 45, 7, 407740].




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