I solved one problem in our school math competition. And I think I find the answer in more general case. But I can't prove this. I need to solve the following problem to complete my proof.
Let S=$\{1 < d_1 < d_2 < ... < d_m < n \}$ is the set of all divisors of $n \in \mathbb{N}$. Here $m=\sigma_0(n)-2$. ($\sigma_m(x)$ - Divisor Function. )
Let $k: 1 \leq k < [m/2] $ is an integer and $D_k=d_1 \cdot d_2 \cdot ... \cdot d_k$.
The equation $D_k^4=n^{4k-m}$ can be solved only if m=3, k=1.
The last 3) is my guess. I can't prove this. May be I'm wrong.