1.Introduction

I found a parameter solution of A4+ B4+ C4+ D4 =  E4+ F4.

Proof is as easy as I am surprised at it.


2.Theorem
      
      a,b:integer

　　　The parameter solution of A4+ B4+ C4+ D4 =  E4+ F4 exists.


　　  　A=3b4a
        B=3a4b
        C=2(a4-b4)a
        D=2(a4-b4)b
        E=(2a4+b4)a
        F=(a4+2b4)b
 　


     
Proof.

x4+m4x4+a4+b4-(a+x)4-(b-mx)4..............................................(0)
=(-4a+4bm3)x3+(-6a2-6b2m2)x2+(-4a3+4b3m)x

To do to make it to 0 the coefficient of x, m is decided. 

        m=a3/b3.......................................................... (1)

Next,obtain x that becomes(-4a+4bm3)x3+(-6a2-6b2m2)x2=0

       x=-3/2(a2+b2m2)/(a-bm3)........................................... (2)

Substitute (1) to (2), and obtain x=3/2b4a/(a4-b4)....................... (3)

Substitute (1) and (3) to (0),the following parameter solutions are obtained. 　　
 
　　　

(3b4a)4+ (3a4b)4+ (2(a4-b4)a)4+ (2(a4-b4)b)4= ((2a4+b4)a)4+ ((a4+2b4)b)4


Q.E.D.


　
                  
       
3.Example

I show a few examples.
a,b<=5 


    
          (a,b)=(2,1)    14 +    84 +   104 +    54 =   114 +    34
          (a,b)=(3,1)    94 +  2434 +  4804 +  1604 =  4894 +   834
          (a,b)=(3,2)   724 +  2434 +  1954 +  1304 =  2674 +  1134
          (a,b)=(4,1)    24 +  1284 +  3404 +   854 =  3424 +   434
          (a,b)=(4,3)  4864 + 11524 +  7004 +  5254 = 11864 +  6274
          (a,b)=(5,1)    54 +  6254 + 20804 +  4164 = 20854 +  2094
          (a,b)=(5,2)   404 +  6254 + 10154 +  4064 = 10554 +  2194
          (a,b)=(5,3) 12154 + 56254 + 54404 + 32644 = 66554 + 23614
          (a,b)=(5,4)  6404 + 12504 +  6154 +  4924 = 12554 +  7584