Possible values for this specific line of variables.

Question

I have this line of numbers:

xy + z = xz + y = yz + x 

I need to find out all the possible values of x, y and z in this equation.

Thank you!:)

My usual problem when guessing this:

x = 1
y = 2
z = 2

1 * 2 + 2 = 1 * 2 + 2 != 2 * 2 + 1

And it is like so with any values I try that first 2 equations are true, but the last one isn't.

Answer 1

Taking cue from the results of @user133281, we have

$xy+z=xz+y$ implies $x(y-z) = y-z$, hence $y=z$ or $x=1$

$xz+y=yz+x$ implies $z(x-y) = x-y$, hence $x=y$ or $z=1$

$xy+z=yz+x$ implies $y(x-z) = x-z$, hence $x=z$ or $y=1$

So $x=y=z=1,2,3,4,...$ are the solutions.

However if one of the equalities does not hold, then

1. $x=1=>y\not=z=>$ either $x=z=>(1,y,1)$ or $y=1=>(1,1,z)$.
2. $y=1=>x\not=z>$ either $y=z=>(x,1,1)$ or $x=1=>(1,1,z)$
3. $z=1=>x\not=y>$ either $x=z=>(1,y,1)$ or $y=1=>(x,1,1)$

Hence the final solutions are $(x,y,z),(x,1,1),(1,y,1),(1,1,z)$ where $x,y,z$ are arbitrary..

Answer 2

Hint: $xy+z=xz+y$ implies $x(y-z) = y-z$, hence $y=z$ or $x=1$. Similarly, $x=y$ or $z=1$ and $x=z$ or $y=1$.

Now we distinguish cases based on the number of values that equal $1$.

Suppose none of the numbers equals $1$. Then all numbers are equal (because $y=z$, $x=y$ and $x=z$) and we find the solution $(x,x,x)$.

Suppose that only $x=1$. Then we find $x=y$ and $x=z$, contradiction.

Suppose that two of the numbers are $1$, say $x=y=1$. Then we also find $x=y$ (which is true), so we find the solution $(1,1,z)$.

If all numbers are $1$ we find the solution $(1,1,1)$.

In conclusion, the solutions are $(x,x,x)$ and $(1,1,z)$ with $x$ and $z$ arbitrary (and their permutations).