Estonian Math Competitions

If $a=0$ then the first equation implies $b^{n+2}c=0$. Hence $b=0$ or $c=0$; w.l.o.g., $b=0$. Then the last equation reduces to $c^{n+3}=0$ which implies $c=0$. Thus if $a=0$ then $a=b=c=0$. Analogously we can prove that if $b=0$ or $c=0$ then $a=b=c=0$. The triple $(0,0,0)$ satisfies the equation.

It remains to consider triples $(a,b,c)$ whose all terms are different from $0$. Let $p$ be an arbitrary prime number. If $p\mid a$ then the first equation implies $p\mid b^{n+2}c$. Thus $p\mid b$ or $p\mid c$; w.l.o.g., $p\mid b$. Then the last equation gives $p\mid c^{n+1}$ which implies $p\mid c$. Thus if $p\mid a$ then $p\mid a,p\mid b,p\mid c$. Analogously we can prove that if $p\mid b$ or $p\mid c$ then $p\mid a,p\mid b,p\mid c$.

But rewriting $a=p{a}^{\prime },b=p{b}^{\prime },c=p{c}^{\prime }$ enables us to divide both sides of all equations by $p^{n+3}$, giving a similar system of equations having $a^{\prime },b^{\prime },c^{\prime }$ in the role of $a,b,c$. Hence the triple $(a^{\prime },b^{\prime },c^{\prime })$ also satisfies the system of equations. As the numbers other than $0$ cannot be infinitely divided in integers, a finite number of divisions should give us a solution whose terms have no common prime factors. By the previous paragraph, this is possible only if the values of variables have no prime factors, i.e., $a=\pm 1,b=\pm 1,c=\pm 1$. We show now that such solutions do not exist. Indeed, if $n$ is even then the system of equations reduces to $$\{\begin{array}{l}a+c+c+b=0,\\ b+a+a+c=0,\\ c+b+b+a=0.\end{array}$$ Adding all equations results in $4(a+b+c)=0$, implying $a+b+c=0$; but the sum of three odd numbers cannot equal the even number $0$.

If $n$ is odd then the system of equations reduces to $$\{\begin{array}{l}1+bc+1+ab=0,\\ 1+ca+1+bc=0,\\ 1+ab+1+ca=0.\end{array}$$ Adding all equations results in $2(ab+bc+ca)+6=0$, implying $ab+bc+ca=-3$. As $|ab|=|bc|=|ca|=1$, the only possibility is $ab=bc=ca=-1$. This demands that $a,b,c$ have pairwise opposite signs which is impossible. Consequently, no solutions except $(0,0,0)$ exist.