Equation with divisors 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. 
 A: My comment above was incorrect; I relied on Wolfram Alpha which gave me a false answer. There are in fact infinitely many solutions, for example $k = 35$ which gives $m = 119$. Indeed, take $n = 2^{120}$, $m = 119$ and $k = 35$.  This gives a counterexample to 3, since $$D_k^4 = (2*...*2^{35})^4 = 2^{2*35*36} = 2^{2520} = 2^{120*(4*35 - 119)} = n^{4k - m}$$
A: Thanks to cocopuffs it is clear now how to construct counterexamples.
Let $n = p^{a+1}$ here p is any prime number. It is easy to see in this case, that
$m=a$, $D_k^4=(p \cdot p^2 \cdot ... \cdot p^k)^4=p^{2k(k+1)}$ and $n^{4k-m}=(p^{a+1})^{4k-m}= p^{(m+1)(4k-m)}$. If $D_k^4=n^{4k-m}$ then 
$(m+1)(4k-m)=2k(k+1)$ and we have Diophantine equation: $2k - m + 4 k m- 2 k^2 - m^2 = 0$ with condition $k<m$ . First solution is $(k=1, m=3)$, second solution is $(k=6, m=20)$. 
So $p=2, \ a=m=20, \ k=6, \ n=p^{21}$ is the smallest counterexamples.
By the way $k=35, m=119$ is the 3rd solution. 
$$
k_n=\frac{(3 + 2\sqrt{2})^n-(3 - 2\sqrt{2} )^n}{4 \sqrt{2}},
$$ 
$$
m_n=\frac{1}{4} \left(\left(1+\sqrt{2}\right) \left(3+2 \sqrt{2}\right)^m+\left(1-\sqrt{2}\right)\left(3-2 \sqrt{2}\right)^m -2\right).
$$
$k_1=1, \ k_2=6, \ k_3=35, \ k_4=204$ and $m_1=3, \ m_2=20, \ m_3=119, \ m_4=696$.
$k_n=6k_{n-1} - k_{n-2}, \ \ m_n=6m_{n-1} - m_{n-2} + 2$.
