1.Introduction

We show that ax^4+by^4=cz^2 has infinitely many integer solutions
                 
                 
2.Theorem
     
Diophantine equation ax^4+by^4=cz^2 has infinitely many integer solutions.

condition: c=a+b.
a,b are arbitrary.


 
Proof.

ax^4+by^4=cz^2..................................(1)

Let U=x/y, V=cz/y^2,the equation (1) becomes to below equation.

V^2=acU^4+bc....................................(2)
Let c=a+b.

Since quartic equation (2) has a rational point Q(U,V)=(1,a+b), then this quartic equation is birationally equivalent to an elliptic curve below.

Y^2+4aYX+(8a^3+16a^2b+8ab^2)Y = X^3+(2a^2+6ab)X^2+(-4a^4-12a^3b-12a^2b^2-4ab^3)X-8a^6-48a^5b-96a^4b^2-80a^3b^3-24a^2b^4
Transformation is given, 

U = (4a^3+16a^2b+2aX+12ab^2+Y+2bX)/Y 
V = (24a^2b^5+56a^6b+144a^5b^2+176a^4b^3+104a^3b^4+4Xab^4+8Yab^3+8a^7+X^3b+72a^4bX+6a^3X^2+12Xa^5+aX^3+88a^2b^3X-8a^3Yb+144a^3b^2X+24a^2X^2b+18aX^2b^2)/(Y^2) 
X = (2aV-2a^2+2bV+2b^2+4a^2U+4abU)/(U^2-2U+1)
Y = (4a^3U^2+12a^2b-16a^2bU+16a^2bU^2+4a^2V+12ab^2U^2-16ab^2U+16ab^2+8abV+4b^3+4b^2V)/(U^3-3U^2+3U-1)

The point corresponding to point Q is P(X,Y)=(-2a^2-6ab, 8a^2b-8ab^2).
Hence we get 2Q(U,V)=( -(a^2-6ab-3b^2)/(3a^2+6ab-b^2), (a+b)(a^4+28a^3b+6a^2b^2+28ab^3+b^4)/((3a^2+6ab-b^2)^2) ).

This point P is of infinite order, and the multiples mP, m = 2, 3, ...give infinitely many points.

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

Case : m=2

x = -a^2+6ab+3b^2
y = 3a^2+6ab-b^2
z = a^4+28a^3b+6a^2b^2+28ab^3+b^4

Case : m=3

x = (5b^2+2ab+a^2)(b^4+12ab^3-26a^2b^2-52a^3b+a^4)
y = (b^2+2ab+5a^2)(b^4-52ab^3-26a^2b^2+12a^3b+a^4)
z = (b^4-4ab^3+70a^2b^2-4a^3b+a^4)(b^8+216ab^7+860a^2b^6+744a^3b^5+454a^4b^4+744a^5b^3+860a^6b^2+216a^7b+a^8)

       
Q.E.D.