Finding a couple of numbers which satisfy $(a + b)|(a + c)$ and $(a + b)|(2c)$ using a "systematic" method I want to find a couple of integers ($a$, $b$, $c$) which satisfy the following:
$(a+b) |(a + c)$ - "$(a+c)$ divisible by $(a + b)$"
$(a+b)|(2c)$ - "$(2c)$ divisibile by $(a+b)$"
I can try and test random numbers and eventually get something out of it, but I want to take a systematic (not random) approach towards it. 
I tried to use the definition of divisibility ($a|b \Rightarrow b=k.a$) to no avail. I'm aware that there probably is an endless number of answers to this question, but I want to get a couple of distinct answers NOT in a hit or miss way.
How do I do that?
P.S: There actually might not be any answers to such expressions so I'm also interested in knowing how to find whether there is an answer or not.
 A: Hint $ $ The statement $\,P\,$ has scaling symmetry $\,P(an,bn,cn)\iff P(a,b,c),\,$ thus cancelling $\,(a,b,c)$ reduces to case $\,(a,b,c) = 1$. Let $\, d = a\!+\!b,\,$ $\,\color{#c00}{(d,a,c)} = (a\!+\!b,a,c) = (b,a,c) = 1,\,$ so
$$d\mid a\!+\!c,2c\,\Rightarrow\, d\mid2a = 2(a\!+\!c)\!-\!2c\,\Rightarrow\, d\mid(2a,2c) = 2(a,c) \,\Rightarrow\, d\mid 2,\ \ {\rm by} \  \color{#c00}{(d,a,c)}  = 1$$
Thus we have reduced to the case of $\,(a,b,c) = 1\,$ and $\,a\!+\!b\mid 2,\,$ which is easy to  finish.
A: If $(a,b,c)$ is a solution, then $a+b\mid 2c-(a+c)=c-a$.
On the other hand, if $a+b\mid a+c$ and $a+b\mid c-a$, then also $a+b\mid (c-a)+(a+c)=2c$.
So pick integers $u,v,w$ arbitrary, except that $w$ must be even if $u,v$ have different parity, and let
$$ a=\frac{(u-v)w}2,\qquad b=w-a,\qquad c=\frac{(u+v)w}2.$$
Then $a+b=w$, $a+c=uw$ and $2c=(u+v)w$ so that the divisibility condition holds. 
(We recognize that $u=\frac{a+c}{a+b}$ and $v=\frac{c-a}{a+b}$).
A: One thing to note is that if you have one solution, you can multiply $a,b,c$ by any number and get another.  You can subtract the two to get $(a+b)|(c-a)$ and again to get $(a+b)|2a$.  This shows we need $a=b$ 
The easiest is to take $a=b$ and $c=ka$ for $k$ odd.  You didn't say they have to be distinct, but may have been thinking it.
