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How do I manually solve $x+y=xyz-1$ assuming that $x, y$ and $z$ are positive integers? I was able to guess all possible solutions, but I do not know how to show that these are the only ones:
$x=1, y=1, z=3$
$x=1, y=2, z=2$
$x=2, y=1, z=2$
$x=2, y=3, z=1$
$x=3, y=2, z=1$

Any hints would be appreciated.

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    $\begingroup$ $x+y=z*xy-1$ is not linear. $\endgroup$ – uniquesolution Oct 17 '15 at 12:07
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We have $$z=\frac{x+y+1}{xy}=\frac 1y+\frac 1x+\frac{1}{xy}\le 1+1+1$$$$\Rightarrow z=1,2,3$$

Also, $$zxy-x-y=1\iff z^2xy-zx-zy+1=z+1\iff (zx-1)(zy-1)=z+1$$

So, for $z=1$, we have $(x-1)(y-1)=2$.

For $z=2$, we have $(2x-1)(2y-1)=3$.

For $z=3$, we have $(3x-1)(3y-1)=4$.

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WLOG let $x\le y$, and we need $$\frac1x+\frac1y+\frac1{xy}=z$$

Clearly if $x\ge3$, LHS $<1$, so we need to check only for $x=1,2$. Similarly it is not hard to bound possible $y$ hence you only get a few values to test out.

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Note that for $x,y$ positive integers $$xy\ge x+y+1$$ for $x>2$ and $y\ge 3$ (or vice verse). Therefore you can control only $y=1$(or vice verse) and $y=x=2$

If $x=1$ the equation becomes: $$y(z-1)=1$$ and the solutions are $z=2$ $y=2$ and $z=3$, $y=1$. The same thing for $y=1$. While for $y=x=2$ there aren't solutions.

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Alternative solution:

$x\mid y+1$ and $y\mid x+1$, so $x-1\le y\le x+1$. Simply check cases:

  • If $x-1=y$, then $y\mid x+1$ gives $x-1\mid x+1$, i.e. $x-1\mid 2$, i.e. $x\in\{3,2\}$, so $y\mid x+1\in\{4,3\}$, so $y\in\{1,2,3,4\}$.

  • If $x=y$, then $x\mid 1, y\mid 1$.

  • If $x+1=y$, then $x\mid y+1=x+2$, so $x\mid 2$, so $(x,y)\in\{(1,2),(2,3)\}$.

It's left to check $11$ cases (knowing $(x,y)$ find $z$).

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