How to find all integer solutions (that is, $x,y,z \in Z$) to equation:
$$(x + y)(y + z)(x + z) = 13xyz$$
It is known that there are zero-value solutions, but how to find non-zero ones? Could anyone give some tips?
More updates:
It is hard to find all non-zero integer resolutions, however, if we assume: $$\gcd(x,y)=\gcd(x,z)=\gcd(y,z)=1$$
it is very simple.
$ 13xyz = (x + y)(y + z)(x + z) = (x + y)(xy + xz + yz + z^2) = (x + y)[(xy + xz + yz) + z^2] = (x + y)(xy + xz + yz) + (x + y)z^2 = (x + y)(xy + xz + yz) + z(xz + yz) = (x + y)(xy + xz + yz) + z(xy +xz+yz) - xyz = (x + y + z)(xy + xz + yz) - xyz \iff (x + y + z)(xy + xz + yz) = 14xyz \implies xyz | (x + y + z)(xy + xz + yz) \\ Then \ (x,y)=(x,z)=1 \implies (x,yz)=1 \implies (x,xy + xz + yz)=1 \ Similarly, \ (y,xy + xz + yz) = (z,xy + xz + yz) = 1 \implies (xyz,xy + xz + yz) = 1 \ but \ xyz | (x + y + z)(xy + xz + yz), \ then \ xyz |(x + y + z) \implies |xyz| \le |x + y + z| \ assume \ |x| \le |y| \le |z|, \ then |x|.|y|.|z| \le |x + y + z| \le 3|z| \implies |x|.|y| \le 3$
So ${x, y} \in \{\pm 1, \pm 2, \pm 3\} $ then we can enumerate the solutions.
This is an 1995 Romanian olypiad training exercise, see here, number $10$: $$ (x+y)(y+z)(z+x)=txyz. $$ It has solutions available (search for a link!). To exclude infinitely many "trivial" solutions (and others), one assumes there that $(x,y)=(x,z)=(y,z)=1$. Here we do not need to assume this. It makes sense, though, to assume that the solution is primitive, i.e., that $gcd(x,y,z)=1$. A primitive solution, for $t=13$ is for example $(x,y,z)=(3,2,10)$, see Jyrki's comment. It produces infinitely many solutions (non-primitive) $(2k,3k,10k)$ for all $k$.
Edit: Without pairwise coprime $x,y,z$ it is nontrivial to find all solution. We need to use techniques from algebraic number theory for such Diophantine equations, e.g. for plane algebraic curves and rational points on it.