1.Introduction

Ajai Choudhry gave a parametric solution of a0x^5+a1x^4y+a2x^3y^2+a3x^2y^3+a4xy^4+a5y^5 = a0u^5+a1u^4v+a2u^3v^2+a3u^2v^3+a4uv^4+a5v^5.

We give two parametric solutions 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.Theorem
        
     

    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 has parametric solutions.
    
    Case 1:
    
    If r^2=((10a2-3a3+36a0-21a1)p^2+(-2a2-18a0+8a1)qp+(6a0-a1)q^2)(a3-4a2-20a0+10a1) has a rational solution, (1) has a quadratic parametric solution.
    
    condition: a6 = a4-4a1+3a2-2a3+5a0
               a5 = 2a4-5a1+4a2-3a3+6a0
               a4 = 3a3-6a2-15a0+10a1
               a0, a1, a2, a3, p, and q are arbitrary integers.
    
    Case 2:
    
    x = (-a1+6a0)t+1 
    y = (24a0+2a2-9a1)t-1
    u = (-a1+6a0)t-1
    v = (24a0+2a2-9a1)t+1
    
    condition: a6 = a4-4a1+3a2-2a3+5a0
               a5 = 2a4-5a1+4a2-3a3+6a0
               a4 = 6a2-20a1+45a0
               a3 = 4a2-10a1+20a0
               a0, a1, a2, and t are arbitrary integers. 
 
Proof.

Case 1:

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)

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

Substitute (2) to (1) , and simplifying (1), we obtain

(2q^5a1-12q^5a0-6a3p^3q^2+8a4p^3q^2-2a5p^5-8a2p^2q^3+6a3p^2q^3+4a2pq^4+10a5p^4q-10a1pq^4-4a4p^4q+12a6p^5)t^5
+(20a5p^2q+40a6p^3-8q^3a2+8a4pq^2-20a5p^3-40q^3a0+20q^3a1-18a3pq^2+2q^3a3-20a1pq^2-2a3p^3+24a2pq^2+8a4p^3-8a2p^2q-24a4p^2q+18a3p^2q)t^3
+(4a2p+12a6p-8qa2-2a1p-10a5p+2qa5+10qa1+8a4p-4qa4-12qa0-6a3p+6qa3)t.......................................................................(3)

Equating to zero the coefficient of t in (3),  then we obtain

a6 = a4-4a1+3a2-2a3+5a0

a5 = 2a4-5a1+4a2-3a3+6a0

To obtain a solution of (3), we have to find the rational solution of (4).

s^2 = -4(p+q)^4(-19p^3a1+14p^3a2-9a3p^3+24p^3a0+4a4p^3+13p^2qa1-8a2p^2q+3a3p^2q-18p^2qa0-7a1pq^2+2a2pq^2+12pq^2a0+q^3a1-6q^3a0)
              *(4a4p+20a2p+40pa0-11a3p-30a1p-4qa2+qa3+10qa1-20qa0)........................................................................(4)

Take a4 = 3a3+10a1-15a0-6a2, then (4) becomes to (5).

s^2 = -4(p+q)^6(-10p^2a2+3p^2a3-36p^2a0+21p^2a1+2pqa2+18pqa0-8pqa1-6q^2a0+a1q^2)(a3-4a2-20a0+10a1).......................................(5)

Hence, we only have to find a rational solution of quadratic equation r^2 = ((10a2-3a3+36a0-21a1)p^2+(-2a2-18a0+8a1)qp+(6a0-a1)q^2)(a3-4a2-20a0+10a1).

Accordingly, if r^2 = ((10a2-3a3+36a0-21a1)p^2+(-2a2-18a0+8a1)qp+(6a0-a1)q^2)(a3-4a2-20a0+10a1) has a rational solution, (1) has a quadratic parametric solution.



Case 2:

In the same way as the case 1, equating to zero the coefficient of t in (3),  then we obtain

a6 = a4-4a1+3a2-2a3+5a0

a5 = 2a4-5a1+4a2-3a3+6a0

Equating to zero the coefficient of t^3 in (3),  then we obtain

a4 = 6a2-20a1+45a0

a3 = 4a2-10a1+20a0

Similarly, equating to zero the coefficient of t^5 in (3),  then we obtain

p = -a1+6a0

q = 24a0+2a2-9a1

Consequently, we obtain a parametric solution of (1).

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


   
Q.E.D.　
 
　　                  
       
3.Example


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

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

a is arbitrary integer.

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

x= 4t+1
y= 12t-1
u= 4t-1
v= 12t+1

t is arbitrary integer.

 


4.Reference


[1].Ajai Choudhry: On the Solvability of Quintic and Sextic Diophantine Equations of the Typef(x, y)=f(u, v), Journal of Number Theory 88(2001)