Prime Power Divisors (PPD) Problem

Question

The Prime Power Divisors (PPD) of a positive integer are the largest prime powers that divide it. For example, the PPD of 450 are 2,9,25. Which numbers are equal to the sum of their PPD?

This is my attempt at the solution.

Let $m_1=p_1^{e_1}$, where $p_1$ is prime, and $e_1\in \mathbb{N}^+$. In this case it is obvious that $m_1$ is equal to the sum of its PPD.

Let $m_2=p_1^{e_1}p_2^{e_2}$, where $p_1,p_2 (p_1<p_2)$ are prime and $e_1,e_2\in \mathbb{N}^+$.

$$p_1^{e_1}p_2^{e_2}=p_1^{e_1}+p_2^{e_2}=p_1^{e_1}p_2^{e_2}\left(\frac{1}{p_2^{e_2}}+\frac{1}{p_1^{e_1}}\right)$$ Therefore $$\frac{1}{p_2^{e_2}}+\frac{1}{p_1^{e_1}}=1$$ $$\frac{1}{p_2^{e_2}}=1-\frac{1}{p_1^{e_1}}=\frac{p_1^{e_1}-1}{p_1^{e_1}}$$ $$p_2^{e_2}=\frac{p_1^{e_1}}{p_1^{e_1}-1}=1+\frac{1}{p_1^{e_1}-1}$$ Since $p_1^{e_1},p_2^{e_2}\in \mathbb{N}^+$, we must have that $$1+\frac{1}{p_1^{e_1}-1}\in \mathbb{N}^+$$ This is true only if $p_1^{e_1}=2=2^1$, i.e. $p_1=2$ and $e_1=1$. This would give us $$p_2^{e_2}=1+\frac{1}{2-1}=2^1$$ i.e. $p_2=2$ and $e_2=1$, but this contradicts our assumption that $p_1<p_2$, therefore $m_2$ cannot be equal to the sum of its PPD.

I was wondering how this can be proved in general, for $m_k, k=1,2,3,...$

Answer 1

I like dxiv's answer—but for the present purposes, the proof is even simpler than the proof of "Sum Equals Product". It's easy to check that if $a\ge2$ and $b\ge2$, then $ab\ge a+b$, with equality if and only if $a=b=2$. By induction it follows for $k\ge2$ that if $a_1,\dots,a_k\ge2$, then $a_1\times\cdots\times a_k > a_1+\cdots+a_k$ unless $a_1=\cdots=a_k=2$. Applying this to distinct prime powers, we see that $p_1^{e_1}\times\cdots\times p_k^{e_k} > p_1^{e_1}+\cdots+p_k^{e_k}$ whenever $k\ge2$; so non-prime-powers are never the sum of their prime-power-divisors.

Answer 2

No such numbers exist with $k \gt 1$ PPDs. This follows from the stronger result proved at Sum Equals Product:

Let $a_1 < a_2 < \cdots < a_k$ be positive, unequal integers. Then solutions to $$a_1+a_2+\cdots+a_k = a_1a_2\cdots a_k$$ only exist for $k=1,3$ and they are $\{n\}$ for all positive integers $n$, and $\{1,2,3\}$ respectively.

In your case the $m_j$ PPDs are all distinct and $\prod m_j = \sum m_j = n$ (so they satisfy the premise of the previous proposition), but $1$ is excluded (so the solution for $k=3$ does not apply). This leaves $k=1$ and $n=p^e$ as the only solutions.