diophantine equation ax^6 + by^6 + cz^6 = w^2 dioph143e

1.Introduction


We show some solutions of ax^6 + by^6 + cz^6 = w^2 and ax^6 + by^6 + cz^6 = w^3.


2.Theorem
        
     

    1. a^2x^6 + b^2y^6 + 2abz^6 = w^2 where pq=n^2.
    
       (x, y, z, w)=( p, q, pq, ap^3+bq^3 )
       a, b, p, q, n: arbitrary
    
    2. 3ax^6 - 3ay^6 - 3a^4z^6 = w^2
    
       (x, y, z, w)=( p^2+aq^2, p^2-aq^2, 2pq, 6a(p^2-aq^2)pq(p^2+aq^2) )
       a, p, q  : arbitrary
       
    3. 3ax^6 - 3ay^6 - 48a^4z^6 = w^2
    
       (x, y, z, w)=( p^2+aq^2, p^2-aq^2, pq, 6pqa(p^4+a^2q^4 ))
       a, p, q  : arbitrary

    4. 12a^2x^6 - 3a^2y^6 - 12a^5z^6 = w^2
    
       (x, y, z, w)=( p^2+aq^2, p^2-aq^2, 2pq, 3a(aq^2-p^2)(a^2q^4+6p^2aq^2+p^4))
       a, p, q  : arbitrary

    5. x^6 + y^6 - z^6 = w^3
    
      (x, y, z, w)=(3t^2, s, 1, 9t^3-1 )  where s^2=3t-9t^4.
     
        
    6. 2x^6 - 2y^6 + z^6 = w^3
    
       (x, y, z, w)=( (3p-n)(-p+2q-n), -6pn-4pq+13p^2+n^2+q^2, q^2-pq-11p^2-qn+8pn-n^2,
        -q^4+(14p-2n)q^3+(3n^2-6pn-33p^2)q^2+(-114p^2n+158p^3+42pn^2-6n^3)q+160p^3n+n^4-42p^2n^2-199p^4)
       n, p, q  : arbitrary
       
    7. x^8 + y^8 + z^8 = 2w^2
       (x, y, z, w)=(a^2+b^2, a^2-b^2, 2ab, a^8+14a^4b^4+b^8 )
       a, b  : arbitrary

    
 
Proof.

1. a^2(p)^6 + b^2(q)^6 + 2ab(pq)^3 = (ap^3 + bq^3)^2

The simplest identity!
Expand (ap^3 + bq^3)^2, then we obtain above identity.
If pq is square, we obtain a solution for a^2x^6 + b^2y^6 + 2abz^6 = w^2.


2. 3a(p^2 + aq^2)^6 - 3a(p^2 - aq^2)^6 - 3a^4(2pq)^6 = (6a(p^2 - aq^2)pq(p^2 + aq^2))^2

Expand (s^2 + at^2)^3 = s^6 + 3as^2t^2(s^2 + at^2) + a^3t^6
If s^2+at^2 is square, we obtain a solution for 3ax^6 - 3ay^6 - 3a^4z^6 = w^2.

3. 3a(p^2 + aq^2)^6 - 3a(p^2 - aq^2)^6 - 48a^4(pq)^6 = (6pqa(p^4 + a^2q^4))^2

It's obvious, since 48a^4(pq)^6 + (6pqa(p^4 + a^2q^4))^2 = 3a^4(2pq)^6 + (6a(p^2 - aq^2)pq(p^2 + aq^2))^2.
Thus, it's a solution of 12ax^6 - 12ay^6 - 3a^4z^6 = w^2 too.

4. 12a^2(p^2 + aq^2)^6 - 3a^2(p^2 - aq^2)^6 - 12a^5(2pq)^6 = (3a(aq^2 - p^2)(a^2q^4 + 6p^2aq^2 + p^4))^2

We use an identity 12(X+1)^6 - 3(X)^6 - 12(2X+1)^3 = (3X(X^2+4X+2))^2.
If 2X+1 is square, we obtain a solution for 12a^2x^6 - 3a^2y^6 - 12a^5z^6 = w^2.

5. x^6 + y^6 - z^6 = w^3

Piezas[1] gaved the solution by using a famous identity (3t^2)^6 + (3t - 9t^4)^3 - 1 = (9t^3 - 1)^3.
The curve s^2 = 3t - 9t^4 has rank 1. One solution (t,s) = (3/13, 138/169) leads to 27^6 + 138^6 - 169^6= (-25402)^3.


6. 2x^6 - 2y^6 + z^6 = w^3

We use an identity 2((p + q)t + 3)^6 - 2(3tp + 3)^6 + ((-q^2 + 11p^2 + pq)t^2 + (-3q + 24p)t + 9)^3 = ((5pq - 5p^2 + q^2)t^2 + 9tq + 9)^3
s^2 = (-q^2 + 11p^2 + pq)t^2 + (-3q + 24p)t + 9 is a genus 0 curve.
One solution (t,s) = (0,3) leads to the parametric solution.
   
7. x^8 + y^8 + z^8 = 2w^2

We use an identity A^4 + B^4 + (A+B)^4 = 2(A^2 + AB + B^2)^2.
m^8 + n^8 + (m^2 + n^2)^4 = 2(m^4 + m^2n^2 + n^4)^2
Let m = a^2 - b^2, n = 2ab, then (a^2 - b^2)^8 + (2ab)^8 + (a^2 + b^2)^8 - 2(a^8 + 14a^4b^4 + b^8)^2.
   
Q.E.D.　
 
　　                  
       
3.Example


1. a^2(p)^6 + b^2(q)^6 + 2ab(pq)^3 = (ap^3+bq^3)^2

Let p=u^2, q=v^2, then we obtain

a^2(u^2)^6 + b^2(v^2)^6 + 2ab(uv)^6 = (au^6+bv^6)^2,

a^2(u)^12 + b^2(v)^12 + 2ab(uv)^6 = (au^6+bv^6)^2.

For example, (a,b)=(1,-1), (u,v)=(n+1,n-1), then
((n+1)^2)^6 + ((n-1)^2)^6 -2(n+1)^6(n-1)^6 = (12n^5+40n^3+12n)^2



 
 4.Reference

[1]. Tito Piezas, https://sites.google.com/site/tpiezas/023
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