1.Introduction

This time, we show a parametric solution of m(x1^4+x2^4+x3^4)=n(y1^4+y2^4+y3^4).

2.Theorem
     
    There is a parametric solution of m(x1^4+x2^4+x3^4)=n(y1^4+y2^4+y3^4).
    
    x1 = a^3+(-d-c+b)a^2+(b^2+(-c-d)b)a+(-c-d)b^2+(c^2+d^2+cd)b
    x2 = (-d-c+b)a^2+(b^2+(-c-d)b+c^2+d^2+cd)a+b^3+(-c-d)b^2
    x3 = a^3+(-2d-2c+2b)a^2+(2b^2+(-2d-2c)b+c^2+d^2+cd)a+b^3+(-2d-2c)b^2+(c^2+d^2+cd)b
    y1 = -c^3+(a+b-d)c^2+(-d^2+(a+b)d)c+(a+b)d^2+(-ab-b^2-a^2)d
    y2 = (a+b-d)c^2+(-d^2+(a+b)d-ab-b^2-a^2)c-d^3+(a+b)d^2
    y3 = (-c-d)a^2+((-c-d)b+2cd+2c^2+2d^2)a+(-c-d)b^2+(2cd+2c^2+2d^2)b-c^3-2c^2d-2cd^2-d^3
    
    m = (c^2+d^2+cd)^2, n = (b^2+a^2+ab)^2.
    a,b,c,d are arbitrary.

 
Proof.

m(x1^4+x2^4+x3^4)=n(y1^4+y2^4+y3^4)....................................(1)

Let x3=x1+x2, y3=y1+y2, then we obtain,

x1^4+x2^4+x3^4 = 2(x1^2+x1x2+x2^2)^2

y1^4+y2^4+y3^4 = 2(y1^2+y1y2+y2^2)^2.

Hence equation (1) becomes to following equation.

m(x1^2+x1x2+x2^2)^2=n(y1^2+y1y2+y2^2)^2................................(2)

Let m=p^2 and n=q^2, the we obtain,

p(x1^2+x1x2+x2^2)=q(y1^2+y1y2+y2^2)....................................(3)

Substitute x1=t+a, x2=t+b, y1=t+c, y2=t+d to equation (3).

(3p-3q)t^2+(p(3b+3a)-q(3c+3d))t+p(b^2+a^2+ab)-q(c^2+d^2+cd)=0..........(4)

Equating to zero the coefficient of t^0, then we obtain

(p,q,t)=( c^2+d^2+cd, b^2+a^2+ab, -(-c^2b-c^2a-d^2b-d^2a-cdb-cda+b^2c+b^2d+a^2c+a^2d+abc+abd)/(-c^2-d^2-cd+b^2+a^2+ab)).

Finally, we obtain a parametric solution. 

Q.E.D.