TITLE: Math contest integer triplet problem: $x + y + z + xy + yz + xz = xyz − 1$
QUESTION [8 upvotes]: Can any one help me with this?
Determine all integer triples $(x,y,z)$ such that $1 ≤ x ≤ y ≤ z$ and
$x + y + z + xy + yz + xz = xyz − 1$.
I thought of Vieta's formula but don't let me lead you into a dead end.

REPLY [7 votes]: The equation can be rewritten as
$$(1+x)(1+y)(1+z)=2xyz$$
and this rewrites as
$$\left(1+{1\over x}\right)\left(1+{1\over y}\right)\left(1+{1\over z}\right)=2$$
This already shows that all three numbers are greater than $1$, since otherwise the product on the left would be greater than $2$.  Since we're assuming $x$ is the smallest of the three positive integers, we must have
$$1+{1\over x} \ge 2^{1/3}\approx 1.2599$$
hence $x=2$ or $3$.  If we let $x=2$, the same idea shows that
$${4\over3}\gt 1+{1\over y} \ge \sqrt{4\over3}\approx1.1547$$
which implies $4\le y\le6$, and if we let $x=3$, it gives
$${3\over2}\gt1+{1\over y} \ge \sqrt{3\over2}\approx1.2247$$
which implies $3\le y \le 4$.  Out of all this, we get the triples
$$(2,4,15), (2,5,9), (2,6,7), (3,3,8),\text{ and } (3,4,5)$$