1. Introduction

I will explain parameter solution of A4+ B4+ C4= D4+ E4+ F4.
I found interesting identity in Dickson's book([1]).


(23x+57y)4+(40x-6y)4+(17x-63y)4=(23x-57y)4+(40x+6y)4+(17x+63y)4

It is surprising thing that it is symmetry as beautiful as the above-mentioned, and 
such a simple relational expression consists. 
Oppositely, saying as approval because of symmetry might be accurate. 

I prove that there are infinitly many parameter solutions.

[1]. Dickson: HISTORY OF THE THEORY OF NUMBERS VOL.U

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

   The parameter solution ofA4+ B4+ C4= D4+ E4+ F4 exists infinitly.


     A=a1x+c1
        B=a2x+c2
        C=(a1+a2)xc1+c2
        D=a1x-c1
        E=a2x-c2
        F=(a1+a2)x-c1-c2
  

   Condition

         c1=-c2(a1+2a2)/(a2+2a1)
         a1 and a2 are chosen for c1 to become an integer.
         
     Moreover, the following identity is obtained if it is especially assumed
     c1=-(2a2+a1) and c2=2a1+a2.
      
     (a1x-a1-2a2)4+(a2x+a2+2a1)4+((a1+a2)x+a1-a2)4=(a1x+a1+2a2)4+(a2x-a2-2a1)4+((a1+a2)x-a1+a2)4
   
     
Proof.

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

To do to make it to 0 the coefficient of x, c1 and c2 are decided. 
When expanding and simplifying it
  
coefficient of x3

16a13c1+8a23c1+24a12a2c1+16a23c2+24a1a22c1+24a1a22c2+8a13c2+24a12a2c2
=8(a22+a1a2+a12)*(2a1c1+a1c2+a2c1+2a2c2)


coefficient of x

8a2c13+24a2c12c2+16a2c23+24a2c1c22+8a1c23+16a1c13+24a1c12c2+24a1c1c22
=8(c22+c1c2+c12)*(2a1c1+a1c2+a2c1+2a2c2)


If the simultaneous equations about c1 and c2 is solved, it becomes the following. 

    c1=-c2(a1+2a2)/(a2+2a1)

Therefore, the parameter solution will exist infinitly because c1 becomes an integer
if a1 and a2 are suitably given. 

Q.E.D.

                  
       
3.Example

I show a few examples.

    
1<=a1<=7,a2=1


        (x-5)4+(2x+4)4+(3x-1)4=(x+5)4+(2x-4)4+(3x+1)4

        (2x-4)4+(x+5)4+(3x+1)4=(2x+4)4+(x-5)4+(3x-1)4

        (3x-5)4+(x+7)4+(4x+2)4=(3x+5)4+(x-7)4+(4x-2)4

        (4x-6)4+(x+9)4+(5x+3)4=(4x+6)4+(x-9)4+(5x-3)4

        (5x-7)4+(x+11)4+(6x+4)4=(5x+7)4+(x-11)4+(6x-4)4

        (6x-8)4+(x+13)4+(7x+5)4=(6x+8)4+(x-13)4+(7x-5)4

        (7x-9)4+(x+15)4+(8x+6)4=(7x+9)4+(x-15)4+(8x-6)4




 


Well, though it is(23x+57y)4+(40x-6y)4+(17x-63y)4=(23x-57y)4+(40x+6y)4+(17x+63y)4 enumerated in opening,

Let X=x/y then it becomes (23X+57)4+(40X-6)4+(17X-63)4=(23X-57)4+(40X+6)4+(17X+63)4


(a1x-a1-2a2)4+(a2x+a2+2a1)4+((a1+a2)x+a1-a2)4=(a1x+a1+2a2)4+(a2x-a2-2a1)4+((a1+a2)x-a1+a2)4

Let a1=17 and a2=23 by above identity, and an opening identity is obtained.
(17x-63)4+(23x+57)4+(40x-6)4=(17x+63)4+(23x-57)4+(40x+6)4  












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