63rd Czech and Slovak Mathematical Olympiad

If for example $x=0$, we get a system $0=y{z}^2=2y^{2}z$ which means that one of the unknowns $y$ and $z$ vanishes and the other can be arbitrary. The cases $y=0$ or $z=0$ are discussed similarly. Thus we have obtained three groups of solutions $(x,y,z)$ which are formed by triples $(t,0,0)$, $(0,t,0)$ and $(0,0,t)$ respectively, where $t$ is any real number. Moreover, we have observed that all the other solutions satisfy the condition $xyz\ne 0$, which is supposed to hold in what follows.

Factorizing the equation $x(y^2+2z^2)=y(z^2+2x^2)$ yields $(2x-y)(z^2-xy)=0$. Thus we distinguish two cases (depending on the fact which of the two factors vanishes).

i. $2x-y=0$. After setting $y=2x$ the given system is reduced to the only equation $$2x(2x^2+z^2)=9x^{2}z,$$ which can be simplified (by dividing $x\ne 0$) to $$4x^2+2z^2-9xz=0\text{or}(x-2z)(4x-z)=0.$$ Thus the case (i) yields exactly two groups of solutions $(2t,4t,t)$ and $(t,2t,4t)$, where $t$ is any real number.

ii. $z^2-xy=0$. Substituting $z^2=xy$ into the given system, we now get the only equation $$xy(2x+y)=z(x^2+2y^2),$$ which is (because of the inequality $x^2+2y^2>0$) equivalent to $$z=\frac{xy(2x+y)}{x^2+2y^2}.$$ At this moment we have to find when such a $z$ obeys the condition $z^2=xy$. After direct substitution we get the following condition on the unknowns $x$ and $y$: $$\frac{x^2{y}^2(2x+y{)}^2}{(x^2+2y^2{)}^2}=xy.$$ Dividing by $xy\ne 0$ and removing the fraction yields $$xy(2x+y{)}^2=(x^2+2y^2{)}^2\text{or}(4y-x)(x^3-y^3)=0.$$ Thus we conclude that either $x=4y$, or $x^3=y^3$, i.e. $x=y$. Returning to the formula for $z$, we obtain $z=2y$ or $z=x$, according as $x=4y$ or $x=y$. Consequently, there are two groups of solutions in the case (ii), namely triples $(4t,t,2t)$ and $(t,t,t)$, where $t$ is any real number.

Answer. All the solutions are $(t,0,0)$, $(0,t,0)$, $(0,0,t)$, $(t,t,t)$, $(4t,t,2t)$, $(2t,4t,t)$ and $(t,2t,4t)$, where $t$ is any real number.

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Alternative solution.

To avoid unnecessary repetition from the above solution, we will solve the problem under the condition that $xyz\ne 0$.

Dividing both sides of the given equations by $xyz$ we obtain $$\frac{y}{z}+\frac{2z}{y}=\frac{z}{x}+\frac{2x}{z}=\frac{x}{y}+\frac{2y}{x},(3)$$ which can be read as a coincidence of values of a function $f(s)=s+2/s$ in three nonzero points $s_1=y/z$, $s_2=z/x$ and $s_3=x/y$. Thus we first find when $f(s)=f(t)$ for two nonzero real numbers $s$ and $t$. It follows from the identity $$f(s)-f(t)=s+\frac{2}{s}-t-\frac{2}{t}=\frac{(s-t)(st-2)}{st}$$ that $f(s)=f(t)$ if and only if $s=t$ or $st=2$. Consequently, the system (3) holds if and only if the introduced numbers $s_1,s_2,s_3$ possess the following property: $s_i=s_j$ or $s_i{s}_j=2$, for any indices $i$ and $j$. However, if there exists a permutation $(i,j,k)$ of $(1,2,3)$ such that $s_i{s}_j=2$, then the identity $s_i{s}_j{s}_k=1$ implies that $s_k=\frac{1}{2}$ and hence $s_i\in \{\frac{1}{2},4\}$ (because $s_i=s_k$ or $s_i{s}_k=2$). Thus the assumption $s_i{s}_j=2$ leads to the conclusion that $(s_1,s_2,s_3)$ is a permutation of $(\frac{1}{2},\frac{1}{2},4)$. It is easy to check that exactly three such permutations are satisfactory and yield the solutions $(4t,t,2t)$, $(2t,4t,t)$ and $(t,2t,4t)$ of the given system. In the remaining case when $s_1=s_2=s_3$, the identity $s_1{s}_2{s}_3=1$ implies that $s_i=1$ for each $i$, which yields the solutions $(t,t,t)$.