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.

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 $n=2^2\cdot 7=28$. 
