1.Introduction


I will show that 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=1,3,5,7 has infinetly many integer solutions.

I used following Piezas's identity and Theorem.

Piezas's identity.([1])

(c+bp+aq)^k + (c-bp-aq)^k + (d+ap-bq)^k + (d-ap+bq)^k  = (c+bp-aq)^k + (c-bp+aq)^k + (d+ap+bq)^k + (d-ap-bq)^k,
for k=2,4,6, where (4q^2-p^2)b^2+(4p^2-q^2)a^2=15c^2 and (4p^2-q^2)b^2+(4q^2-p^2)a^2=15d^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+x3^k+x4^k+x5^k+x6^k = y1^k+y2^k+y3^k+y4^k+y5^k+y6^k
  for k=1,3,5,7 has infinetly many solutions,
  where (4q^2-p^2)b^2+(4p^2-q^2)a^2 = 15c^2 and (4p^2-q^2)b^2+(4q^2-p^2)a^2 = 15d^2.
  
        x1=-c+d+ap-bq       y1=-c+d+ap+bq
        x2=-c+d-ap+bq       y2=-c+d-ap-bq
        x3=-2c-bp-aq        y3=-2c-bp+aq
        x4=-2c+bp+aq        y4=-2c+bp-aq
        x5=-c-d-ap+bq       y5=-c-d-ap-bq
        x6=-c-d+ap-bq       y6=-c-d+ap+bq

     
     
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................(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 Piezas's identity to above Theorem, we obtain 


[T+c+bp+aq, T+c-bp-aq, T+d+ap-bq, T+d-ap+bq, T-c-bp-aq, T-c+bp+aq, T-d-ap+bq, T-d+ap-bq]^k =

[T+c+bp-aq, T+c-bp+aq, T+d+ap+bq, T+d-ap-bq, T-c-bp+aq, T-c+bp-aq, T-d-ap-bq, T-d+ap+bq]^k,for k=1,3,5,7,
where (4q^2-p^2)b^2+(4p^2-q^2)a^2=15c^2 and (4p^2-q^2)b^2+(4q^2-p^2)a^2=15d^2.

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


[-c+d+ap-bq, -c+d-ap+bq, -2c-bp-aq, -2c+bp+aq, -c-d-ap+bq, -c-d+ap-bq]^k =

[-c+d+ap+bq, -c+d-ap-bq, -2c-bp+aq, -2c+bp-aq, -c-d-ap-bq, -c-d+ap+bq]^k, for k=1,3,5,7,

where (4q^2-p^2)b^2+(4p^2-q^2)a^2=15c^2 and (4p^2-q^2)b^2+(4q^2-p^2)a^2=15d^2.

By transform simultaneous equation {(4q^2-p^2)b^2+(4p^2-q^2)a^2=15c^2 , (4p^2-q^2)b^2+(4q^2-p^2)a^2=15d^2} to an elliptic curve,
we can prove the infinity of the solutions of this equation.

Since Piezas's identity has infinetly many integer solutions,
this identity also has infinetly many integer solutions.


Q.E.D.　
 
　　                  
       
3.Examples


 (p,q)=(3,2)

    [-c+d+3a-2b, -c+d-3a+2b, -2c-3b-2a, -2c+3b+2a, -c-d-3a+2b, -c-d+3a-2b]^k =

    [-c+d+3a+2b, -c+d-3a-2b, -2c-3b+2a, -2c+3b-2a, -c-d-3a-2b, -c-d+3a+2b]^k

    for k=1,3,5,7, where 7b^2+32a^2=15c^2 and  32b^2+7a^2=15d^2.

    Numeric example,
    {a,b,c,d}={1,13,9,19}: [ -13, 33, -59, 23, -5, -51]^k = [ 39, -19, -55, 19, -57, 1]^k
    {a,b,c,d}={466,607, 797,942}: [ 329, -39,-4347,1159,-1923,-1555]^k = [2757,-2467,-2483,-705,-4351,873]^k
    {a,b,c,d}={607,466, 942,797}: [ 372, -517,-2248, 364,-1314, -425]^k = [ 1304,-1449,-1034, -850,-2246, 507]^k 



 (p,q)=(4,1)

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

   [-c+d+4a+b, -c+d-4a-b, -2c-4b+a, -2c+4b-a, -c-d-4a-b, -c-d+4a+b]^k

   for k=1,3,5,7, where -12b^2+63a^2=15c^2 and  63b^2-12a^2=15d^2.

   {a,b,c,d}={89,82,167,148}: [255, -293, -751, 83, -589, -41]^k = [ 419, -457, -573, -95, -753, 123]^k
   {a,b,c,d}={82,89,148,167}: [129, -110, -367, 71, -277, -38]^k = [ 218, -199, -285, -11, -366, 51]^k






4.References

[1].Tito Piezas:http://sites.google.com/site/tpiezas/025