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


We show diophantine equation x^3 + y^3 + z^3 - 3xyz = 1 has infinitely many parametric solutions.


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

Let z=s-x-y, equation (1) reduces to

3sx^2+(3sy-3s^2)x+s^3+3sy^2-3s^2y-1=0.

Since this quadratic in x must have rational solutions, the discriminant must be square number.

v^2 = -3s^4+18ys^3-27y^2s^2+12s...................................................(2)

Let y = -2(-1+n)/(3+n^2), we know Q(s,v)=(1,3(n+1)(n-3)/((3+n^2) is a solution of equation (2). 
n is arbitrary.

Hence this quartic equation (2) is birationally equivalent to an elliptic curve below.

Y^2-36(n^2-1)YX/((3+n^2)(n-3))-72n(-9n+n^2+n^3-9)Y/((3+n^2)^2)
= X^3-18(n^4-6n^2+24n-3)X^2/((3+n^2)(n-3)^2)+108(n^4-4n^3-2n^2+12n+9)X/((3+n^2)^2)-1944(n^6+2n^5-5n^4+12n^3+39n^2+18n-3)/((3+n^2)^3)

The corresponding point is P(X,Y)=( 18(n^4-6n^2+24n-3)/((3+n^2)(n-3)^2), 72(n^3+n^2+3n+3)/((n-3)^3) ).

Using P(X,Y) we obtain a parametric solution of equation (1) below.

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

Using 2P(X,Y) we obtain other parametric solution of equation (1) below.

(x,y,z)=(1/6(n+1)(n^5-7n^4+34n^3-78n^2+93n-75)/((-1+n)(n^2-4n+7)(3+n^2)),
        -2(-1+n)/(3+n^2),
        -1/6(n+1)(n^5-7n^4+22n^3-66n^2+105n-87)/((-1+n)(n^2-4n+7)(3+n^2)) ).

This point P is of infinite order, and the multiples kP, k = 2, 3, ...give infinitely many points.

Thus we can obtain infinitely many parametric solutions for equation (1).