1.Introduction

Many integer solutions of x3+y3+z3=w3 are known for a long time. 

      33+43+53=63

            13+63+83=93

            73+143+173=203
         .
                  .
                  .
Some parameter solutions are known besides an individual numeric solution. 
For instance, Ramanujan(genius mathematician in India) obtained the parameter solution of
x3+y3+z3=w3 as follows.  

    x=3a2+5ab-5b2, y=4a2-4ab+6b2, z=5a2-5ab-3b2, w=6a2-4ab+4b2  ([1]. Hardy)

Moreover, to another

   x=7a2-16ab-3b2, y=14a2+4ab+6b2, z=-14a2+4ab-6b2, w=7a2+16ab-3b2  ([2]. Dickson)

   x=a2-7ab+63b2, y=8a2-20ab-42b2, z=6a2+20ab-56b2, w=9a2-7ab+7b2



[1]. Hardy and Wright:  An introduction to the theory of numbers
[2]. Dickson:           History of the theory of numbers V.2

I studied the solution like the form of the  Ramanujan's one. 
I found many parameter solutions of x3+y3+z3=w3.


  
2.Lemma

      a,b,p,q,r,s: integer
      gcd(p,q,r,s)=1
      

   Solution of x3+y3+z3=w3 is given 
                    
      x= (q2+pq)a2+(-r2+s2)ba+(pr-ps)b2

          y= (p2+pq)a2+(-s2+r2)ba+(qr-qs)b2

          z= (qr+pr)a2+(-p2+q2)ba+(s2-rs)b2

          w= (ps+qs)a2+(-p2+q2)ba+(rs-r2)b2


      Condition

      p3+q3+r3=s3



Proof.


   (pt+a)3+(qt-a)3+(rt+b)3-(st+b)3=0 

      Expanding and simplifying


   (p3+q3+r3-s3)t3+(-3q2a+3p2a-3s2b+3r2b)t2+(3rb2+3qa2+3pa2-3sb2)t=0 
   

      By the condition,p3+q3+r3-s3=0 then obtain

      t=-(rb2+qa2+pa2-sb2)/(-q2a+p2a-s2b+r2b)

      Let x=pt+a, y=qt-a, z=rt+b, w=st+b then 

        x= (q2+pq)a2+(-r2+s2)ba+(pr-ps)b2

          y= (p2+pq)a2+(-s2+r2)ba+(qr-qs)b2

          z= (qr+pr)a2+(-p2+q2)ba+(s2-rs)b2

          w= (ps+qs)a2+(-p2+q2)ba+(rs-r2)b2


     We know that coefficient of a2 is the multiple of (p+q) and 
     coefficient of b2 is the multiple of (r-s).

Q.E.D.



3.Theorem
    
      x,y,z,w: integer

   There are infinitely many parameter solutions of x3+y3+z3=w3
 
Proof.
   
      By lemma,we can obtain the parameter solution of x3+y3+z3=w3 if there is the solution of
      p3+q3+r3=s3.
    By the way, the parameter solution of x3+y3+z3=w3 can be infinitely obtained,
      because the solution of p3+q3+r3=s3 can be infinitely obtained from the solution
      that Ramanujan gave.

     Permutation of one solution (p,q,r,s) gives 6 parameter solutions.
      But they are 3 different solutions essentially.
          
Q.E.D.                    
       
4.Example

      Example 1.

        x= (q2+pq)a2+(-r2+s2)ba+(pr-ps)b2.................(1)

          y= (p2+pq)a2+(-s2+r2)ba+(qr-qs)b2.................(2)

          z= (qr+pr)a2+(-p2+q2)ba+(s2-rs)b2.................(3)

          w= (ps+qs)a2+(-p2+q2)ba+(rs-r2)b2.................(4)
     
     When we substitute (p,q,r,s) = (1,6,8,9) to (1),(2),(3) and (4),and obtain


          x=(42a2+17ba-b2)
          y=(7a2-17ba-6b2)
          z=(56a2+35ba+9b2)
          w=(63a2+35ba+8b2)
          


          Let's show the x, y, and z>0 from among the example of the numerical 
          value made from this parameter solution. 
     
            7393+  383+ 10453= 11563

          16133+ 1443+ 22353= 24863

          37043+ 2373+ 52023= 57653

                   43693+ 5243+ 59593= 66583

                   47013+ 1363+ 67313= 74223
                .
                                .
                                .

        Of course (p,q,r,s) = (1,8,6,9),(6,1,8,9),(6,8,1,9),(8,1,6,9),(8,6,1,9) give 
        parameter solution,but they are omitted here.
   
       Example 2.
      
     No.1: Substitute (p,q,r,s) = (3,4,5,6) to (1),(2),(3) and (4),and obtain


               x=(28a2+11ba-3b2)
               y=(21a2-11ba-4b2)
               z=(35a2+7ba+6b2)
               w=(42a2+7ba+5b2)
           
         A few numerical values

               1313+ 583+1603=1873
               1413+ 763+1713=2023
               1633+ 963+1983=2353
               3063+1073+3813=4403
               3763+2333+4583=5453
               4193+1623+5173=6003
               5533+1683+6983=8013
               5933+ 163+8503=9373


         No.2: Substitute (p,q,r,s) = (3,5,4,6) to (1),(2),(3) and (4),and obtain

              x=20a2+10ba-3b2
              y=12a2-10ba-5b2
              z=16a2+8ba+6b2
              w=24a2+8ba+4b2
  
               493+ 153+ 423= 583
               573+  73+ 543= 703
               893+ 153+ 823=1083
               973+ 233+ 863=1163
               
               Let substitute -2b to b at above parameters and simplify it,then obtain
               Ramanujan's identity below.

               x=5a2-5ba-3b2
               y=3a2+5ba-5b2
               z=4a2-4ba+6b2
               w=6a2-4ba+4b2


         No.3: Substitute (p,q,r,s) = (4,3,5,6) to (1),(2),(3) and (4),and obtain

              x=21a2+11ba-4b2
              y=28a2-11ba-3b2
              z=35a2-7ba+6b2
              w=42a2-7ba+5b2

 
               143+  73+ 173= 203
               343+ 293+ 443= 533
               963+1073+1413=1703
              1093+1083+1503=1813
              1273+  63+1803=1993 
 

         No.4: Substitute (p,q,r,s) = (4,5,3,6) to (1),(2),(3) and (4),and obtain

              x=15a2+9ba-4b2
              y=12a2-9ba-5b2
              z=9a2+3ba+6b2
              w=18a2+3ba+3b2
       

                743+ 253+ 483= 813
                793+ 383+ 483= 873
                793+ 203+ 543= 873
               1733+ 343+1233=1923


         No.5: Substitute (p,q,r,s) = (5,3,4,6) to (1),(2),(3) and (4),and obtain

              x=12a2+10ba-5b2
              y=20a2-10ba-3b2
              z=16a2-8ba+6b2
              w=24a2-8ba+4b2

 
               213+ 193+ 183= 283
               373+ 273+ 303= 463
               893+ 153+ 823=1083
               953+ 973+ 863=1343

         No.6: Substitute (p,q,r,s) = (5,4,3,6) to (1),(2),(3) and (4),and obtain

              x=12a2+9ba-5b2
              y=15a2-9ba-4b2
              z=9a2-3ba+6b2
              w=18a2-3ba+3b2
 
               613+ 383+ 363= 693
               653+ 343+ 393= 723 
               853+ 323+ 543= 933
              1423+ 653+ 873=1563


  


      









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