See this problem in series marathon.

Are you sure you linked to the right problem? For solutions, go to http://www.artofproblemsolving.com/Forum/viewtopic.php?f=349&t=259636 and follow the links.

Yes, I did, but it seems the search function addresses something else. Anyways, see Series Marathon, page 11, comment # 217, posted by Farenhajt.

Very hard problem for p4

Let $S_{m, r}:=\{a_i|1\leq i\leq n,\ a_i\equiv r\pmod m\}, T_{m, r}:=\{i|1\leq i\leq n,\ i\equiv r\pmod m\}$. Lemma: $\sum_{r=0}^{m-1}\binom{|S_{m, r}|}2\geq\sum_{r=0}^{m-1}\binom{|T_{m, r}|}2$. Proof: We know that if $a+2\leq b$ then $\binom a2+\binom b2>\binom{a+1}2+\binom{b-1}2$. $\therefore\sum_{r=0}^{m-1}\binom{|S_{m, r}|}2\geq\sum_{r=0}^{m-1}\binom{b_r}2$, where $b_0+\cdots+b_{m-1}=|S_{m, 0}|+\cdots+|S_{m, m-1}|=n$ and for all $i, j,\ |b_i-b_j|<2$. $\Rightarrow$ multiset $\{b_0, \cdots, b_{m-1}\}=\{\lfloor \frac nm\rfloor, \lfloor\frac{n+1}m\rfloor, \cdots, \lfloor\frac{n+m-1}m\rfloor\}=\{|T_{m, 0}|, \cdots, |T_{m, m-1}|\}$. $\therefore\sum_{r=0}^{m-1}\binom{|S_{m, r}|}2\geq\sum_{r=0}^{m-1}\binom{b_r}2=\sum_{r=0}^{m-1}\binom{|T_{m, r}|}2$. By the lemma, for any prime $p$, $\nu_p(\prod_{i<j}(a_j-a_i))=\sum_{i=1}^\infty\sum_{j=0}^{p^i-1}\binom{|S_{p^i, j}|}2\geq\sum_{i=1}^\infty\sum_{j=0}^{p^i-1}\binom{|T_{p^i, j}|}2=\nu_p(\prod_{i<j}(j-i))$.