What rules apply for this function result to be always an integer? what ( if any ) are the main conditions/rules for function
a = (b*c)/(b+c)

to be true, when using only positive integers? This is true:
4 = (5*20)/(5+20)
4 = (6*12)/(6+12)

Meaning, is there a specific pattern, how the values form?
 A: The assumption $b + c \mid b c$ implies
$$\tag{formula}
\frac{b+c}{\gcd(b+c, b)}
=
\frac{b+c}{\gcd(c, b)} \mid c.
$$
Write $d = \gcd(b, c), b = s d, c = t d$, so that $\gcd(s,t) = 1$. 
(formula) yields that $s + t$ divides $c$, and similarly $s + t$ divides $b$. It follows that $s + t$ divides $\gcd(b,c)$.
This yields the following general construction rule for $b, c$: take any positive integer $d$, and coprime non-negative integers $s, t$ such that $s + t \mid d$. Define
$$
b = s d, \qquad c = t d,
$$
so that $d = \gcd(b, c)$.
Then
$$
b c = s t d^{2}, \qquad b+c = (s+t) d,
$$
so that $b + c \mid bc$.
The cases listed by OP occur for $d = 5, s = 1, t = 4$, and $d = 6, s= 1, t = 2$. Another example is $d = 5, s = 2, t = 3$, which yields $b = 10, c = 15$.
A: $\begin{eqnarray}\rm B=db,&&\rm (B,C) = d\\ \rm C = dc, &&\rm \color{blue}{(b,c)\ =\  1}\end{eqnarray}\rm   \,\Rightarrow\,\  B\!+\!C\mid BC\!\! \overset{cancel \ d\!\!}\iff b\!+\!c\mid \color{#c00}b\,\color{#0a0}c\,d\!\iff\! b+c\mid d\,$ by $\,\begin{eqnarray}\rm (b\!+\!c,\color{#c00}b) = \color{blue}{(c,b)=1}\\ \rm (b\!+\!c,\color{#0a0}c)\, = \color{blue}{(b,c)= 1}\end{eqnarray}$
A: $b$ and $c$ must have a common divisor.
Suppose they do not. Then $b+c$ cannot divide b, since c does not divide b, and likewise $b+c$ cannot divide c, since b does not divide c.
edit: per the comments, I should say that $gcd(b,c)>1$ and also that $b|b+c$ and $c|b+c$, as opposed to my comment that $b+c|b$ which is logically impossible since $b+c>b$.
