Math contest integer triplet problem: $x + y + z + xy + yz + xz = xyz − 1$

Question

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.

Answer 1

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

$$(2,4,15), (2,5,9), (2,6,7), (3,3,8),\text{ and } (3,4,5)$$

Answer 2

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.

Answer 3

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.

Answer 4

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.