Please change the topic names. The fact that it is RMO 2012 problem is more important than the fact that it is "similar to RMO 2002".

Actually for any $n \in \mathbb{N}$, if $a \mid b^n, b \mid c^n $ and $c \mid a^n$ then $abc \mid (a+b+c)^{n^2 +n+1}$. All terms appearing in the expansion $(a+b+c)^{n^2 +n+1}$ is summation of terms of the form $a^i b^j c^k$, with $i,j,k \ge 0$ and $i+j+k = n^2 + n + 1$. If $i,j,k \ge 1$, then $a^ib^jc^k$ is divisible by $abc$. Otherwise there are two types of cases. (1) When any two of $i,j,k$ is zero. WLOG let $i=j= 0$, then $k = n^2 + n+1$. We show that $abc \mid c^{n^2+n+1}$. Indeed $c \mid c$, $b \mid c^n$ and $a \mid c^{n^2}$. (Since $a \mid b^n$ and $b \mid c^n$, we get $a \mid b^n \mid (c^n)^n = c^{n^2}$. ) Multiplying them $abc \mid c^{n^2 + n+1}$. The other case, type (2) is when exactly one of $i,j,k$, WLOG $i$ is equal to zero. Then $j+k = n^2 + n + 1$. So one of $k$ and $l$ is $\ge \frac{n(n+1)}{2} + 1$ and the other is $\le \frac{n(n+1)}{2} $. If $j \ge \frac{n(n+1)}{2} + 1$ we easily get that $abc \mid b^j c^k$. If $j \le \frac{n(n+1)}{2} $, we start having a problem if $j \le n$, but then $k \ge n^2 + 1$. So $b \mid b^j$, $c \mid c$ and $a \mid c^{n^2}$, so $abc \mid b^jc^k$. So we are done.

Another solution Let $v_p(a)=x,v_p(b)=y,v_p(c)=z$ and fix $x\le y\le z$ Then by $a|b^5,b|c^5,c|a^5$ hence $5y\geq x$ $5z\geq y$ $5x\geq z$ then $25x\geq y$ We have $v_p(abc)=x+y+z$ $v_p(a+b+c)\geq min\{x,y,z\}=x$ hence $v_p((a+b+c)^31)\geq 31x$ We will prove $31x\geq x+y+z \Rightarrow 30x\geq y+z$ But we have proven that $25x\geq y,5x\geq z$ then $30x\geq y+z$ is obvious, then done!

tahanguyen98 wrote: Another solution Let $v_p(a)=x,v_p(b)=y,v_p(c)=z$ and fix $x\le y\le z$ Then by $a|b^5,b|c^5,c|a^5$ hence $5y\geq x$ $5z\geq y$ $5x\geq z$ then $25x\geq y$ We have $v_p(abc)=x+y+z$ $v_p(a+b+c)\geq min\{x,y,z\}=x$ hence $v_p((a+b+c)^31)\geq 31x$ We will prove $31x\geq x+y+z \Rightarrow 30x\geq y+z$ But we have proven that $25x\geq y,5x\geq z$ then $30x\geq y+z$ is obvious, then done! We can not fix $x\le y\le z$,but we can denote $min\{x,y,z\}=x$.

what is RMO???

It's mathematics olympiad stage 1 in India