1.Introduction


We show a solution of ax^6 + by^6 + cz^6 = w^2.


2.Theorem
        
     

    ax^6 + by^6 + cz^6 = w^2 has a solution as follows.
    
    x = (3a+u^2)(108a^2-153u^2a+44u^4)
    
    y = -(3a-4u^2)(108a^2-63u^2a-u^4)
    
    z = 5u^2(27a^2-27u^2a+4u^4)
    
    w = u(139237542a^8-556950168u^2a^7+898942293u^4a^6-747501291u^6a^5+338831181u^8a^4-81602073u^10a^3+9744516u^12a^2-702000u^14a-64u^16)(-u^2+2a)
            
    condition: {a+b+c=m^2, b=a-u^2, c=4a(a-u^2)(54a^2-27u^2a+u^4)(54a^2-81u^2a+28u^4)/(u^10), m = (56u^4a-u^6-162u^2a^2+108a^3)/(u^5)}
    a, u  : arbitrary
 
Proof.

ax^6 + by^6 + cz^6 = w^2..................................................(1)

Let x=t+p, y=t-p, z = t, w = mt^3+nt^2+rt+s...............................(2)

Substitute (2) to (1) and using {a+b+c=m^2, b=a-u^2, s=p^3u} and simplifying (1), we obtain
(-2mn+12ap-6pu^2)t^5+(-2mr-n^2+15p^2u^2)t^4
+(40ap^3-20p^3u^2-2nr-2mp^3u)t^3
+(-2np^3u-r^2+15p^4u^2)t^2
+(-2rp^3u-6p^5u^2+12ap^5)t................................................(3)

Equating to zero the coefficient of t, t^2, and t^3 then we obtain

m = (56u^4a-u^6-162u^2a^2+108a^3)/(u^5)

n = -3p(-u^4-6u^2a+6a^2)/(u^3)

r = 3p^2(-u^2+2a)/u.

Finally, we obtain t as follows

t = 5/6u^2(27a^2-27u^2a+4u^4)p/(54a^3-81u^2a^2+19u^4a+4u^6).

Substitute m, n, r, and t to (2), and obtain a solution.            


   
Q.E.D.　
 
　　                  
       
3.Example


Case1: (a,b,c)=(6,3,112), 6x^6 + 3y^6 + 112z^6  = w^2

x = 3
y = 1
z = 1
w = 67

Case2: (a,b,c)=(30,6,640), 30x^6 + 6y^6 + 640z^6  = w^2

x = 119
y = 129
z = 5
w = 10622734

Case3: (a,b,c)=(64,192,5073), 64x^6 + 192y^6 + 5073z^6  = w^2

x = 70
y = 104
z = 17
w = 15830137