1.Introduction

About ax^4+by^2=c with the Mathoverflow, we show that ax^4+by^2=c has infinitely many integer solutions where c=a+b.
                 
                 
2.Theorem
     
Diophantine equation ax^4+by^2=c has infinitely many integer solutions.

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

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

Let U=x, V=by,the equation (1) becomes to below equation.

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

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

Y^2-4aYX-8b^2aY = X^3+(-6ab-4a^2)X^2+4ab^3X-24a^2b^4-16a^3b^3.

Transformation is given, 

U = (2bX-12b^2a-8a^2b+Y)/Y
V = (bX^3-18b^2X^2a-12bX^2a^2+72b^3a^2X+96b^2a^3X-24a^2b^2Y+32ba^4X-16ba^3Y-8b^3aY-4ab^4X+24a^2b^5+16a^3b^4)/(Y^2)
X = (2bV+2b^2-4abU+4ab)/(U^2-2U+1)
Y = (4b^2V+4b^3+16ab^2U-4b^2a-12ab^2U^2-8a^2bU^2+16a^2bU-8a^2b)/(U^3-3U^2+3U-1).

The point corresponding to point Q is P(X,Y)=( 6ab+4a^2, 24a^2b+16a^3+8b^2a ).

Hence we get 2P(X,Y)=( 1/4(b^4+40a^2b^2+64ba^3+16a^4)/((b+2a)^2), -1/8(-16b^5a-1104a^4b^2-192a^6-768a^5b+b^6-768a^3b^3-260a^2b^4)/((b+2a)^3) ).

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).
       
Q.E.D.


3.Examples

Case : m=2

x = (-3b^2+4a^2)/(b^2+8ab+4a^2)
y = (-b^4+72a^2b^2+96ba^3+48a^4+24ab^3)/((b^2+8ab+4a^2)^2)

Case : m=3

x = (4a^2+8ab+5b^2)(16a^4-32ba^3-56a^2b^2-8ab^3+b^4)/((4a^2+b^2)(16a^4+96ba^3+136a^2b^2+56ab^3+b^4))
y = (80a^4+160ba^3+88a^2b^2+8ab^3+b^4)(256a^8+1024ba^7+2304a^6b^2+3328a^5b^3+2144b^4a^4-64b^5a^3-624b^6a^2-208b^7a+b^8)/((4a^2+b^2)^2(16a^4+96ba^3+136a^2b^2+56ab^3+b^4)^2)
a,b are arbitrary.






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