Estonian Mathematical Olympiad

Solution 1: If $x\ne 0$ or $y\ne 0$, then from the second equation $z>0$. Thus $z=0$ can only hold if $x=y=0$. The triple $(x,y,z)=(0,0,0)$ satisfies all equations. Now assume $z\ne 0$. Squaring the first equation yields $x^2+2xy+y^2=z^2$. Subtracting the second equation yields $$2xy=z^2-4z.(1)$$ Cubing the first equation yields $x^3+3x^{2}y+3x{y}^2+y^3=z^3$. Subtracting the third equation and factoring yields $$3xy(x+y)=z(z^2-18).$$ Using $x+y=z\ne 0$, this can be reduced to $$3xy=z^2-18.(2)$$ Expressing $xy$ from both (1) and (2) yields $3(z^2-4z)=2(z^2-18)$. This simplifies to $z^2-12z+36=0$ or $(z-6{)}^2=0$, from which $z=6$.

Now equation (1) yields $2xy=36-24=12$, from which $xy=6$. On the other hand $x+y=z=6$. Combining by Viète's formulas yields the quadratic equation $x^2-6x+6=0$, from which $x=3\pm \sqrt{3}$ and respectively $y=3\mp \sqrt{3}$. The triples $(x,y,z)=(3+\sqrt{3},3-\sqrt{3},6)$ and $(x,y,z)=(3-\sqrt{3},3+\sqrt{3},6)$ satisfy all three equations.

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Alternative solution.

Solution 2: Like in Solution 1, we show that $z=0$ yields only the solution $(x,y,z)=(0,0,0)$. We also similarly deduce (1). We then use the identity $x^3+y^3=(x+y)(x^2-xy+y^2)$. Substituting $x+y,x^2+y^2$ and $x^3+y^3$ from the given equations and $xy$ from (1) yields $$18z=z\left(4z-\frac{z^2-4z}{2}\right).$$ Dividing both sides by $z\ne 0$ and simplifying yields $z^2-12z+36=0$ or $(z-6{)}^2=0$, from which $z=6$. We proceed like in Solution 1.