# Equation with divisors II

Upd$^{*}$: I've followed Matthew Conroy advice and found "amazing" numbers such as $2^6 \cdot p$, $3^6 \cdot p$.

Upd$^{**}$: If $n=p^6 \cdot q$, $(p,q)$ - primes $\Rightarrow m=12$. Let $k=4 \Rightarrow n^{4k-m}=n^4=(p^6 \cdot q)^4$. In this case from $D_k^4=n^{4k-m}$ follows that $D_4=n=(p^6 \cdot q)$ and sufficient condition: First 4 divisors of $n$ are $\{p,p^2,p^3,q\}$. So, if $p^2<q<p^4$ then $n=p^6\cdot q$ - "amazing" number. And we have a new guess: (I didn't find any other types of "amazing" numbers $<10^9$).

Upd$^{***}$: $$\text{Let } n=(p^\alpha \cdot q^\beta)^4 \Rightarrow D_k= (p^\alpha \cdot q^\beta)^{4k+2-(4\alpha+1)(4\beta+1)}.$$

This give us some other types. I found one more type: $p^{16} \cdot q^8$. And the smallest "amazing" number of this type: $3^{16} \cdot 7^8=248 155 780 267 521$.

Short description

For any $n \in \mathbb{N}$ - we can divide the set of all its divisors (except 1 and n) on 2 subsets $S_1^k(n)=\{ d_1 < d_2 < \dots < d_{m-k} \}$; $S_2^k(n)=\{ d_{m-k+1} < \dots < d_m \}$. If for n exist such $k$ that $\prod\limits_{i=1}^{m-k}d_i=\prod\limits_{j=m-k+1}^{m}d_j$ we call $n$ an "amazing" number. This problem is about: 1. how many "amazing" numbers are in some cases. 2. how to solve equation $\prod\limits_{i=1}^{k}d_i=n^{4k-m}$.

Here I will try to explain a little and very interesting problem (for me). May be it is a well known issue. First I describe all my definitions and variables:

1. Let $S(n)=\{1 < d_1 < d_2 < \dots < d_m < n \}$ is the set of all divisors of $n \in \mathbb{N}$. And $S_1^k(n)=\{ d_1 < d_2 < \dots < d_{m-k} \}$; $S_2^k(n)=\{ d_{m-k+1} < \dots < d_m \}$; $m>2k$; $m=\sigma_0(n)-2$.
2. $D_r=\prod\limits_{i=1}^r d_i =d_1 \cdot d_2 \cdot \ ... \ \cdot d_r ; \ d_i \in S(n); \ 1\leq r \leq m$.
3. $P_1^k(n)=\prod\limits_{i=1}^{m-k}d_i ; \ d_i \in S_1^k(n)$ - product of all elements $S_1^k(n)$. $P_2^k(n)=\prod\limits_{i=m-k+1}^{m}d_i ; \ d_i \in S_2^k(n)$.

We will call the number $n \in \mathbb{N}$ "amazing" if there exist some integer $k \ \ \left(2k<m(n) \right)$ which divides the set $S(n)$ into two subsets $S_1^k(n)$, $S_2^k(n)$ such that $P_1^k(n)=P_2^k(n)$. Let $\mathbb{A}$ - is set of all "amazing" numbers.

Easy to see that all $n=p^4, \ (p$ is any prime) - are "amazing". So we consider a finite subsets of the set $\mathbb{A}$ in two ways:

I. $\mathbb{A_1}(n)$ - all "amazing" numbers which are less then n. $\mathbb{A_1}(n)=\{a : a \in \mathbb{A}, n \in \mathbb{N}, a < n \}$.

II. $\mathbb{A_2}(n)=\{a : a \in \mathbb{A}, n \in \mathbb{N}, d_{m-k}(a) < n < d_{m-k+1}(a)\}$.

Let $F_{1,2}(n)$ - number of all elements in $\mathbb{A_1}(n)$ and $\mathbb{A_2}(n)$.

Questions are:

$1^{*}$) Is it true that if n is "amazing" then $n=p^\alpha$ -? $p$ - prime, $\alpha \in \mathbb{N}$

2) Find the formula for $F_1(n)$ -?

3) Find the formula for $F_2(n)$ -?

$1$) New guess: Equation $D_k^4=n^{4k-m}$ has an infinite number of different types of solutions. How can I prove it?

What I have on this moment: We can write that $d_m=\frac{n}{d_1}, \dots ,d_{m-k+1}=\frac{n}{d_k}$. So if $P_1^k(n)=P_2^k(n)$ then

$$P_2^k=\frac{n^k}{D_k} \Rightarrow \ P_1^k \cdot P_2^k = D_m=\frac{n^{2k}}{D_k^2},$$

$$D_m=d_1 \cdot \ ... \ \cdot d_m=\frac{n}{d_m} \cdot \ ... \ \cdot \frac{n}{d_1} = \frac{n^m}{D_m} \Rightarrow D_m^2=n^m.$$

$$\frac{n^{4k}}{D_k^4}=n^m \Rightarrow D_k^4=n^{4k-m}.$$

$$\text{So we can say that} \ \ n \in \mathbb{A} \Leftrightarrow \ \text{if} \ \exists k \geq 1 \ \ \left(2k<m(n) \right) \ \text{such that} \ D_k^4=n^{4k-m}.$$

This give us, that if $n=p^\alpha$ then $\alpha-1=q_n$ here $q_n$ is the integer sequence which can be define by recurence formula $q_n=6q_{n-1} - q_{n-2} + 2, \ q_1=3, \ q_2=20, \ q_3=119$. This was proved here.

I have 3 types of "amazing" numbers: $p^{q_n}$; Upd$^{**}$: and Upd$^{***}$:

4) Any advices (links) about equation $D_k^4=n^{4k-m}$? Does anybody know some related equation or problem?

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Have you tried finding some "amazing" numbers? For instance, have you tried finding all amazing numbers less than 10000? If you do that, I think you'll find an answer to question 1. – Matthew Conroy Aug 19 '12 at 0:32
You right. $2^6 \cdot 5, 2^6 \cdot 7, 2^6 \cdot 11, 2^6 \cdot 13, 3^6 \cdot 11, 3^6 \cdot 13$ - "amazing numbers". Thank you. – Mike Aug 19 '12 at 8:30