XXVII Olimpiada Matemática Rioplatense

Assume $\{a,b,c\}$ is a triple satisfying the required conditions. For every positive integer $r$, let $S_r=a^r+b^r+c^r$. We denote $S=S_1$.

First, let us show that $S$ divides $S_{11k+1}$ for every non-negative integer $k$. We proceed by induction on $k$. For $k=0,1,2$, the claim is true by our assumptions on $\{a,b,c\}$. Let $k\ge 2$ and assume the result holds for $k-2,k-1$ and $k$; we will see it is true for $k+1$. We have that $$\begin{array}{rl}{S}_{11k+1}\cdot S_{11}& =(a^{11k+1}+b^{11k+1}+c^{11k+1})(a^{11}+b^{11}+c^{11})\\ & =a^{11(k+1)+1}+b^{11(k+1)+1}+c^{11(k+1)+1}+(ab{)}^{11}(a^{11(k-1)+1}+b^{11(k-1)+1})\\ & +(ac{)}^{11}(a^{11(k-1)+1}+c^{11(k-1)+1})+(bc{)}^{11}(b^{11(k-1)+1}+c^{11(k-1)+1})\\ & =S_{11(k+1)+1}+((ab{)}^{11}+(ac{)}^{11}+(bc{)}^{11})S_{11(k-1)+1}-\\ & (abc{)}^{11}(a^{11(k-2)+1}+b^{11(k-2)+1}+c^{11(k-2)+1})\\ & =S_{11(k+1)+1}+((ab{)}^{11}+(ac{)}^{11}+(bc{)}^{11})S_{11(k-1)+1}-(abc{)}^{11}{S}_{11(k-2)+1}.\end{array}$$ By the induction assumption, $S$ divides $S_{11k+1}$, $S_{11(k-1)+1}$ and $S_{11(k-2)+1}$; therefore, the above equality implies that $S$ divides $S_{11(k+1)+1}$, as we wanted to prove.

Consider the factorization $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$. If $x,y,z$ are integers, it implies that every divisor of $x+y+z$ also divides $x^3+y^3+z^3-3xyz$. By taking $x=a^{11004}$, $y=b^{11004}$ and $z=c^{11004}$, and recalling that $S$ divides $S_{11004}$, we deduce that $S$ divides $S_{33012}-3(abc{)}^{11004}$. Since $33012\equiv 1(\mathrm{mod}11)$, we know that $S$ divides $S_{33012}$; therefore, $S$ divides $3(abc{)}^{11004}$.

Assume $p>3$ is a prime factor of $S$. Since $p$ divides $3(abc{)}^{11004}$, then, it divides $a,b$ or $c$. With no loss of generality, assume $p$ divides $a$; then, $p$ divides $b+c$, as it divides $S=a+b+c$. But we also have that $p$ divides $a^{12}+b^{12}+c^{12}$, and looking modulo $p$, we get that $a^{12}+b^{12}+c^{12}\equiv 0^{12}+b^{12}+(-b{)}^{12}\equiv 2b^{12}(\mathrm{mod}p)$. It follows that $p$ divides $b$ and, as a consequence, it divides $c$, contradicting the fact that $a,b$ and $c$ are coprime. We conclude that $S$ does not have a prime divisor greater than 3.

Hence, $S=2^x{3}^y$ for non-negative integers $x$ and $y$. Finally, we will show that $x,y\le 1$. If $x\ge 2$, we have that $a^{12}+b^{12}+c^{12}\equiv 0(\mathrm{mod}4)$. As the quadratic residues modulo 4 are 0 and 1, the only possibility is that $a\equiv b\equiv c\equiv 0(\mathrm{mod}2)$, contradicting the coprimality of $a,b,c$. Similarly, if $y\ge 2$, we have that $a^{12}+b^{12}+c^{12}\equiv 0(\mathrm{mod}9)$ but, taking into account that for an integer $m$, the possible residues of $m^6$ modulo 9 are 0 and 1, this implies that $a\equiv b\equiv c\equiv 0(\mathrm{mod}3)$, which is again a contradiction.

Therefore, the possible values of $S=a+b+c$ are 3 and 6, since $a,b,c$ are positive integers and, consequently, the possible triples $\{a,b,c\}$ are $\{1,1,1\}$, $\{1,2,3\}$, $\{1,1,4\}$ and $\{2,2,2\}$. It is clear that the first one satisfies the conditions and that the last one is not a solution because $a,b,c$ are not coprime. Now, $\{1,2,3\}$ is not a solution either, since $1+2+3=6$ does not divide $1^{12}+2^{12}+3^{12}$ (this number has residue 2 modulo 3). To check that $\{1,1,4\}$ satisfies the conditions, it suffices to note that $1^r+1^r+4^r\equiv 0(\mathrm{mod}2)$ and $1^r+1^r+4^r\equiv 0(\mathrm{mod}3)$ for every positive integer $r$.

We conclude that the solutions are $\{1,1,1\}$ and $\{1,1,4\}$.