Finding values that make $xy=x+y+z$ true.

Question

Let $x,y,z \in \mathbb{Z^+}$ and $x \neq y \neq z$

Find all values that make this true:

$$xy=x+y+z$$

I am trying to solve a way to make this true but clearly, I cannot find any values that make this equation true. My approach has been trial and error and I haven't been successful.

So I'm convinced that there is no solutions in $\mathbb{Z^+}$. Of course this probably wouldn't be the case in the reals.

If there is any solution that points this out to be true please point out how you got the solution. If not, any ideas on how to start the proof?

Answer 1

$$ \begin{array}{c|rrrrrrrrrrrr} & 2 & 3 & 4 & 5 & 6 & 7 & \cdots \\ \hline 2 & 0 & 1 & 2 & 3 & 4 & 5 & \cdots \\ 3 & 1 & 3 & 5 & 7 & 9 & 11 & \cdots \\ 4 & 2 & 5 & 8 & 11 & 14 & 17 & \cdots \\ 5 & 3 & 7 & 11 & 15 & 19 & 23 & \cdots \\ 6 & 4 & 9 & 14 & 19 & 24 & 29 & \cdots \\ 7 & 5 & 11 & 17 & 23 & 29 & 35 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \end{array} $$ Let $x$ and $y$ be the numbers in the left and upper margins; let $z$ be the number in the interior, calculated by $z = xy-x-y$. Eliminate the main diagonal since you want $x\ne y$.

Answer 2

Rewrite the equation as $(x-1)(y-1)=z+1$. Now it is easy to find infinitely many solutions. One can give a general description of the solutions in terms of the prime power factorization of $z+1$.

Answer 3

Hint: Express z in terms of x and y, then add $1$ to both sides, and factor the entire expression.