1.Introduction

      I show that there are infinitely many solutions for x1^k+x2^k+x3^k+x4^k+x5^k+x6^k = y1^k+y2^k+y3^k+y4^k+y5^k+y6^k,
      for k=1,2,4,6,8 by Sinha's Theorem.


      We apply Sinha's Theorem( See the previous problem [61]) to
      x1^k+x2^k+x3^k+x4^k+x5^k+x6^k+x7^k = y1^k+y2^k+y3^k+y4^k+y5^k+y6^k+y7^k, for k=1,2,4,6,8.

      If x1+y4=0, we can obtain the solution for k.6.6, for k=1,2,4,6,8.

2. Theorem


       There are infinitely many solutions for x1^k+x2^k+x3^k+x4^k+x5^k+x6^k =
       y1^k+y2^k+y3^k+y4^k+y5^k+y6^k, for k=1,2,4,6,8


     
      x1 =  2bx-38a-14b
      x2 = -2ax+14a+6b
      x3 = -6ax-4b-4a-2bx
      x4 = -2ax+28a+12b-2bx
      x5 = -4ax-24a-8b
      x6 =  4ax+10a+2b+2bx

      y1 = -2ax+8b+24a-2bx
      y2 = -4ax-12b-28a
      y3 = -6ax+4b+4a-2bx
      y4 =  2ax+38a+14b
      y5 =  2bx-6b-14a
      y6 =  4ax-2b-10a+2bx

      where (8x^2-39x-539)a^2+(24x+7x^2-406)ba+(-75+15x)b^2 = 0.


     

              
Proof.

     x1^k+x2^k+x3^k+x4^k+x5^k+x6^k = y1^k+y2^k+y3^k+y4^k+y5^k+y6^k,for k=1,2,4,6,8.........(1)


     1. Solving for a1^2 +a2^2 +a3^2 = b1^2 +b2^2 +b3^2.


           a1=ax+s1, a2=bx+s2, a3=-3(ax+s1)-(bx+s2), b1=ax-s2, b2=bx-s3, b3=-(3a+b)x+3s1+s2


           Take s1=3b+7a, s2=-19a-7b then


           a1^2 +a2^2 +a3^2 - ( b1^2 +b2^2 +b3^2) = 0



      2. Solving for a1^4 +a2^4 +a3^4 = b1^4 +b2^4 +b3^4.

      
           a1^4 +a2^4 +a3^4 -( b1^4 +b2^4 +b3^4)=162x(a+b)(3a+b)f

       
           f=(8x^2-39x-539)a^2+(24x+7x^2-406)ba+(-75+15x)b^2=0.............................(2)

      

     We must find the rational value (a,b,x) for (2).
     y^2=49x^4-144x^3-368x^2+1152x+3136....................................................(3)

     Transform (3) to weierstrass form (4).

     V^2 = U^3 + U^2 -3225U + 61623........................................................(4)

     We find that Rank of (4) is 2 and generator is (23,-12) and (21,-60) by using MWRANK(Cremona).

     Since (4) has infinitely many rational points (U,V), (3) has infinitely many rational points (x,y).

     So, substitute (x,a,b) to (5), we obtain infinitely many solutions for (1).

      
      
      x1 = 2a2       =  2bx-38a-14b
      x2 = b1+b2+b3  = -2ax+14a+6b
      x3 = 2a3       = -6ax-4b-4a-2bx
      x4 = b1-b2+b3  = -2ax+28a+12b-2bx
      x5 = -b1+b2+b3 = -4ax-24a-8b
      x6 = b1+b2-b3  =  4ax+10a+2b+2bx
      

      y1 = a1-a2+a3  = -2ax+8b+24a-2bx
      y2 = -a1+a2+a3 = -4ax-12b-28a
      y3 = 2b3       = -6ax+4b+4a-2bx
      y4 = 2b1       =  2ax+38a+14b
      y5 = 2b2       =  2bx-6b-14a
      y6 = a1+a2-a3  =  4ax-2b-10a+2bx.....................................................(5)


     

    Q.E.D.



3. Example

        (8x^2-39x-539)a^2+(24x+7x^2-406)ba+(-75+15x)b^2=0

        height(x)<50

        [x, a, b  ]        [  x1   x2   x3   x4   x5   x6] [  y1   y2   y3   y4   y5   y6]

        [ 2/3    9   -31], [  19,  27,  35,  34,   3,   4],[   1,  36,  31,  30,   7,  17] 
        [ 2/3   65  -163], [ 152,  58, 131,   2, 161, 105],[ 145,  14, 163, 103,  56, 138] 
        [  -3   24   -35], [ 106, 135, 133,  93,   4,  46],[ 115,  18,  89, 139,  42, 124] 
        [   5   37  -178], [ 347, 460, 617, 155, 102, 513],[ 437, 180,  53, 358, 615, 527] 
        [ 8/3   63  -211], [ 106, 135, 133,   4,  93,  46],[ 115,  18,  89,  42, 139, 124] 
        [  11    1     0], [  19,   4,  35,   3,  34,  27],[   1,  36,  31,  30,   7,  17] 
        [  11    6   -47], [ 302, 165, 401, 253,  16, 402],[ 335,  66, 237, 149, 418, 368] 
        [ -12   15   -23], [  19,  27,  35,   3,  34,   4],[   1,  36,  31,   7,  30,  17] 
        [ -12   17    47], [ 152,  58, 131, 161,   2, 105],[ 145,  14, 163,  56, 103, 138] 
        [16/3    9   187], [ 181, 216, 575,  76, 357, 497],[  71, 504, 281, 573, 140, 323] 
        [17/15   18  -67], [ 766,1431,1691,1417, 168, 182],[  53,1638,1249,1599,  14, 872]