Estonian Mathematical Olympiad

The given equation is equivalent to the equation $$x^3+y^3+z^3=2025+3xyz.$$ As $3\mid 2025$ and $3\mid 3xyz$, also $3\mid x^3+y^3+z^3$. Thus $3\mid x+y+z$ as an integer and its cube are congruent modulo 3. Let $x+y+z=3m$.

W.l.o.g., let $x\le y\le z$ and $x=m-a$, $y=m+b$, $z=m+c$, where $a\ge 0$, $c\ge 0$ and $b=a-c$. Substituting for $x$, $y$, $z$ in the given equation and simplifying leads to $$3m(a^2+b^2+c^2+ab+ac-bc)-a^3+b^3+c^3+3abc=2025.$$ Substituting also $b=a-c$ now gives $3m(3a^2-3ac+3c^2)=2025$ or, equivalently, $$m(a^2-ac+c^2)=225.$$ As $a^2-ac+c^2=(a-c{)}^2+ac\ge 0$, the number $m$ is a positive divisor of 225. The positive divisors of 225 are 1, 3, 5, 9, 15, 25, 45, 75, 225.

If $a=0$ then $m{c}^2=225$; we obtain 1, 9, 25, 225 as possible values of $m$, giving 3, 27, 75, 675 as the corresponding values of $x+y+z$. If $a=2c$ then $m{c}^2=75$; now $m$ can be either 3 or 75, giving 9 and 225 as the corresponding values of $x+y+z$. In the remaining cases, $5\mid m$ but $25\nmid m$; then also $5\mid a^2-ac+c^2$ but $25\nmid a^2-ac+c^2$. Thus $5\mid (a^2-ac+c^2)(a+c)=a^3+c^3$. Cubing integers that are incongruent modulo 5 produces results incongruent modulo 5. As $a^3=-c^3\equiv (-c{)}^3(\mathrm{mod}5)$, we must have $a\equiv -c(\mathrm{mod}5)$. Hence $0\equiv a^2-ac+c^2\equiv (-c{)}^2-(-c)c+c^2=3c^2(\mathrm{mod}5)$, implying that $5\nmid c$. But then also $5\mid a$ and $25\mid a^2-ac+c^2$. The contradiction shows that there are no other solutions.

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Alternative solution.

It is easy to see that $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx).(15)$$ As $3\mid 2025$, either $3\mid x+y+z$ or $3\mid x^2+y^2+z^2-xy-yz-zx$. On the other hand, the representation $$x^2+y^2+z^2-xy-yz-zx=(x+y+z{)}^2-3(xy+yz+zx)$$ shows that $3\mid x^2+y^2+z^2-xy-yz-zx$ if and only if $3\mid x+y+z$. Hence $3\mid x+y+z$ anyway.

Let $x+y+z=3m$. W.l.o.g., let $x\le y\le z$ and $x=m-a$, $y=m+b$, $z=m+c$, where $a\ge 0$, $c\ge 0$ and $a=b+c$. Using the equality (15) along with the equality $$x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}((x-y{)}^2+(y-z{)}^2+(z-x{)}^2),$$ as well as the equations about $x$, $y$ and $z$, the initial equation can be converted to $$3m((a+b{)}^2+(b-c{)}^2+(a+c{)}^2)=4050.$$ Substituting also $a=b+c$ gives $3m(6b^2+6bc+6c^2)=4050$ which reduces to $$m(b^2+bc+c^2)=225.$$ Note that $b^2+bc+c^2=\frac{1}{2}((b+c{)}^2+b^2+c^2)\ge 0$; hence $m$ is a positive divisor of 225. The positive divisors of 225 are 1, 3, 5, 9, 15, 25, 45, 75, 225. If $b=0$ then $m{c}^2=225$; we obtain 1, 9, 25, 225 as possible values of $m$. If $b=c$ then $m{c}^2=75$; now $m$ can be either 3 or 75. It remains to check the possibilities $m=5$, $m=15$ and $m=45$. In these cases $b^2+bc+c^2=d$ where $d$ is 45, 15 or 5, respectively. As $b^2+bc+c^2=d$ is equivalent to $3b^2+(b+2c{)}^2=4d$, the number $4d-3b^2$ is a perfect square. If $d=5$ then $20-3b^2$ must be a perfect square which is impossible. If $d=15$ then we have $3\mid 4d-3b^2$, hence $60-3b^2=(3k{)}^2$, implying $20-3k^2=b^2$ for some integer $k$ which is impossible. If $d=45$ then $3\mid 4d-3b^2$ again, hence $180-3b^2=(3k{)}^2$, implying that $60-3k^2=b^2$ for some integer $k$ which is impossible.