Prime factorization number theory Let $n$ be a positive integer, and let
$ 1=d_1<d_2<\dots <d_6=n $
be all of its divisors. Find all $n$ that satisfy
$ \frac 1{d_1} +\frac 1{d_2} + \dots + \frac 1{d_6 } = 2. $
I started by noting $n=a^2b$ where $a,b$ are primes, and tried to manipulate the second equation. Not getting anywhere
solution would help.
 A: As you wrote, $n$ can be expressed as $n=a^2b$ where $a,b$ are distinct primes. Now we have
$$\frac{1}{1}+\frac{1}{a}+\frac{1}{b}+\frac{1}{a^2}+\frac{1}{ab}+\frac{1}{a^2b}=2,$$
i.e.
$$a^2b+ab+a^2+b+a+1=2a^2b,$$
i.e.
$$b=1+\frac{2a+2}{a^2-a-1}$$
Here, since $(2a+2)/(a^2-a-1)\lt 1$ for $a\ge 4$, we have $a=2,3$. 
Thus, the answer is $n=2^2\cdot 7=28$. 
A: The numbers $\frac{n}{d_1}$, $\frac{n}{d_2}$, and so on range (backwards) over all the divisors of $n$. So multiplying by $n$ we find that $\sigma(n)=2n$, where $\sigma$ is the sum of divisors function.
So $n$ is perfect and has $6$ divisors.
Recall that the even perfect numbers have the shape $2^{p-1}(2^p-1)$ where $2^p-1$ is a Mersenne prime. The only way we get $6$ divisors is if $p=3$, that is, $n=28$.
But maybe $n$ could be an odd perfect number. In fact, it could not be, for an odd perfect number must have more than $6$ divisors, for it cannot be of shape $p^5$ or $p^2q$, where $p$ and $q$ are prime.
A: Note that $d_1 d_6 = d_2 d_5 = d_3 d_4 = n$. It can be very useful.
