Obviously the general case also holds. We may assume $a_i\neq0$ and $a_i+a_{i+1}\neq0$. For n=1, we have $a_1=1$. Suppose $a_n=n$ for all n. Then\[1^3+...+n^3+a_{n+1}^3=(1+...+n+a_{n+1})^2\]Solving this gives $a_{n+1}=n+1$.

I see most of the problems where taken from Russian olympiad!

Where can I find them?

I dont know, I have them in a book: "Problems and solutions from arround the world" i think they are from 2000-2001 . About a web site with them as i understand you can only find them in Russian, but i am not very sure!

Yes,I'm pretty sure I've already seen this problem

Yes Russia 2000. Pierre.

The problem was proposed in 1978 at the county level of RMO (I was a competitor at that time...and I solved it ) under the more restrictive condition all $a_i$'s are positive reals (proposed by L. Panaitopol). However, if this condition is dropped, there are infinitely many sequences satisfying the given equality. mecrazywong wrote: We may assume $a_i\neq0$ and $a_i+a_{i+1}\neq0$. We cannot assume this WLOG, otherwise, we miss solutions like $1,-1,1,-1,\ldots$ or like $1,2,-2,2,3,-3,\ldots$. The problem, as stated at ARO 2000 can be solved by induction, using the following (strenghted) induction hypothesis: for each $n$, the numbers $a_1,a_2,\ldots ,a_n$ are integers and their sum is a triangular number, that is, for each $n$ there exists $k_n$ such that $a_1+a_2+\ldots a_n=\frac {k_n(k_n+1)}{2}$.

very nice problem i have seen this problem in "problems from national and regional olympiad 2000-01"(maa) and it was the problem of our imo selection test2002