Math contest integer triplet problem: $x + y + z + xy + yz + xz = xyz − 1$ 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.
 A: 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
$$(2,4,15), (2,5,9), (2,6,7), (3,3,8),\text{ and } (3,4,5)$$
A: Presumably there is a clever way, since this is a contest problem. The following approach is unpleasantly unclever, and the details are somewhat tedious.  
Note first that $x$ cannot be $1$, for if it is the left-hand side is bigger than the right-hand side.
Note also that $x\lt 4$. Suppose to the contrary that all our numbers are $\ge 4$. Divide both sides by $xyz$. Then the terms on the left have sum $\le \frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{15}{16}$, while the right is $\ge 1-\frac{1}{64}$.
So $x=2$ or $x=3$. We end up with two accessible problems in $2$ variables. 
A: You can rewrite the equation as $2(x+y+z)=(x-1)(y-1)(z-1)$.  Since the right side is cubic and the right is quadratic, the variables must be small or the right will be too big.  Let us set $u=x-1,v=y-1,w=z-1$ for convenience, making the equation $2(u+v+w)+6=uvw$  Starting from small $u$, if $u=1$ we get $2(v+w)+8=uv, (u-2)(v-2)=12$, leading to solutions $(2,4,15),(2,5,9),(2,6,7)$.  If $u=2$ we get $2(v+w)=2vw, 24=(2v-2)(2w-2)$ leading to $(3,3,8),(3,4,5)$.  As André Nicolas has shown, that is it.
A: Your formula leads to $(x+1)(y+1)(z+1) = 2xyz$.  Note none of $x,y,z$ equal $1$. Also if all $x,y,z$ are greater than or equal to $4$, then $(x+1)(y+1)(z+1) \leq (125/64)xyz < 2xyz$, a contradiction.  Similarly assuming $x = 2$ or $x = 3$, then you cannot have $y$ and $z$ both greater than or equal to $7$, because otherwise $(x+1)(y+1)(z+1) \leq (3/2)x(y+1)(z+1) \leq (3/2)(64/49)xyz < 2xyz$.  So in the worst case, you only need to check a small number of possible values of $x,y$ to see if there is integer solution for $z$.  If you assume $x = 2,3$ and $2 \leq y \leq 6$, the remaining solutions follow by symmetry.
