1.Introduction


I will show that x1^k+x2^k+...+x12^k = y1^k+y2^k+...y12^k   
for k=1,3,5,7,9,11 has infinetly many integer solutions.

I used following Theorem and Wroblewski-Piezas identity.

Wroblewski-Piezas identity.([1])

(8b)^k + (-5a-4b)^k + (5a-4b)^k +  (-a-2d)^k + (a-2d)^k + (-12b+4c)^k + (12b+4c)^k  =
(-16b)^k + (-4a+8b)^k + (4a+8b)^k + (-a+4c)^k + (a+4c)^k + (-3a-2d)^k + (3a-2d)^k  
,for 1,2,4,6,8,10,where a^2+b^2 = c^2 and a^2+52b^2 = d^2.


Theorem

If a1^k+a2^k+....+am^k = b1^k+b2^k+....+bm^k,for k=2,4,...2n,then

(T+a1)^k+(T+a2)^k+...+(T+am)^k,(T-a1)^k+(T-a2)^k+...+(T-am)^k =

(T+b1)^k+(T+b2)^k+...+(T+bm)^k,(T-b1)^k+(T-b2)^k+...+(T-bm)^k,
for k=1,2,3,...2n+1,where T is arbitrary integer.









2. Theorem

  x1^k + x2^k +...+ x12^k = y1^k + y2^k +...y12^k,for k=1,3,5,7,9,11
  has infinetly many solutions where a^2+b^2 = c^2 and a^2+52b^2 = d^2.
         
     x1=  a+8b
     x2= -4a-4b
     x3=  6a-4b
     x4=  2a-2d
     x5=  a-12b+4c
     x6=  a+12b+4c
     x7=  a-8b
     x8=  6a+4b
     x9= -4a+4b
     x10= 2a+2d
     x11= a+12b-4c
     x12= a-12b-4c


     y1=   a-16b
     y2=  -3a+8b
     y3=   5a+8b
     y4=   2a+4c
     y5=  -2a-2d
     y6=   4a-2d
     y7=   a+16b
     y8=   5a-8b
     y9=  -3a-8b
     y10=  2a-4c
     y11=  4a+2d
     y12= -2a+2d
     
Proof.

x1^k+x2^k+...+x12^k = y1^k+y2^k+...y12^k............................................(1)

Let [a1,a2,...,am]=[b1,b2,...,bm](k=1,2,...,n) denote a1^k+a2^k+....+am^k = b1^k+b2^k+....+bm^k(k=1,2,...,n).

First we apply Wroblewski-Piezas identity to above Theorem,we obtain 


[t+8b, t-5a-4b, t+5a-4b, t-a-2d, t+a-2d, t-12b+4c, t+12b+4c, t-8b, t+5a+4b, t-5a+4b, t+a+2d, t-a+2d, t+12b-4c, t-12b-4c]^k =

[t-16b, t-4a+8b, t+4a+8b, t-a+4c, t+a+4c, t-3a-2d, t+3a-2d, t+16b, t+4a-8b, t-4a-8b, t+a-4c, t-a-4c, t+3a+2d, t-3a+2d]^k,
for k=1,3,5,7,9,11,where a^2+b^2 = c^2 and a^2+52b^2 = d^2.

Here,set T=a,we can reduce four terms and obtain


[a+8b, -4a-4b, 6a-4b, 2a-2d, a-12b+4c, a+12b+4c, a-8b, 6a+4b, -4a+4b, 2a+2d, a+12b-4c, a-12b-4c]^k =

[a-16b, -3a+8b, 5a+8b, 2a+4c, -2a-2d, 4a-2d, a+16b, 5a-8b, -3a-8b, 2a-4c, 4a+2d, -2a+2d]^k,for k=1,3,5,7,9,11,

where a^2+b^2 = c^2 and a^2+52b^2 = d^2..............................................(2)

Since Wroblewski-Piezas identity has infinetly many solutions,(2) also has infinetly many solutions.




Q.E.D.　
 
　　                  
       
3.Small Solutions

[a,   b,  c,  d]            [ x1 ,    x2 ,  x3  ,  x4  , x5  ,  x6  , x7 , x8 ,  x9,    x10,  x11,   x12 ]
                            [ y1 ,    y2 ,  y3  ,  y4  , y5  ,  y6  , y7 , y8 ,  y9,    y10,  y11,   y12 ]

[3,   4,  5, 29]            [ 35,    -28,    2,  -52,   -25,   71,  -29,   34,     4,    64,   31,    -65]
                            [-61,     23,   47,   26,   -64,  -46,   67,  -17,   -41,   -14,   70,     52]

[783,56,785,881]            [1231, -3356, 4474, -196,  3251, 4595,  335, 4922, -2908,  3328, -1685, -3029]
                            [-113, -1901, 4363, 4706, -3328, 1370, 1679, 3467, -2797, -1574,  4894,   196]


 






4.References

[1].Tito Piezas:http://sites.google.com/site/tpiezas/030