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.
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.
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.