1.Introduction

We show that X^4 + Y^4 + Z^8 = 2W^4 has the parametric solutions.
                 
                 
2.Theorem
     
Diophantine equation X^4 + Y^4 + Z^8 = 2W^4 has the parametric solutions.

a,b are arbitrary.
 
Proof.

X^4 + Y^4 + Z^8 = 2W^4..........................................................(1)

We consider the Proth's identity: A^4 + B^4 + (A+B)^4 = 2(A^2 + AB + B^2)^2.....(2)

Let A^2 + AB + B^2 = u^2........................................................(3)

We can obtain a parametric solution of equation (3) easily.
(A,B,u)=(a^2-b^2, b^2+2ab, b^2+ab+a^2).

By using this solution, we can obtain three parametric solutions of equation (1) as follows. 

Case 1: A is square number

X^8 + Y^4 + Z^4 - 2W^4=0........................................................(4)

We can obtain a parametric solution of a^2-b^2=v^2.
(a,b,v)=(p^2+q^2, 2pq, p^2-q^2).

Hencw, we obtain a parametric solution of equation (4) below.

X = p^2-q^2
Y = 4pq(-pq+p^2+q^2)
Z = (p^2+q^2)(q^2-4pq+p^2)
W = q^4-2pq^3+6p^2q^2-2p^3q+p^4
p,q are arbitrary.

Case 2: B is square number

X^4 + Y^8 + Z^4 - 2W^4=0............................(5)

We can obtain a parametric solution of b^2+2ab=v^2.
(a,b,v)=(1/2(2p+1)/p, p, p+1).

Hencw, we obtain a parametric solution of equation (5) below.

X = (2p^2-2p-1)(2p^2+2p+1)
Y = 2(p+1)p
Z = (2p+1)(4p^2+2p+1)
W = 1+4p+6p^2+4p^3+4p^4
p is arbitrary.

Case 3: A+B is square number

X^4 + Y^4 + Z^8 - 2W^4=0...........................(6)

We can obtain a parametric solution of a^2+2ab=v^2.
(a,b,v)=(1, 2p(-1+p), -1+2p).

Hencw, we obtain a parametric solution of equation (6) below.

X = (2p^2-2p+1)(2p^2-2p-1)
Y = 4p(-1+p)(p^2-p+1)
Z = -1+2p
W = 1-2p+6p^2-8p^3+4p^4
p is arbitrary.

       
Q.E.D.