1.Introduction

We show a parametric solution of a(1)x1^n+a(2)x2^n+...+a(n)xn^n=0.
This problem is reduced to solving the linear simultaneous equation, so we can apply this method for any n>=3.


2.Method
        

a(1)x1^n + a(2)x2^n +...+ a(n)xn^n =0..............................(1)

Let x1=b(1)t+1, x2=b(2)t+1,..., xn=b(n)t+1.........................(2)
n and b(n) is arbitrary.

Substitute (2) to (1), we obtain the equation (3).

c(n)t^(n) + c(n-1)t^(n-1) + .... + c(0)=0..........................(3)

Find the solution of simultaneous equation {c(0)=0,c(1)=0,...,c(n-2)=0} for {a(2),...,a(n)}.

Hence we obtain a parametric solution of (1).

     
3.Results


Case 1: ax^4 + by^4 + cz^4 + dw^4=0

Let x=pt+1, y=qt+1, z=rt+1, w=st+1.

[a,b,c,d] = [(-r+q)(s-q)(s-r), (-r+p)(p-s)(s-r), -(-q+p)(p-s)(s-q), -(-r+q)(-r+p)(-q+p)]

[x,y,z,w] = [-3p+q+s+r, -3q+p+s+r, -3r+p+q+s, -3s+p+q+r]

p, q, r, and s are arbitrary.

Case 2: ax^5 + by^5 + cz^5 + dw^5 + et^5=0

Let x=pu+1, y=qu+1, z=ru+1, w=su+1, t=vu+1.

[a,b,c,d,e] = [-(q-r)(-v+q)(q-s)(-s+r)(-v+r)(-v+s), (-v+s)(-v+r)(-s+r)(p-r)(p-s)(-v+p), -(-v+s)(q-s)(-v+q)(p-q)(p-s)(-v+p), (-v+p)(p-r)(p-q)(q-r)(-v+q)(-v+r), -(p-s)(p-r)(p-q)(q-r)(q-s)(-s+r)]

[x,y,z,w,t] = [-4p+r+v+q+s, -4q+p+r+v+s, -4r+p+v+q+s, -4s+p+r+v+q, -4v+p+r+q+s]

p, q, r, s, and v are arbitrary.