1.Introduction

I proved that there was always a parameter solution of A15+A25+A35+...+An5=B15+B25+B35+...+Bn5.



2.Theorem
      
     
    
    n:integer

　　If n > 2, there is always a parameter solution of A15+A25+A35+...+An5=B15+B25+B35+...+Bn5.

     
Proof.

a1,a2,...,an:integer
p,q,k,r:integer
k:numbers of p
r:numbers of q
Let n=k+r+1,k=[n/2],r=n-k-1

Case n=even number
(a1+px)5+(a2+px)5+...+(ak+px)5+(ak+1+qx)5+(ak+2+qx)5+...+(ak+r+qx)5+(an+x)5
=(ak+1+px)5+(ak+2+px)5+...(ak+r+px)5+(an+px)5+(a1+qx)5+(a2+qx)5+...+(ak+x)5

Case n=odd number
(a1+px)5+(a2+px)5+...+(ak+px)5+(ak+1+qx)5+(ak+2+qx)5+...+(ak+r+qx)5+(an+x)5
=(ak+2+px)5+(ak+3+px)5+...(ak+r+px)5+(an+px)5+(a1+qx)5+(a2+qx)5+...+(ak+1+x)5........(0)


Case n=3
     k=[3/2]=1,r=1
     (a1+px)5+(a2+qx)5+(a3+x)5=(a3+px)5+(a1+qx)5+(a2+x)5

Case n=4
     k=[4/2]=2,r=1
     (a1+px)5+(a2+px)5+(a3+qx)5+(a4+x)5=(a3+px)5+(a4+px)5+(a1+qx)5+(a2+x)5

Case n=5
     k=[5/2]=2,r=2
     (a1+px)5+(a2+px)5+(a3+qx)5+(a4+qx)5+(a5+x)5=(a4+px)5+(a5+px)5+(a1+qx)5+(a2+qx)5+(a3+x)5



Make it the same.
Expanding and simplifying (0),we obtain
f=f4x4+f3x3+f2x2+f1x
fi:coefficient of xi


First,solve the simultaneous equation of {f1=0,f2=0} for (p,q).
We know easily that f1(p,q) and f2(p,q) = 0 at (p,q)=(1,1).
So,first root of above simultaneous equation is (p,q)=(1,1).
Cause another root must be rational number,then we always obtain a rational solution (p,q).

     p=g1(a1,...,an)
     q=g2(a1,...,an)..............................(1)

Next,solve f4x4+f3x3=0 for x,and obtain
     x=-f3/f4....................................(2)  
            
Substitute (1) to (2).
Substitute p,q,x to (0),and obtain a parameter solution.
       
I omit each parameter solution because there is much number of the items.

   
Q.E.D.　
 
　　                  
       
3.Example

I show a few numerical examples.
a1,a2,a3,a4,a5<=3


    
Case n=4

(a1,a2,a3,a4)


(1 -2 -3 0):    46375+    34975+   25235+   12875=   31175+   42575+   40435+    5275
                          
(1 -1 -3 2):    99895+   168275+  150995+   50135=   34195+  166935+  121795+  146375

(1 0 -3 -2):    40095+    41995+   52115+   23695=   47695+   45795+   44515+   19895

(1 0 3 2):      46675+    82135+  345135+  309675=  371875+  175975+   19935+  215835



Case n=5

(a1,a2,a3,a4,a5)

(1 -3 2 -1 -1):  1883175+  3801475+  688755+  688755+ 2628335=
                   56415+  3428895+ 2515515+ 3260675+  428995

(2 -3 0 -1 3):   1290535+   280135+  294375+ 1304775+  992035=
                  684295+  1492615+  698535+  900615+  385795

(2 -1 1 0 3):     682715+   412415+  430515+  340415+  740395= 
                  502515+   772815+  520615+  250315+  560195

(2 0 1 -2 -2):    331375+   610575+  960235+  960235+   31675=
                  358135+  1000375+  933475+  287675+  314435

(3 -2 2 -1 1):    462075+   259875+  503415+  382095+   14535=
                  300315+   381195+  543855+  341655+   54975  

(3 2 0 -1 -2):   1048935+   542035+ 2068535+  471775+  978675=
                 1486735+  1993635+  540875+   33975+ 1054735