Prove that the equation $ xyz=xy+xz+yz+x+y+z$ has finite numbers of natural solutions.

Question

Prove that the equation $$\begin{equation} xyz=xy+xz+yz+x+y+z\tag{*} \end{equation}$$ has finitely many natural solutions.

My attempt. Suppose that $x \leq y \leq z$ then $$ xyz\leq xz+xz+yz+3z \implies xy \leq 2x+y+3 \implies x \leq \frac{y+3}{y-2} $$ Since $$x \leq \frac{y+3}{y-2}=\frac{y-2+5}{y-2}=1+\frac{5}{y-2} \implies (x-1)(y-2) \leq 5,$$ we can conclude that the last inequality has finite number of natural solutions, so there are only finite numbers for $x,y.$ For any fixed $x,y$ the equation (*) reduces to linear equation for $z$ with finite numbers of solutions.

Is it ok? Is there any other solutions?

Answer 1

If $xyz = xy + xz + yz + x + y + z$ is a natural solution.

$(xy - x - y - 1)z = xy + x + y > 0$

$z = \frac {xy + x + y}{xy - x - y - 1}$

$= 1 + \frac {2x + 2y + 1}{xy - x-y -1}\in \mathbb N$ so

$xy - x - y -1\le 2x + 2y + 1$

$xy \le 3x + 3y + 2$

Likewise $xz \le 3x + 3z + 2$

And $yz \le 3y + 3z + 2$.

If we relabel $c = \max(x,y,z); a = \min(x,y,z)$ and $b$ is the median of $x,y,z$.

then

$ab \le 3a + 3b + 2 \le 6b + 2$ so $a \le 6 + \frac 2b$ has only finitely many possible values.

And $bc \le 3b + 3c +2 \le 6c + 2$ so $b \le 6 + \frac 2c$ has only finitely many possible values and

$c =\frac {ab + a + b}{ab - a - b - 1}$ has only finitely many possible values.

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We can work our way through these: $a,b < 6$ and

If $a = 1$ then $c = \frac {2b + 1}{-2}$ which is not possible.

$a = 2$ then $c = \frac {3b + 2}{b-3}$ if $b = 4$ then $c = 14$

If $b = 5$ we have $\frac {odd}{even}$ which is not an integer .

If $a=3$ then $c = {4b +3}{2b -4}$ which is an impossible $\frac {odd}{even}$

If $a = 4$ then $c = {5b + 4}{3b -5}$. If $b = 4;c={24}{7}$. No good. $b=5$ yields $\frac {odd}{even}$.

If $a =5; b=5$ then $c = \frac {29}{10}$. No go.

So, unless I made an error $\{x,y,z\} = \{2,4,14\}$ are the only solutions.

Answer 2

Add $xyz+1$ to both sides to obtain $$2xyz+1=xyz+xy+yz+zx+x+y+z+1=(x+1)(y+1)(z+1)$$

The left-hand side is odd, so the right-hand side must also be odd, whence $x,y,z$ are all even.

Now divide by $xyz$ to obtain $$\left(1+\frac 1x\right)\left(1+\frac 1y\right)\left(1+\frac 1z\right)=2+\frac 1{xyz}\gt 2$$

Now note that $\left(\frac 54\right)^3=\frac {125}{64}\lt 2$, so that one of $x,y,z$ at least must be equal to $2$ - say $x$.

We then have $$\frac 32\left(1+\frac 1y\right)\left(1+\frac 1z\right)=2+\frac 1{2yz}$$ which gives $$\left(1+\frac 1y\right)\left(1+\frac 1z\right)\gt \frac 43$$ from which it follows that one of $x$ and $y$ must be less than $8$. Note that it can't be $6$, because that would not work in the top equation of this answer taken mod $3$. Once $y$ is fixed, there is at most one possibility for $z$ (the equation is linear in $z$) and any answer must be a permutation of one with $x\le y\le z$.

Using this method, you should be able to find the solutions.

Answer 3

Suppose $$xyz = xy + xz + yz + x + y + z $$ is a natural solution with $$x \leq y \leq z$$ then

$z = \frac {xy + x + y}{xy - x - y - 1}= 1 + \frac {2x + 2y + 1}{xy - x-y -1}$

So it must be $$xy - x - y -1\le 2x + 2y + 1$$ or $$xy \le 3x + 3y + 2$$ which actually means at least $x$ is smaller than 8. Likewise we can conclude that $y$ is smaller than 8. Now the $$z = \frac {xy + x + y}{xy - x - y - 1}= 1 + \frac {2x + 2y + 1}{xy - x-y -1}$$ gives us finitely many possibilities for $z$.