1. Introduction

      I show the parametric solution of diophantine system {X^2 + Y^2 + Z^2 + W^2 + S^2 = U^2, X^4 + Y^4 + Z^4 + W^4 + S^4 = V^4}.



2.Theorem
            
　　　  The parametric solution of X^k + Y^k + Z^k + W^k + S^k = {U^2,V^4} for k=2,4,

　　  　X = 24x^2 + 24yx - 2y^2
        Y = 24x^2 - 24yx - 2y^2
        Z = 48x^2 - 4y^2
        W = 48x^2 + 4y^2
        S = 36x^2 + 3y^2
        U = 7(12x^2 + y^2)
 　     V = 5(12x^2 + y^2)

      

Proof.

Case 1. X^4 + Y^4 + Z^4 + W^4 + S^4 = V^4
  
(a1x^2+b1x-c1)^4+(a1x^2-b1x-c1)^4+(2a1x^2-2c1)^4+(2a1x^2+2c1)^4+(a2x^2+c2)^4-((a1+a2)x^2+(c1+c2))^4......(1)
 

Expanding and simplifying above equation,
　　
(-4a1a2^3-4a1^3a2+33a1^4-6a1^2a2^2)x^8
+(-4a1^3c2-12a1^2a2c2-12a1a2^2c1-12a1^2a2c1-12a1a2^2c2-4a2^3c1+12a1^2b1^2-12a1^3c1)x^6
+(2b1^4-12a1a2c1^2-12a1a2c2^2-6a2^2c1^2-12a1^2c1c2-24a1b1^2c1+198a1^2c1^2-12a2^2c1c2-24a1a2c1c2-6a1^2c2^2)x^4
+(-12a1c1c2^2-12a2c1c2^2-12a1c1^3+12b1^2c1^2-12a2c1^2c2-4a2c1^3-12a1c1^2c2-4a1c2^3)x^2
-6c1^2c2^2-4c1c2^3+33c1^4-4c1^3c2

First of all, let a2=3a1/2 then coefficient of x^8=0.

Similarly, let c2=3c1/2 then coefficient of x^0=0.

Next, let c1=1/12b1^2/a1 then coefficient of x^6=0.

The coefficient of x^4 and x^2 is also similar to x^6. 

Substitute {a2=3a1/2, c2=3c1/2, c1=1/12b1^2/a1} to (1), and change variables a1x to x and b1 to y, then obtain a parametric solution.




Case 2. X^2 + Y^2 + Z^2 + W^2 + S^2 = U^2

Automatically, we obtain a parametric solution.



Q.E.D.



3.Example 

[x,y]
                  
[1, 1], 46^2 +  2^2 + 44^2 + 52^2 + 39^2 = 91^2
[1, 1], 46^4 +  2^4 + 44^4 + 52^4 + 39^4 = 65^4

[1, 2], 4^2  +  2^2 +  2^2 +  4^2 +  3^2 = 7^2
[1, 2], 4^4  +  2^4 +  2^4 +  4^4 +  3^4 = 5^4

[1, 3], 26^2 + 22^2 +  4^2 + 28^2 + 21^2 = 49^2
[1, 3], 26^4 + 22^4 +  4^4 + 28^4 + 21^4 = 35^4