First we prove that $n+1$ is even: Let $n+1=a$, from the equation we have: $3a(a^2+2)=m^3$. So $m$ is divisible by $3$. Let $m=3k$, we have: $a(a^2+2)=9k^3$. (1) Suppose that $a$ is odd, so $gcd(a,a^2+2)=1$ (2), (because with any prime divisor $p$ of $a$, $p|a^2$, and $p$ does not divide $a^2+2$). $a^2+2$ is not divisible by 9, so from (1),(2) we have $a=9l^3, a^2+2=q^3$. We have $a^2+2$ have the form of $9k+2$ and $q^3$ have the form of $9k+1$ or $9k-1$, this is a contradiction. So $a$ is even: If $a$ is not divisible by 4, then $a=4k+2$, so $a(a^2+2)=4(2k+1)(8k^2+8k+3)$ not divisible by 8, a contradiction. So $4|n+1$.

hqdhftw wrote: $a^2+2$ is not divisible by 9, $9|5^2+2$ $9|13^2+2$ ...

By multiplying we get $3a(a^2+2)=k^3$ for some $k$. Because $a>0$ we have $gcd(3a,a^2+2)=1,2$ or $3$. If $a$ is even, we have $a^2+2\equiv 2 \pmod 4$, thus $4|a$. If $gcd(3a,a^2+2)=1$ ,then $3a=s^3$ and $a^2+2=t^3$ from where $3|a$ and $t^3\equiv 2 \pmod 9$ which is not possible. If $gcd(3a,a^2+2)=3$, then $a=s^3$ and $3(a^2+2)=3(s^6+2)=t^3$ so $s^6\equiv 7 \pmod 9$, a contradiction

can someone give a solution for the equation? (distinct from $ n+1=4 $)?

Replacing $n+1$ by $n-1$ we've $n(n^2+2)=9t^3\implies n=9x^3,n^2+2=y^3\implies 81y^6=y^3-2$ (if $n$ is odd) mod $9$ gives no solution. If $n$ is even then it's obvious that $4|n$.

Quite interesting. $n^3+(n+2)^3+(n+1)^3=m^3$ from which we obtain $(2n+2)(n^2-n^2-2n+n^2+4n+4)+(n+1)^3=m^3$ yielding $2(n+1)(n^2+2n+4)+(n+1)^3=m^3$ which further gives $(n+1)|m$ so $m=(n+1)k$ for integral $k$. That further gives, $2(n^2+2n+4)+(n+1)^2=(n+1)^2.k^3$ giving $2(n^2+2n+4)=(n+1)^2(k^3-1)$ from which we get that $2(n+1)^2+6=(n+1)^2(k^3-1)$ giving $(n+1)^2(k^3-3)=6$ which gives $n+1=1$ as the only perfect square dividing $6$ is $1$. What's happening, guys? I spent more than an hour trying to figure out if I am wrong but surprisingly, I could not find any error. Maybe I am going really wrong but then, kindly point it out to me!

See Here http://www.artofproblemsolving.com/Forum/viewtopic.php?p=506193&sid=aeee025586cb54c402f21caa5b0a09eb#p506193 By the way, what does UZMO means?

Aranya your solution is wrong in the part that $(n+1)|m^3$ does not implies $(n+1)|m$

MexicOMM, thank you for pointing out that ridiculous error. I have come up with a correct and new solution. UZMO means Uzbekistan Mathematical Olympiad. Now, I would like to post a follow-up question: try it guys. Show that the only possible positive integer $n+1$ satisfying this equation is $4$ itself i.e. there exists no other solution than $3^3+4^3+5^3=6^3$. I found this interesting and have worked it out. Try it !!

But I can't believe that Uzbekistan actually took that problem from the ARMO and put it in their national Olympiad.

I think there are very few countries that post really ORIGINAL problems. Most problems are copied or variants of each other. What is ARMO by the way?

Aranya wrote: What is ARMO by the way? All-Russian Mathematical Olympiad.