# Number of ordered sets of integers

How many ordered sets of integers $(x,y,z)$ satisfying $$x,y,z \in [-10,10]$$ are solutions to the following system of equations: $$x^2y^2+y^2z^2=5xyz$$ $$y^2z^2+z^2x^2=17xyz$$ $$z^2x^2+x^2y^2=20xyz$$

By a quite cumbersome method, which is too long for me to type down here, I arrived at the answer 5: the ordered sets being $(0,0,0), (8,2,4), (-8,-2,4), (-8,2,-4)$ and $(8,-2,-4)$. But this does not seem to be the right answer. What I basically did was divide the system of equations with $(xyz)$, then by a process of elimination of the variables, found the possible values of x, y and z. However, it seems I've somehow missed out other solutions as well. Can anyone help me out with this?

Thanks!

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If any two of $x,y,z$ are $0$, all the equations say $0 = 0$. On the other hand, if only one is $0$, it's easy to see that another must also be $0$.

So now let's assume none of them are $0$. Then we can cancel a $y$, $z$, or $x$ respectively from each equation, leaving \eqalign{(x^2 + z^2) y &= 5 x z\cr (x^2 + y^2) z &= 17 x y\cr (y^2 + z^2) x &= 20 y z\cr}

Next, notice that changing signs of any two of the variables preserves all three equations. Thus we may assume, say, $x > 0$ and $y > 0$, and get the other solutions by symmetry.

Now $x z \le (x^2 + y^2)/2$ (because $x^2 - 2 x z + y^2 = (x-y)^2 \ge 0$), and similarly $-x z \le (x^2 + y^2)/2$, i.e $|x z| \le (x^2 + y^2)/2$.
So from the first equation we get $$|y| = \frac{5 |x z|}{x^2 + y^2} \le \frac{5}{2}$$ Since at this point $y$ is a positive integer, the only possibilities are $y= 1$ and $y = 2$.

If $y = 1$, the second equation says $z (x^2 + 1) = 17 x$. Since $x^2 + 1$ and $x$ are relatively prime, this implies that $x$ divides $z$. So let $z = m x$. We then have $m (x^2 + 1) = 17$. Since $17$ is prime, one of $m$ and $x^2 + 1$ must be $1$ and the other $17$. But $x \ne 0$ so it must be $x^2 + 1 = 17$ (i.e. $x=4$), and $m= 1$ so $z=4$. But $x=4,y=1,z=4$ doesn't satisfy the first or third equation.

The remaining possibility is $y=2$. Then the second equation says $z (x^2 + 4) = 34 x$. If $x$ is odd, so is $x^2 + 4$, and that is relatively prime to $x$ so it divides $34$. The only odd factors of $34$ are $1$ and $17$, so we'd need $x^2 + 4 = 17$, but that doesn't work since $17-4=13$ is not a square. So $x$ must be even: say $x = 2 t$. Now the second equation, after dividing by $4$, becomes $z (t^2 + 1) = 17 t$. As in the previous paragraph, we conclude that $t=4$ (so $x=8$) and $z=4$. Thus we have $(x,y,z) = (8,2,4)$, which is a solution. By symmetry (changing any two signs) we also have solutions $(-8,-2,4), (-8,2,-4)$ and $(8,-2,-4)$.

Summing up, we have the solutions $(x,y,z) = (0,0,z)$, $(0,y,0)$, $(x,0,0)$, $(8,2,4)$, $(-8,-2,4), (-8,2,-4)$ and $(8,-2,-4)$.

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Thank you all for your answers! They were all quite helpful. I found Robert's answer the most elegant though, and hence my "accepted answer" goes for the same. I've upvoted all of you. Thanks again! – Anton Schigur Dec 12 '12 at 10:35
I think joriki's is more elegant. Maybe mine is more elementary. – Robert Israel Dec 12 '12 at 10:54

The discriminant of the first equation with respect to $z$ is $x^2y^2(25-4y^2)$. This must be a square, so we must have $y\in\{-2,0,2\}$ or $x=0$. If one variable is $0$, one of the others must also be $0$, and we get the solutions that Johan pointed out. If $y=\pm2$, we need $x=\pm8$ or $x=\pm9$ to get a factor of $17$ from $x^2+y^2$ in the second equation. Substituting this into the second equation yields integer $z$ only for $x=\pm8$, and this yields the solutions that you found. Thus the solutions found by Johan and you are the only ones.

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If two of the variables are zero, then the third one can be anything. So $(0,0,z)$, where $z$ is anything, is a a set of solutions, as is $(0,y,0)$ and $(x,0,0)$.

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Let $(x,y,z)$ be a solution. Then the three numbers

$$u_1=x^2y^2+y^2z^2-5xyz, \ u_2=y^2z^2+z^2x^2-17xyz, \ u_3=z^2x^2+x^2y^2=20xyz$$

are all zero.

Now

$$\begin{array}{lcl} (1) \ -u_1+u_2+u_3 &=& 2x^2z^2-32xyz=32xz\bigg( \frac{xz}{16}-y\bigg) \\ (2) \ u_1-u_2+u_3 &=& 2x^2y^2-8xyz=8xy\bigg( \frac{xy}{4}-z\bigg) \\ (3) \ u_1+u_2-u_3 &=& 2y^2z^2-2xyz=2yz\bigg( yz-x\bigg) \end{array}$$

If $x,y$ and $z$ are all nonzero, we see that $x=yz, y=\frac{xz}{16}$ and $z=\frac{xy}{4}$. Multiplying, we deduce $xyz=\frac{(xyz)^2}{64}$, so $xyz=64$. Using $x^2=xyz, y^2=\frac{xyz}{16}$ and $z^2=\frac{xyz}{4}$, we deduce $x^2=64,y^2=4,z^2=16$.

This produces four solutions : $(8\varepsilon_1,2\varepsilon_2,4\varepsilon_1\varepsilon_2)$ where each of $\varepsilon_1$ and $\varepsilon_2$ can be $+1$ or $-1$.

If $z=0$, then by (2) we have $2(xy)^2=0$ and hence $x=0$ or $y=0$. Similarly when $x=0$ or $y=0$ : this will produce the solutions found by Johan.

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I had suggested a slight edit in your answer, but it doesn't seem to have come through; didn't you mean 4ε1ε2, and not 16ε1ε2? – Anton Schigur Dec 12 '12 at 10:39
@JobinIdiculla : Corrected, thanks. – Ewan Delanoy Dec 12 '12 at 11:44