Chartrand Mathematical Proofs 3e Exercise 5.46 I am self-studying this book, and am stuck on this question:
Prove that there exist four distinct positive integers such that each integer divides the sum of the remaining integers
This is what I have so far, but I am not sure where to go next:
Let a, b, c  and d be distinct positive integers. Without loss of generality, let $a<b<c<d$.
Then $$a\mid(b+c+d)$$ $$b\mid(a+c+d)$$ $$c\mid(a+b+d)$$ $$d\mid(a+b+c)$$
Since a,b,c and d are positive,
$$ap = b + c + d, p\in \mathbb Z^+ $$
$$bq = a + c + d, q\in \mathbb Z^+ $$
$$cr = a + b + d, r\in \mathbb Z^+ $$
$$ds = a + b + c, s\in \mathbb Z^+ $$
so
$$ ap + bq + cr + ds = 3(a+b+c+d)$$
One solution to this equation is when $p=q=r=s=3$
...and I'm stuck. 
Am I even going in the right direction?
 A: Consider the positive integers $1$, $2$, $3$, and $6$.
A: Here is a systematic approach: note that $a \mid b+c+d$ is equivalent to $a \mid a+b+c+d$.  So we can rephrase the problem as finding $a,b,c,d$ that simultaneously divide $a+b+c+d$.  If we define $n=a+b+c+d$ for a putative solution then we can rephrase this yet again as "find an integer $n$ and four distinct divisors of $n$ which add up to $n$.
Now, any divisor of $n$ can be written as $n/A$ for some positive integer $A$.  So then we have yet another (equivalent, though less obviously so) form:
$$\frac{n}{A} + \frac{n}{B} + \frac{n}{C} + \frac{n}{D} = n,$$
whereupon the $n$ factors away and leaves the simple Egyptian fraction problem $1/A + 1/B + 1/C + 1/D = 1$.  (For the equivalence to be precise, we also need $n$ to be divisible by $A,B,C,D$ but this is easily done by taking LCMs).  This is a very well-studied problem, and it is not hard to see there are only finitely many solutions $(A,B,C,D)$ (see for instance, this similar question if you're unfamiliar with the finiteness argument).
In particular, assuming $A<B<C<D$ we have the six solutions
$$(A,B,C,D) \in \{(2,3,7,42), (2,3,8,24), (2,3,9,18), (2,3,10,15), (2,4,5,20), (2,4,6,12) \},$$
which give rise to the following primitive (in the sense of no common factor) solutions to the original equation:
$$(1,6,14,21),\\(1,3,8,12),\\(1,2,6,9),\\(2,3,10,15),\\(1,4,5,10),\\(1,2,3,6).$$
