x^3 + y^3 + z^3 - 3xyz = 1


We show diophantine equation x^3 + y^3 + z^3 - 3xyz = 1 has a parametric solution.


x^3 + y^3 + z^3 - 3xyz = 1.....................................................(1)

Let x=pt+1, y=qt, z=t, then equation (1) reduces to

(p^3+q^3+1-3qp)t^2+(3p^2-3q)t+3p = 0...........................................(2)

Let t = 3q/(1+q^3).

Then we obtain a parametric solution of equation (1) below.

(x,y,z)=(1, 3q^2/(1+q^3), 3q/(1+q^3)).

q is arbitrary.