1. Introduction

I will explain about the parameter solution of A4+ B4+ C4= D4+ E4+ F4

I prove that there are infinitly many parameter solutions.



2. Theorem
      
      a1,a2,c1,c2,c3,c4:integer

　　　There are infinitly many parameter solutions of A4+ B4+ C4= D4+ E4+ F4


　　  　A=a1x+c1
        B=a2x+c2
        C=(a1+a2)x+c1+c2
        D=a1x+c3
        E=a2x+c4
        F=(a1+a2)x+c3+c4
　　

      Condition
    
      c1=(a12c3+a12c4+2a1c4a2-a22c3)/(a22+a1a2+a12)
      c2=(a22c3+2a1c3a2+a22c4-a12c4)/(a22+a1a2+a12)
      c3,c4:Choose it so that c1, c2 may become integers.

     

Proof.

(a1x+c1)4+(a2x+c2)4+((a1+a2)x+c1+c2)4-(a1x+c3)4-(a2x+c4)4-((a1+a2)x+c3+c4)4

Decide c1, c2, c3, c4 to make the coefficient of x equal to 0.
Expanding and simplifying
　　
coefficient of x3=(12a12a2+12a1a22+8a13+4a23)c1+(12a1a22+8a23+4a13+12a12a2)c2
                 +(-12a1a22-12a12a2-8a13-4a23)c3+(-8a23-12a12a2-4a13-12a1a22)c4

coefficient ofx2=(12a1a2+12a12+6a22)c12+(12a22+24a1a2+12a12)c2c1+(12a22+12a1a2+6a12)c22
                +(-12a12-12a1a2-6a22)c32+(-24a1a2-12a22-12a12)c4c3+(-12a1a2-12a22-6a12)c42

coefficient ofx=(4a2+8a1)c13+(12a1+12a2)c2c12+(12a1+12a2)c22c1+(8a2+4a1)c23
               +(-4a2-8a1)c33+(-12a2-12a1)c4c32+(-12a2-12a1)c42c3+(-8a2-4a1)c43

Constant:-2c34+2c14+4c13c2+6c12c22-4c3c43-4c33c4-2c44-6c32c42+2c24+4c1c23

Solve a simultaneous equation about c1, c2.


c1=(a12c3+a12c4+2a1c4a2-a22c3)/(a22+a1a2+a12)
c2=(a22c3+2a1c3a2+a22c4-a12c4)/(a22+a1a2+a12)
c1, c2 become integers if a1, a2, c3, c4 are given to it suitably.
So there are infinitly many parameter solutions of A4+B4+C4=D4+E4+F4


 
                  
       
3. Example

I show a few examples.



(2x+7)4+x4+(3x+7)4=(2x+3)4+(x+5)4+(3x+8)4

(3x+26)4+x4+(4x+26)4=(3x+16)4+(x+14)4+(4x+30)4

(3x+11)4+(2x+1)4+(5x+12)4=(3x+4)4+(2x+9)4+(5x+13)4
　　　　　　　　　　　　　　　　　　　　
(4x+63)4+x4+(5x+63)4=(4x+45)4+(x+27)4+(5x+72)4

(4x+15)4+(3x+2)4+(7x+17)4=(4x+5)4+(3x+13)4+(7x+18)4

(5x+61)4+(2x+1)4+(7x+62)4=(5x+34)4+(2x+37)4+(7x+71)4

(6x+215)4+x4+(7x+215)4=(6x+175)4+(x+65)4+(7x+240)4