1.Introduction

We show a parametric solution for x1^n + x2^n + x3^n = y1^3+ + y2^3 + y3^3.

2.Theorem
 
There is a parametric solution of x1^n + x2^n + x3^n = y1^3+ + y2^3 + y3^3,

      x1 = a^(m/n)(2c^(2m)-a^(2m)-3a^(4/3m)b^(2/3m)-3a^(2/3m)b^(4/3m)-b^(2m))^(m/n)
      x2 = b^(m/n)(2c^(2m)-a^(2m)-3a^(4/3m)b^(2/3m)-3a^(2/3m)b^(4/3m)-b^(2m))^(m/n)
      x3 = c^(m/n)(2c^(2m)-a^(2m)-3a^(4/3m)b^(2/3m)-3a^(2/3m)b^(4/3m)-b^(2m))^(m/n)
      y1 = -(c^(2m)a^(1/3m)+3c^(2m)b^(1/3m)+3c^ma^(4/3m)+6c^ma^(2/3m)b^(2/3m)+3c^mb^(4/3m)+a^(1/3m)a^(2m)+3a^(1/3m)a^(4/3m)b^(2/3m)+3a^(1/3m)a^(2/3m)b^(4/3m)+a^(1/3m)b^(2m))(2c^(2m)-a^(2m)-3a^(4/3m)b^(2/3m)-3a^(2/3m)b^(4/3m)-b^(2m))^(1/3m-1)
      y2 = -(3c^(2m)a^(1/3m)+c^(2m)b^(1/3m)+3c^ma^(4/3m)+6c^ma^(2/3m)b^(2/3m)+3c^mb^(4/3m)+b^(1/3m)a^(2m)+3b^(1/3m)a^(4/3m)b^(2/3m)+3b^(1/3m)a^(2/3m)b^(4/3m)+b^(1/3m)b^(2m))(2c^(2m)-a^(2m)-3a^(4/3m)b^(2/3m)-3a^(2/3m)b^(4/3m)-b^(2m))^(1/3m-1)
      y3 = (2c^(2m)-a^(2m)-3a^(4/3m)b^(2/3m)-3a^(2/3m)b^(4/3m)-b^(2m))^(1/3m-1)(3a^mc^(4/3m)+3c^(4/3m)a^(2/3m)b^(1/3m)+2c^(1/3m)a^(2m)+6c^(1/3m)a^(4/3m)b^(2/3m)+6c^(1/3m)a^(2/3m)b^(4/3m)+3c^(4/3m)b^(2/3m)a^(1/3m)+3c^(4/3m)b^m+2c^(1/3m)b^(2m)+2c^(7/3m))
    
 condition: 
    
     m=lcm(3,n)
     a, b, c, n: arbitrary.

     
Proof.

x1^n + x2^n + x3^n = y1^3+ + y2^3 + y3^3......................................................(1)
Let m = lcm(3,n).
Set x1=a^(m/n), x2=b^(m/n), x3=c^(m/n), y1=t+a^(m/3), y2=t+b^(m/3), y3=pt+c^(m/3) ............(2)

Then we obtain p =-((a^(1/3m))^2+(b^(1/3m))^2)/((c^(1/3m))^2), t = -3(c^(2m)a^(1/3m)+c^(2m)b^(1/3m)+c^ma^(4/3m)+2c^ma^(2/3m)b^(2/3m)+c^mb^(4/3m))/(2c^(2m)-a^(2m)-3a^(4/3m)b^(2/3m)-3a^(2/3m)b^(4/3m)-b^(2m)).
Substitute p and t to (2), and we obtain a solution.
 
Q.E.D.




HOME