1.Introduction

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

Proof is as easy as I am surprised again.

2.Theorem
      
      a,b:integer

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


　　  　A=6ab8
        B=a(a8-4b8)
        C=b(a8-4b8+3a4b4)
        D=a(a8+2b8)
        E=b(a8-4b8-3a4b4)
       
 　


     
Proof.

x4+a4+(b+mx)4-(a+x)4-(b-mx)4..............................................(0)
=(8bm3-4a)x3-6a2x2+(8b3m-4a3)x


To do to make it to 0 the coefficient of x, m is decided. 
        m=a3/(2b3)....................................................... (1)
Next,solve (8bm3-4a)x3-6a2x2=0 about x


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

Substitute (1) to (2),and obtain x=6ab8/(a8-4b8)......................... (3)

Substitute (1) and (3) to (0),and obtain　　

(6ab8)4 + (a(a8-4b8))4 + (b(a8-4b8+3a4b4))4 = (a(a8+2b8))4 + (b(a8-4b8-3a4b4))4


Q.E.D.
　
                  
       
3.Example

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


    
          (a,b)=(2,1)       14 +      424 +      254 =      434 +      174
          (a,b)=(3,1)      184 +   196714 +    68004 =   196894 +    63144
          (a,b)=(3,2)    46084 +   166114 +   188504 =   212194 +    32984
          (a,b)=(4,1)       24 +   218444 +    55254 =   218464 +    53974
          (a,b)=(4,3)   393664 +   392924 +   761254 =   786584 +   171874
          (a,b)=(5,1)      104 +  6510354 +  1308324 =  6510454 +  1295824
          (a,b)=(5,2)    25604 +  6493354 +  2797344 =  6518954 +  2397344
          (a,b)=(5,3)  1968304 + 18219054 + 15487684 = 20187354 +  6375184
          (a,b)=(5,4)  6553604 +  2141354 +  8113084 =  8694954 +  4686924