If a,b are amicable numbers show that $(\sum_{d|a} d^{-1})^{-1} + (\sum_{d|b} d^{-1})^{-1}=1$ Amicable numbers are two different numbers so related that the sum of the proper divisors of each is equal to the other number. (A proper divisor of a number is a positive integer divisor other than the number itself. For example, the proper divisors of 6 are 1, 2, and 3
.)
For example, the smallest pair of amicable numbers is (220,284)
; for the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110, of which the sum is 284; and the proper divisors of 284 are 1, 2, 4, 71, and 142, of which the sum is 220. (from the Wikipedia article) 
They are used in astrology so I guess they are pretty useful.

This is problem 8.4.51 from Elementary Number Theory from Koshy.

 A: I'm pretty sure the sum is over all divisors, not just the proper ones.
Because the divisors of $a$ are the same numbers as the quotients of $a$ by each divisor, we can re-index the sum:
$$ \sum_{d\mid a} \frac 1d = \sum_{d'\mid a} \frac1{a/d'} = \sum_{d'\mid a} \frac{d'}{a} = \frac1a \sum_{d'\mid a} d' $$
And since the sum of $a$'s divisors except $a$ itself is $b$ we get further
$$ \sum_{d\mid a} \frac1d = \cdots = \frac1a(a+b) $$
Can you take it from here?
A: Want to show
$(\sum_{d|a} d^{-1})^{-1} + (\sum_{d|b} d^{-1})^{-1}=1
$.
If $a$ and $b$ are amicable,
then
$b
=\sum_{d|a, d<a} d
$
and
$a
=\sum_{d|b, d<b} d
$.
We have
$\begin{array}\\
\sum_{d|a} d^{-1}
&=\dfrac1{a}\sum_{d|a} \dfrac{a}{d}\\
&=\dfrac1{a}\sum_{d|a} d\\
&=\dfrac1{a}(a+\sum_{d|a, d<a} d)\\
&=1+\dfrac1{a}\sum_{d|a, d<a} d\\
&=1+\dfrac{b}{a}\\
&=\dfrac{a+b}{a}\\
\text{so}\\
\dfrac1{\sum_{d|a} d^{-1}}
&=\dfrac{a}{a+b}\\
\text{and}\\
\dfrac1{\sum_{d|b} d^{-1}}
&=\dfrac{b}{a+b}\\
\end{array}
$
Adding these,
$\dfrac1{\sum_{d|a} d^{-1}}
+\dfrac1{\sum_{d|b} d^{-1}}
=\dfrac{a}{a+b}
+\dfrac{b}{a+b}
=1
$.
A: One thing to observe is that $\frac{a}{d}=d^{'}$ which is also a divisor. Thus
$$\left(\sum_{d|a}\frac{1}{d}\right)^{-1} + \left(\sum_{d|b}\frac{1}{d}\right)^{-1}=
\left(\frac{1}{a}\sum_{d|a}\frac{a}{d}\right)^{-1} + \left(\frac{1}{b}\sum_{d|b}\frac{b}{d}\right)^{-1}=\\
\left(\frac{1}{a}\sum_{d^{'}|a}d^{'}\right)^{-1} + \left(\frac{1}{b}\sum_{d^{'}|b}d^{'}\right)^{-1}=
\left(\frac{a+b}{a}\right)^{-1} + \left(\frac{a+b}{b}\right)^{-1}=\\
\frac{a}{a+b}+\frac{b}{a+b}=1$$
