$ i)x\geq y\geq z \geq 0$ $ \implies$ $ \frac {1}{z} + \frac {1}{x} + \frac {1}{y} = 1 + \frac {2}{xyz}$ $ \frac {1}{z} + \frac {1}{x} + \frac {1}{y}\leq\frac {3}{z}$ $ \implies$ $ z\leq 3$ $ z = 3$ $ \implies$ $ xy + 3y + 3x - 3xy = 2$ $ \implies$ $ 3x + 3y = 2xy + 2$ $ \implies$ $ 6x + 6y = 4xy + 4$ $ \implies$ $ 2x(3 - 2y) - 3.(3 - 2y) = - 5$ $ \implies$ $ (2x - 3)(2y - 3) = 5$ $ \implies$ $ 2y - 3 = 1,2x - 3 = 5$ $ \implies$ $ z > y$ contradiction! $ z = 2$ $ \implies$ $ xy + 2y + 2x - 2xy = 2$ $ \implies$ $ 2x + 2y = xy + 2$ $ x(2 - y) - 2(2 - y) = - 2$ $ \implies$ $ (x - 2)(y - 2) = 2$ $ \implies$ $ x = 4,y = 3$ ,$ (4,3,2)$ is a solution $ z = 1$ $ \implies$ $ xy + x + y - xy = 2$ $ \implies$ $ x + y = 2$ $ (x\geq y\geq z)$ $ (1,1,1)$ is a only solution. $ z = 0$ $ \implies$ $ xy = 2$ $ \implies$ $ (2,1,0)$ is a solution $ ii)x\geq y \geq 0$ ,$ z < 0$ $ \implies$ $ z = - r$. $ xy - yr - rx + xyr = 2$ $ \implies$ $ \frac {1}{r} + 1 = \frac {1}{x} + \frac {1}{y} + \frac {2}{xyr}$ $ \implies$ $ x,y,r > 2$ $ \implies$ $ \frac {1}{x} + \frac {1}{y} + \frac {2}{xyr} < 1$ contradiction! $ y = 2$ $ \implies$ $ 2x - 2r - rx + 2rx = 2$ $ \implies$ $ 2x - 2r + rx = 2$ $ \implies$ $ x(2 - r) - 2(2 - r) = - 2$ $ \implies$ $ (x - 2)(r - 2) = 2$ $ \implies$ $ x = 4,r = 3$ and $ x = 3,r = 4$ $ y = 1$ $ \implies$ $ x - r - rx + rx = 2$ $ \implies$ $ x - r = 2$ $ x = r + 2$ $ \implies$ $ (r + 2,1,r)$ similarly $ r = 2,r = 1$ similarly, $ x\geq 0$ $ y\leq z < 0$ $ \implies$ $ y = - p,z = - r$ $ \implies$ $ - px + pr - rx - prx = 2$ but $ - px + pr - rx - prx < 0$ for $ x\geq 1$ $ \implies$ $ x=0$ $ pr=2$ $ \implies$ $ (2,1,0)$ and $ z\leq y \leq x < 0$ $ \implies$ $ z = - r,y = - p,x = - q$ $ \implies$ $ pr + rq + qp + pqr = 2$ $ p,q,r > 0$ $ \implies$ $ pr + qr + qp + pqr\geq 4$contradiction! $ r = 0$ $ \implies$ $ (2,1,0)$ is a solution

lambruscokid wrote: Find all integers solutions for $ xy + yz + zx - xyz = 2$ See here: http://www.mathlinks.ro/viewtopic.php?t=265297

$xyz-xy-yz-zx=-2$ then add both side $x+y+z-1$ we will get $xy(z-1)+y(z-1)+x(z-1)+z-1=-3+x+y+z$ then $(x-1)(y-1)(z-1)=x-1+y-1+z-1$ let $x-1$=a, $y-1=b$ $z-1=c$ so we get $abc=a+b+c$ let $a \geq b\geq c$ then $c^{2}a\leq3a$ it means $a=1$ or $a=-1$ if a=1 $bc=b+c+1$, $ba-b-a+1=2$, $(b-1)(a-1)=2$ or $-ab=a+b-1$ $ab+a+b+1=2$c $(a+1)(b+1)=2$ and solutions are $ (2,4,3);(2,3,4);(1,0,2);(1,2,0);(1,1,1);(0,2,1);(0,1,2)$