1.Introduction



I show that there are many parameter solutions of 6.5.5.




2. Theorem


             There are many parameter solutions of A16+ A26+ A36+ A46+ A56 = B16+ B26+ B36+ B46+ B56.



             A16+ A26+ A36+ A46+ A56 = B16+ B26+ B36+ B46+ B56.................(1)

             Case 1.

             A1=-18a12-8a1kb2-2k2b22
             A2=22a12+10a1kb2+3k2b22
             A3=-6a12+8a1kb2+5k2b22
             A4=-14a12-4a1kb2+4k2b22
             A5=-10a12-12a1kb2+k2b22
             B1=6a12+4a1kb2+4k2b22
             B2=-10a12-6a1kb2-5k2b22
             B3=-14a12+4a1kb2+3k2b22
             B4=-22a12-8a1kb2+2k2b22
             B5=-18a12-16a1kb2-k2b22

             Case 2.

             A1=-8a12-2a1kb2-2k2b22
             A2=-4(-a1+kb2)a1
             A3=10a12+12a1kb2+2k2b22
             A4= 6a12+14a1kb2+4k2b22
             A5=4kb2(2a1+kb2)
             B1=10a12+10a1kb2+4k2b22
             B2=-8a12-12a1kb2-4k2b22
             B3=4a1(a1+2kb2)
             B4=2kb2(5a1+kb2)
             B5=4a1kb2+2k2b22-6a12


Proof.           




             A1=a1a-pc2
             A2=na1a+qc2
             A3=a1a-b2b+c2
             A4=ma1a+c2
             A5=ma1a+b2b+c2  
             B1=a1a+pc2
             B2=na1a-qc2
             B3=a1a-b2b-c2
             B4=ma1a-c2
             B5=ma1a+b2b-c2


      A16+ A26+ A36+ A46+ A56 -( B16+ B26+ B36+ B46+ B56).............................(2)

      As for this equation (2), the factorization is done as follows.
 
      |(m,n,p,q)|<10

      (m,n,p,q)

     (3,-2,3,2)  240a1ac2(5c22+b22b2+2a1ab2b+7a12a2)(-3c22+b22b2+2a1ab2b+3a12a2)
     (3,-1,3,4)  240a1ac2(9c22+b22b2+2b2aa1b+6a12a2)(-7c22+b22b2+2b2aa1b+4a12a2)
     (3,-1,4,3)  240a1ac2(9c22+6a12a2+2b2a1ab+b22b2)(-7c22+4a12a2+2b2a1ab+b22b2)
     (3,1,3,-4)  240a1ac2(9c22+6a2a12+2b2a1ab+b22b2)(-7c22+4a2a12+2b2a1ab+b22b2)
     (3,1,4,-3)  240a1ac2(9c22+6a2a12+2b2a1ab+b22b2)(-7c22+4a2a12+2b2a1ab+b22b2)
     (3,2,3,-2)  240a1ac2(5c22+7a2a12+2a1ab2b+b22b2)(-3c22+3a2a12+2a1ab2b+b22b2)


    Take c2=1
    
    Case 1. 4a2a12+2b2a1ab+b22b2=7.....................................(3)
    

    We can find infinitely many rational solutions of (3),then we obtain infinitely parameter solutions of (2).

    Take x=2a1a,y=b2b 

    Then (3) becomes to x2+xy+y2=7........................................(4)

    (x,y)=(2,1) is a solution of (4).

    We obtain parameter solution of (3) by using (a,b)=(1/a1,1/b2).

                  

             A1=-18a12-8a1kb2-2k2b22
             A2=22a12+10a1kb2+3k2b22
             A3=-6a12+8a1kb2+5k2b22
             A4=-14a12-4a1kb2+4k2b22
             A5=-10a12-12a1kb2+k2b22
             B1=6a12+4a1kb2+4k2b22
             B2=-10a12-6a1kb2-5k2b22
             B3=-14a12+4a1kb2+3k2b22
             B4=-22a12-8a1kb2+2k2b22
             B5=-18a12-16a1kb2-k2b22

   
   Case 2. 3a2a12+2b2a1ab+b22b2=3.........................................(5)
    

    We can find infinitely many rational solutions of (5),then we obtain infinitely parameter solutions of (2).

    Take x=a1a,y=b2b 

    Then (5) becomes to 3x2+2xy+y2=3.........................................(6)

    (x,y)=(1,-2) is a solution of (6).

    We obtain parameter solution of (5) by using (a,b)=(1/a1,-2/b2).
                 

                  

             A1=-8a12-2a1kb2-2k2b22
             A2=-4(-a1+kb2)a1
             A3=10a12+12a1kb2+2k2b22
             A4= 6a12+14a1kb2+4k2b22
             A5=4kb2(2a1+kb2)
             B1=10a12+10a1kb2+4k2b22
             B2=-8a12-12a1kb2-4k2b22
             B3=4a1(a1+2kb2)
             B4=2kb2(5a1+kb2)
             B5=4a1kb2+2k2b22-6a12


   
3. Example


       Case 1: (a1,b2)=(1,1)

            k

            1     46+   56+   16+   26+   36 =    26+   36+   16+   46+   56 
            2     76+   96+   56+   16+   56 =    56+   76+   16+   56+   96
            3    606+  796+  636+  106+  376 =   546+  736+  256+  286+  756
            4    416+  556+  536+  176+  216 =   436+  576+  256+  116+  496
            5    366+  496+  536+  226+  156 =   426+  556+  276+   46+  416
            6    696+  956+ 1116+  536+  236 =   876+ 1136+  596+   16+  756
            7   1726+ 2396+ 2956+ 1546+  456 =  2306+ 2976+ 1616+  206+ 1796
            8     56+   76+   96+   56+   16 =    76+   96+   56+   16+   56
            9   2526+ 3556+ 4716+ 2746+  376 =  3666+ 4696+ 2656+  686+ 2436
           10   1496+ 2116+ 2876+ 1736+  156 =  2236+ 2856+ 1636+  496+ 1396




       Case 2: (a1,b2)=(1,1)

            k
 
            1    16+  06+  26+  26+  16 =   26+  26+  16+  16+  06   
            2   106+  26+ 216+ 256+ 166 =  236+ 246+ 106+ 146+  56
            3    86+  26+ 166+ 216+ 156 =  196+ 206+  76+ 126+  66
            4    86+  26+ 156+ 216+ 166 =  196+ 206+  66+ 126+  76
            5   176+  46+ 306+ 446+ 356 =  406+ 426+ 116+ 256+ 166
            6   466+ 106+ 776+1176+ 966 = 1076+1126+ 266+ 666+ 456
            7   106+  26+ 166+ 256+ 216 =  236+ 246+  56+ 146+ 106
            8   766+ 146+1176+1876+1606 = 1736+1806+ 346+1046+ 776
            9   476+  86+ 706+1146+ 996 = 1066+1106+ 196+ 636+ 486
           10   386+  66+ 556+ 916+ 806 =  856+ 886+ 146+ 506+ 396


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