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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.

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    If you multiply the equation with $n$, does that remind you of something? – Daniel Fischer Jul 19 '15 at 22:34
  • Clearly $n$ cannot be a prime power, that indeed leaves you with the case $n = a^2b$. Rewrite $\sum_{d_i | n} d_i = 2n$ as $(b+1)(a^2-a-1) =2a^2$. Then $a^2 - a-1 = 1,2$ (since, $(a^2,a^2 -a-1) =1$). That gives you $b$ too. – r9m Jul 19 '15 at 22:43

3 Answers3

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

mathlove
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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.

André Nicolas
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Note that $d_1 d_6 = d_2 d_5 = d_3 d_4 = n$. It can be very useful.

Lefundes
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