I have to find the positive integer solutions of the equation $$xy+xz+yz-2xyz=0.$$ Note: If there are solutions, they should be finite in number because $xyz$ is of third degree.
$\begingroup$
$\endgroup$
8
-
$\begingroup$ source of the problem???? $\endgroup$– Will JagySep 4, 2015 at 17:27
-
$\begingroup$ solve the given equation for $x$ $$x=\frac{y z}{-2 y z+y+z}$$ $\endgroup$– Dr. Sonnhard GraubnerSep 4, 2015 at 17:30
-
$\begingroup$ @Will Jagy: A friend of my son engineering student, wants it resolved. I think I found the solution but should check if I'm not mistaken. (If the given answer is correct, I was wrong because I found just one solution) $\endgroup$– PiquitoSep 4, 2015 at 17:43
-
$\begingroup$ @Clayton: $x,y,z >0$ $\endgroup$– PiquitoSep 4, 2015 at 17:45
-
$\begingroup$ @Ataulfo But you said non-negative in the question body, not positive. $\endgroup$– user236182Sep 4, 2015 at 17:47
|
Show 3 more comments
1 Answer
$\begingroup$
$\endgroup$
4
If wlog $x=0$, then $(x,y,z)=(0,0,z),(0,y,0),y,z,\in\Bbb Z_{\ge 0}$ are all the solutions.
Wlog $x\ge y\ge z\ge 1$.
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\le \frac{3}{z}\implies z\le 1.5\implies z=1$$
$$\frac{1}{x}+\frac{1}{y}=1\iff (x-1)(y-1)=1\iff x=y=2$$
Answer: $(x,0,0),(0,y,0),(0,0,z), x,y,z\in\Bbb Z_{\ge 0}, (x,y,z)=(2,2,1),(2,1,2),(1,2,2)$ are all the solutions.
answered Sep 4, 2015 at 17:30
-
$\begingroup$ Yes, I forget the permutations with my only solution $(1,2,2)$. Regards. $\endgroup$– PiquitoSep 4, 2015 at 17:48
-
$\begingroup$ The question has been changed to "positive integers". I'll just leave it solved in non-negative integers. $\endgroup$ Sep 4, 2015 at 18:16
-
$\begingroup$ How did you get that $2 \leq \frac{3}{z}$? $\endgroup$ Nov 29, 2015 at 22:09
-
1$\begingroup$ @1110101001 Because $\frac{1}{x}\le \frac{1}{z}$ and $\frac{1}{y}\le \frac{1}{z}$, so $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le \frac{3}{z}$, and it's given that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. $\endgroup$ Nov 29, 2015 at 22:10