Chinese Logic... tough nut to crack :( #1

Not so elegant of a solution, but I posted anyways.

Firstly, it can be checked that all variables must be non-zero. Then, we should get the signs right. the possibilities are $(+,+,+)$ or $(-,-,+)$ or $(-,+,+)$ for $(x,y,z)$ . $(-,-,-)$ cannot be cause the signs of RHS and LHS would not match.

Note that $x=2$ or $x=-2$ , otherwise if $x=pk$ , where $p$ is a prime greeter than $2$ and $k$ is an integer, the LHS would have $p^3$ but RHS would have at most $p$ (in case $p=503$ ). therefore the equation simplifies to

$(y^3+z^3)=503(yz+1)$

if $x=2$

or

$(y^3+z^3)=503(yz-1)$

if $x=-2$

In both cases above, if $y,z$ are positive, $z< 503$ . The reason is given for $(y^3+z^3)=503(yz-1)$ . Given $z$ , $z^3+1$ is the minimum of $z^3+y^3$ , when the variable is $y$ . Given $z$ $503((z-1)z-1)=503(z^2-z-1)$ maximises $503(yz-1)$ , when $y$ is the variable. So, if we solve $z^3+1=503(z^2-z-1)$ for $z$ we get the positive solution $z\approx 503$ . if $z > 503$ then the sides of $(y^3+z^3)=503(yz-1)$ would definitely be different. So we can check only for $z,y<503$ . also, $-2 \leq y,z$ (cause we know $x$ now)

I did the rest using a simple program to check all the possibilities. $z=252$ and $y=251$ and $x=2$