1.Introduction


We show a parametric solution of a6x^6+a5x^5y+a4x^4y^2+a3x^3y^3+a2x^2y^4+a1xy^5+a0y^6 =
                                 a6u^6+a5u^5v+a4u^4v^2+a3u^3v^3+a2u^2v^4+a1uv^5+a0v^6.


2.Method
        

a6x^6+a5x^5y+a4x^4y^2+a3x^3y^3+a2x^2y^4+a1xy^5+a0y^6 =
a6u^6+a5u^5v+a4u^4v^2+a3u^3v^3+a2u^2v^4+a1uv^5+a0v^6..........................................(1)

Let x = pt + 1, y = qt - 1, u = pt - 1, v = qt + 1............................................(2)

Substitute (2) to (1) and simplifying (1), we obtain
(8a4p^3q^2+2a1q^5-4a4p^4q+4a2pq^4-6a3p^3q^2-2a5p^5+12a6p^5-12a0q^5-10a1pq^4-8a2p^2q^3+6a3p^2q^3+10a5p^4q)t^5
+(-40a0q^3-20a1pq^2+20a1q^3-8a2q^3+2a3q^3-20a5p^3-24a4p^2q+20a5p^2q+8a4pq^2+8a4p^3+18a3p^2q+24a2pq^2-18a3pq^2-8a2p^2q+40a6p^3-2a3p^3)t^3
+(4a2p-4a4q-8a2q-2a1p+6a3q-12a0q+10a1q+2a5q+8a4p-10a5p+12a6p-6a3p)t...........................(3)

Equating to zero the coefficient of t in (3),  then we obtain
a5 = -3a3+2a4+4a2+6a0-5a1
a6 = -2a3+a4+3a2+5a0-4a1
And furthermore, set a4 = 3a3+10a1-15a0-6a2.

To obtain the rational solution t of (3), discriminant must be square.
4(p+q)^6((-21a1+36a0+10a2-3a3)p^2+(-2a2-18a0+8a1)qp+(6a0-a1)q^2)(a3+10a1-20a0-4a2)= square....(4)
  
Thus substitute the parametric solution of (4) to (2), then we obtain the parametric solution of (1).            

In addition, we obtain the parametric solution of degree one as follows.

Set a3 = 4a2-10a1+20a0 and a4 = 6a2-20a1+45a0.
a0,a1,and a2 are arbitrary.

x = (6a0-a1)t+1
y = (-9a1+2a2+24a0)t-1
u = (6a0-a1)t-1
v = (-9a1+2a2+24a0)t+1

　　                  
       
3.Examples


Case 1: (a6,a5,a4,a3,a2,a1,a0)=(1, 4, 5, 2, 1, 2, 1)
        x^6+4*x^5*y+5*x^4*y^2+2*x^3*y^3+x^2*y^4+2*x*y^5+y^6 = u^6+4*u^5*v+5*u^4*v^2+2*u^3*v^3+u^2*v^4+2*u*v^5+v^6

x= 4*a^2-6*a+2
y= 3*a^2+2*a-1
u= 2*a*(a+1)
v= 5*a^2-6*a+1


Case 2: (a6,a5,a4,a3,a2,a1,a0)=(1, 4, 6, 5, 4, 3, 1)
        x^6+4*x^5*y+6*x^4*y^2+5*x^3*y^3+4*x^2*y^4+3*x*y^5+y^6 = u^6+4*u^5*v+6*u^4*v^2+5*u^3*v^3+4*u^2*v^4+3*u*v^5+v^6
        
x= 5*a^2-14*a+9
y= -2*a^2+2*a+4
u= -a^2+2*a+3
v= 4*a^2-14*a+10


Case 3: (a6,a5,a4,a3,a2,a1,a0)=(1, 4, 6, 6, 7, 6, 2)
        x^6+4*x^5*y+6*x^4*y^2+6*x^3*y^3+7*x^2*y^4+6*x*y^5+2*y^6 = u^6+4*u^5*v+6*u^4*v^2+6*u^3*v^3+7*u^2*v^4+6*u*v^5+2*v^6

x= 3*a^2-5*a+2
y= -a^2-a+4
u= -a+5
v= 2*a^2-5*a+1


Case 4: (a6,a5,a4,a3,a2,a1,a0)=(5, 18, 23, 12, 3, 2, 1)
        5*x^6+18*x^5*y+23*x^4*y^2+12*x^3*y^3+3*x^2*y^4+2*x*y^5+y^6 = 5*u^6+18*u^5*v+23*u^4*v^2+12*u^3*v^3+3*u^2*v^4+2*u*v^5+v^6
x= 4*a+1
y= 12*a-1
u= 4*a-1
v= 12*a+1





 




4.Reference


[1]. Ajai Choudhry: On the Quartic Diophantine Equationf(x, y)=f(u, v), Journal of Number Theory 75,(1999)