I am having trouble with the following question:

For which integers $x$, $y$ and $z$ is the following statement correct?


Obviously, $x=3$, $y=3$ and $z=3$ is one answer.

How do I prove that it is the only correct one?

  • 1
    $\begingroup$ No! It isn't the only correct one. You can have $x,y,z=\pm{3}$ $\endgroup$ – crskhr Jul 2 '18 at 9:17

This is more of a hint and general approach:

WLOG we can assume $x,y,z \geq 0$ - otherwise consider $(-x)$ etc.

Note that none of them can equal $0$.

We have $x \geq 2, y \geq 2 , z \geq 3$ by using that $x,y,z \geq 1$

If $x,y,z \geq 4$ then plugging in we get a contradiction. Hence at least one of the $3$ is $\leq 3$

We can now consider again a case by case approach:

Case $1$: $x=2$

Case $2$: $x=3$

Case $3$: $y=2$

Case $4$: $y=3$

Case $5$: $z=3$

Then repeat the idea of finding an upper bound to get all solutions. Once you've done that go back to the WLOG bit and say that if $(x,y,z)$ is a solution then we can make $7$ more solutions by just changing signs


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