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I was going through one of my Mathematics books and I came to this problem:

Find all integer solutions to the equation: $\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}=15$

I tried a few things to begin with but none went in the right direction, any suggestions?

Another thing, if we have the left side of the equation to be equal to 9, there are supposed to be 16 solutions in integers. What way can they be found?

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  • $\begingroup$ No, no other set of constraints is mentioned. $\endgroup$ – madlin Aug 24 '16 at 14:22
  • $\begingroup$ First, if $(x,y,z)$ is a solution, then $(|x|,|y|,|z|)$ is a solution also (Use @Bacon comment for example, taking the absolute value). We may suppose that $1\leq x\leq y\leq z$. We have $xy\leq 15z$, etc.. Mutiplying these inequalities and simplifying by $xyz$, we get that $xyz\leq 15$. Now we have $x^3\leq 15$, this show that $x=1$ or $x=2$, etc. $\endgroup$ – Kelenner Aug 24 '16 at 14:39
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    $\begingroup$ @kelenner but (5,5,5) is a solution. $\endgroup$ – N.S.JOHN Aug 24 '16 at 14:53
  • $\begingroup$ My comment has a typo - obviously I mean that $$xyz \left( \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right)=15$$ $\endgroup$ – complexmanifold Aug 24 '16 at 14:56
  • $\begingroup$ @N.S.John Yes you are true ...Thank you for your comment. When multiplying, we get $xyz\leq 15^3$, so $x^3\leq 15^3$ and $x\leq 15$. This works, but this is much more complicated.... $\endgroup$ – Kelenner Aug 24 '16 at 15:00
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  1. First, let us suppose that $x,y,z$ are positive integers satisfying the proposed equality. Then $$ xyz(15-x-y-z)=\frac{1}{2}\left((xy-yz)^2+(yz-zx)^2+(zx-xy)^2\right)\ge0$$ So we have $x+y+z\le 15$.

  2. Now, note that squares modulo 5 are $\{0,-1,1\}$. Thus, if the sum of three squares modulo $5 $ is $0$ then either all three of them are multiples of $5$, (the case $0+0+0$) or just exactly one of them is a multiple of $5$ and the other two are equal to $+1$ and $-1$ modulo $5$, ( the case $0+1-1$). In both cases $5$ divides at least one of the three numbers. The equality $$(xy)^2+(yz)^2+(zx)^2=15xyz\equiv 0\mod 5\tag1$$ shows then that one of the numbers $xy,yz,zx$ is a multiple of $5$. So, we may and will suppose that $x=5a$. But then $(1)$ proves that $5$ must divide $yz$ and again we may suppose that $ y=5 b$.

  3. Now, $5+5+1\le 5a+5b+z\le15$ implies that $ a=b=1$, that is $x=y=5$ and the equation implies then that $z=5$. We conclude that $(5,5,5)$ is the only solution consisting of positive integers.
  4. The general case. Suppose that $x=sx'$,$y=ty'$ and $z=uz'$ is a solution to the proposed equation where $x',y',z'$ are positive integers, and $s,t,u$ are $\pm1$. It follows from $$xyz\left(\frac{1}{x^2}+ \frac{1}{y^2}+\frac{1}{z^2}\right)=15$$ That $$stu x'y'z'\left(\frac{1}{x'^2}+ \frac{1}{y'^2}+\frac{1}{z'^2}\right)=15$$ This implies that $stu=1$ and $$ x'y'z'\left(\frac{1}{x'^2}+ \frac{1}{y'^2}+\frac{1}{z'^2}\right)=15$$ So, $x'=y'=z'=5$ and exactly two of $s,t,u$ are equal $-1$, or they are all equal to $+1$. This gives the following set of solutions $$\{(5,5,5),(-5,-5,5),(-5,5,-5),(5,-5,-5)\}.$$
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Since obviously $xyz\ne 0$, you have $$(xy)^2+(xz)^2+(yz)^2=15xyz\iff (y^2+z^2)x^2-(15yz)x+(yz)^2=0$$ hence the discriminant must be non-negative. $$(15yz)^2-4(y^2+z^2)(yz)^2\ge 0$$ It follows $$y^2+z^2\le 56$$ This involves only the set $\mathcal S$ of the first seven squares $$\mathcal S=\{1,4,9,16,25,36,49\}$$ Searching for $x=y=z$ one has $3x^4=15x^3$ whose only solution is $x=5$ which can be done with $y=\pm 5$ and $z=\pm 5$.

We make then a table in which we discard the cases $x=y=z$ and also $y=z$ because in those cases the equation $2X^2-15X+1=0$ has no rational solution.

$$\begin{array}{|c|c|}\hline x & y^2+z^2 & yz\\\hline 1&5&4\\\hline1&10&9\\\hline 1&17&16\\\hline1&26&25\\\hline1&37&36\\\hline1&50&49\\\hline2&13&36\\\hline2&20&64\\\hline2&29&100\\\hline2&40&144\\\hline2&53&196\\\hline 3&25&144\\\hline3&34&225\\\hline 3&45&324\\\hline4&41&400\\\hline4&52&576\\\hline\end{array}$$ Because of the congruence $$x^2(y^2+z^2)+(yz)^2\equiv 0\pmod 5$$ the discarding of this sixteen it is straightforward. Thus there are only four solutions basically refered to the $(|x|,|y|,|z|)=(5,5,5)$ changing1 signes in two of the variables.

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Translating the equation to form $$(xy)^2+(yz)^2+(zx)^2=15xyz,\qquad(1)$$ easy to see that $xyz>0.$ That means that all solutions with different signs can be obtained from positive solutions by sign replacing of two unknowns at the same time. So we can search for positive solutions, $$x,y,z>0.$$

All rational roots of equation with integer coefficients $$\left(\dfrac{yz}x\right)^2-15\dfrac{yz}x+y^2+z^2=0$$ must be integer, so $x\ |\ yz$ and similarly $y\ |\ zx,\ z\ |\ xy.$
Let $$u=\dfrac{xy}z,\quad v=\dfrac{yz}x,\quad w=\dfrac{zx}y,$$ then we get $$u+v+w=15,\quad u,v,w,\sqrt{uv},\sqrt{vw},\sqrt{wu}\in\mathbb N$$ with the single solution $$u=v=w=5,$$ $$x=\sqrt{uw}=5,\quad y=\sqrt{uv}=5,\quad z=\sqrt{vw}=5.$$ So the solution set is $$\boxed{(x,y,z)\in\{(5,5,5),(5,-5,-5),(-5,5,-5),(-5,-5,5)\}.}$$

Note

For the equation $LHS=9$ we have $$u+v+w=9,\quad u,v,w,\sqrt{uv},\sqrt{vw},\sqrt{wu}\in\mathbb N$$ with the solutions $$(u,v,w)=\{(4,4,1), (4,1,4), (1,4,4), (3,3,3)\},$$ so the positive solutions of $LHS=9$ are $$(x,y,z)=\{(2,4,2),(4,2,2),(2,2,4),(3,3,3)\}$$ and with account to pair signs permutations we have 16 solutions.

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