6
$\begingroup$

I am trying to solve $\frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} = 3$ over integers, but I have no idea what's the best way to do it. I have tried multiplying both sides by $xyz$ and then figuring out that $z$ divides $xy$, $x$ divides $yz$, etc, but with no effect.

$\endgroup$
  • $\begingroup$ Are you trying to find just 1 solution, because $x=y=z=1$ is a solution. $\endgroup$ – Stephen Meskin Nov 27 '17 at 17:58
  • $\begingroup$ I'm trying to solve this equation so I'm looking for every triple $(x, y, z)$ satisfying this equation. $\endgroup$ – Gilbert Posner Nov 27 '17 at 18:01
1
$\begingroup$

It appears there are no non-trivial (i.e. not $x,y,z = \pm 1$) solutions: sorry, no number theory to do here. (Just some analysis.) Dividing everything by $xyz$ puts the equation in a nice form:

$\frac{1}{x^2} + \frac{1}{y^2}+\frac{1}{z^2} = \frac{3}{xyz}$.

Staring at that for a moment, it seems like the LHS is always larger than the RHS, and this is easy to show. Suppose WLOG $x \leq y \leq z$.

Case 1: $x < z^{2/3}$. Then

$\frac{1}{x^2} + \frac{1}{y^2}+\frac{1}{z^2} \geq \frac{3}{z^2} >\frac{3}{x^3} \geq \frac{3}{xyz}.$

So there are no solutions with $x < z^{2/3}$.

Case 2: $x \geq z^{2/3}$. Then

$\frac{1}{x^2} + \frac{1}{y^2}+\frac{1}{z^2} \geq \frac{3}{z^2} > \frac{3}{z^{7/3}} \geq \frac{3}{xyz}$,

since the strict inequality $z^{-2} > z^{-7/3}$ holds for $z > 1.$ So there are no solutions in this case either.

$\endgroup$
1
$\begingroup$

Obviously $xyz\ne 0$.

Since $$3xyz = x^2y^2+y^2z^2+z^2x^2\geq 3\sqrt[3]{x^4y^4z^4}$$ we have $$xyz\geq (xyz)^2\Longrightarrow 0\leq xyz\leq 1$$

Now since $x,y,z\in \mathbb{Z}$ we have $xyz =1$. So $(1,-1,-1)$ and all permutation are solution as $(1,1,1)$ is.

$\endgroup$
1
$\begingroup$

If an odd number of the variables $x,y,z$ are negative, (i.e. if exactly 1 or all three are negative) then the LHS is clearly negative, so there can be no solution. If exactly two of them are negative, then we can exchange both their signs to get a solution with all of $x,y,z$ positive, and vice versa. So we can take $x,y,z$ all positive WLOG.

But in this case, by AM-GM we have $$\frac{1}{3}\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right) \ge (xyz)^{1/3}$$ But the RHS is obviously $\ge 1$ if $x,y,z$ are positive integers. Therefore if the LHS equals $1$ (which is your equation), we must have equality throughout. Therefore $(1,1,1)$ is the only solution in positive integers, and by swapping signs we get $(1,-1,-1)$, $(-1,1,-1)$ and $(-1,-1,1)$ as the others.

$\endgroup$
0
$\begingroup$

your equation can factorized ot $$x^2y^2+x^2z^2+z^2x^2=3xyz$$ and then use $$(xy)^2+(xz)^2+(yz)^2\geq x^2yz+xy^2z+xyz^2$$ and we can write $$3xyz\geq xyz(x+y+z)$$ can you proceed?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.