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

kiemkhach · P versus NP Offline Joined: 16 Feb 2005 Posts: 25

find integer numbers $x;y;z$ satisfies $x^3+y^3+z^3=3xyz+1$

**Posted:** Fri Feb 18, 2005 2:42 am

darij grinberg

Just a tip: Your equation $x^3+y^3+z^3=3xyz+1$ is equivalent to $x^3+y^3+z^3-3xyz=1$ . But since $x^3+y^3+z^3-3xyz=\frac12\left(x+y+z\right)\left(\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\right)$ , this transforms into $\frac12\left(x+y+z\right)\left(\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\right)=1$ . And noting that the number $\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2$ is always $\geq 2$ unless all three numbers x, y, z are equal, and equals 2 only in the case when two of the numbers x, y and z are equal and the third one differs from them by 1, you can easily solve this equation.

The answer: The solutions of your equation are

x = 1, y = 0, z = 0;
x = 0, y = 1, z = 0;
x = 0, y = 0, z = 1.

darij

**Posted:** Fri Feb 18, 2005 3:09 am