1.Introduction

We show that ax^4+bx^2y^2+cy^4=dz^2 has infinitely many integer solutions
                 
                 
2.Theorem
     
Diophantine equation ax^4+bx^2y^2+cy^4=dz^2 has infinitely many integer solutions.

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

 
Proof.

ax^4+bx^2y^2+cy^4=dz^2..................................(1)

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

V^2 = adU^4+bdU^2+cd....................................(2)
Let d=a+b+c.

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

Y^2+(2b+4a)YX+(8a^3+16a^2b+16a^2c+8ab^2+16abc+8ac^2)Y
= X^3+(2a^2+3ab+6ac+bc)X^2+(-4a^4-12a^3b-12a^3c-12a^2b^2-24a^2bc-12a^2c^2-4ab^3-12ab^2c-12abc^2-4ac^3)X
-8a^6-48a^5c-36a^5b-160a^4bc-60a^4b^2-96a^4c^2-80a^3c^3-216a^3bc^2-180a^3b^2c-44a^3b^3-24a^2c^4-96a^2bc^3
-72a^2b^3c-12a^2b^4-132a^2b^2c^2-12ab^3c^2-4abc^4-12ab^2c^3-4ab^4c
Transformation is given, 

U = (4a^3+10a^2b+16a^2c+12ac^2+2aX+20abc+6ab^2+2bX+Y+2cX+2bc^2+2b^2c)/Y
V = (8ac^2Yb+40ac^3Xb+62ab^2Xc^2+30aX^2bc+28ab^3Xc+2a^2b^2Y+34a^2b^3X+15a^2X^2b+18aX^2c^2+2ab^3Y+9aX^2b^2
    +48a^4bX+88a^2c^3X+158a^2b^2Xc+66a^3b^2X+24a^2X^2c+200a^3cXb+72a^4cX-8a^2bYc+224a^2c^2Xb-8a^3Yc+144a^3c^2X
    +6a^3X^2+12a^2b^5+aX^3+12Xa^5+144a^5c^2+56a^3b^4+104a^4b^3+56a^6c+44a^6b+96a^5b^2+8Yac^3+8a^7+400a^4b^2c
    +296a^3b^3c+528a^3b^2c^2+216a^2b^3c^2+88a^2b^4c+4abc^5+4Xab^4+16ab^4c^2+16ab^2c^4+24ab^3c^3+4ab^5c+244a^5bc
    +472a^4bc^2+4Xac^4+X^3b+X^3c-2b^3cY+2c^3Xb^2+3bc^2X^2+2b^3Xc^2+3X^2b^2c-2c^2Yb^2+240a^2c^3b^2+392a^3c^3b
    +176a^4c^3+104a^3c^4+24a^2c^5+124a^2bc^4)/(Y^2) 
X = (2aV-2a^2-2ab+2bV+2bc+2cV+2c^2+6abU+2b^2U+2bcU+4a^2U+4acU)/(U^2-2U+1)
Y = (6b^2c+10bc^2+2ab^2+16ac^2+12a^2c+2a^2b+4b^3U-16abcU+4c^3+20abc+4a^2V+4b^2V+4c^2V+8abV+8acV+8bcV-16a^2cU
    +4b^2cU-16ac^2U+4ab^2U+10a^2bU^2+16a^2cU^2+2b^2cU^2+12ac^2U^2+6ab^2U^2+2bc^2U^2+20abcU^2+4a^3U^2)/(U^3-3U^2+3U-1).

The point corresponding to point Q is P(X,Y)=( -2a^2-3ab-6ac-bc, 8a^2c-2ab^2-8ac^2+2b^2c ).
Hence we get 2Q(U,V)=( -(a^2-6ac-3c^2-4bc)/(3a^2+4ab+6ac-c^2), (a+b+c)(a^4+4a^3b+28a^3c+28a^2bc+6a^2c^2+16ab^2c+28ac^3+28abc^2+4bc^3+c^4)/((3a^2+4ab+6ac-c^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+6ac+3c^2+4bc
y = 3a^2+4ab+6ac-c^2
z = a^4+(28c+4b)a^3+(6c^2+28bc)a^2+(28c^3+28bc^2+16b^2c)a+4bc^3+c^4

Case : m=3

x = a^6-50a^5c+(-140bc-125c^2)a^4+(-368bc^2-160b^2c-300c^3)a^3+(-240b^2c^2-64b^3c-105c^4-360bc^3)a^2+(62c^5+80bc^4)a+5c^6+20c^5b+16c^4b^2
y = 5a^6+(62c+20b)a^5+(16b^2-105c^2+80bc)a^4+(-360bc^2-300c^3)a^3+(-368bc^3-240b^2c^2-125c^4)a^2+(-160b^2c^3-64b^3c^2-50c^5-140bc^4)a+c^6
z = -a^12+(-12b-212c)a^11+(-16b^2-1012bc-66c^2)a^10+(-4356bc^2-2640b^2c-12420c^3)a^9+(-73212bc^3-17232b^2c^2-4224b^3c-56815c^4)a^8
    +(-135872b^2c^3-20864b^3c^2-186296bc^4-3328b^4c-47784c^5)a^7+(-116864b^3c^3-1024b^5c-8448b^4c^2-241824c^4b^2-27548c^6-128328c^5b)a^6
    +(-185728c^4b^3-128328bc^6-59136b^4c^3-187872c^5b^2-47784c^7)a^5+(-56815c^8-186296bc^7-23552b^5c^3-185728c^5b^3-241824b^2c^6-103936b^4c^4)a^4
    +(-12420c^9-4096b^6c^3-73212c^8b-23552b^5c^4-116864b^3c^6-59136b^4c^5-135872c^7b^2)a^3+(-17232b^2c^8-8448b^4c^6-4356c^9b-20864c^7b^3-66c^10)a^2
    +(-2640c^9b^2-1024c^6b^5-212c^11-3328c^7b^4-1012c^10b-4224c^8b^3)a-12c^11b-c^12-16b^2c^10

       
Q.E.D.