<s>[INST] Dividing squares by sums This question is out of curiosity and came to me thinking about another MO question which is linked below.

Question: Do there exist positive integers $a,b,c$ such that $\gcd(a,b,c) =1 $ and each of $\frac{a^2}{b+c},\frac{b^2}{a+c},$ and $\frac{c^2}{a+b}$
  are also integers?

My question was inspired by this MO question where MAEA2 asked about coprime positive integer solutions to:
$$\begin{equation}\frac{a^2}{b+c} + \frac{b^2}{a+c} + \frac{c^2}{a+b} \in \mathbb{Z}\tag{1}\end{equation}$$
My naive self started looking for solutions to (1) via my question above, but could not find any. Jeremy Rouse has given an excellent answer using elliptic curves, but none of the points produced satisfy my question.
Note if we ask for each of $\frac{a^n}{b+c},\frac{b^n}{a+c},$ and $\frac{c^n}{a+b}$ to be integers there is clearly no solution for $n = 1,$ and for $n \geq 3$ we can take $a = 3, b = 5,$ and $c = 22$ since
$$\begin{align*}
\frac{3^n}{5+22} &= \frac{3^n}{3^3}\\
\frac{5^n}{3+22} &= \frac{5^n}{5^2}\\
\frac{22^n}{3+5} &= \frac{2^n 11^n}{2^3}.
\end{align*}.$$
Also if we remove the positive condition or the $\gcd$ condition we can find solutions like $(1,-2,3)$ or $(2,2,2)$ as well as many others.
 [/INST] First, notice that such $a$, $b$, $c$ must be pairwise coprime (e.g., if prime $p\mid \gcd(a,b)$, then $(a+b)\mid c^2$ implies $p\mid c$, a contradiction to $\gcd(a,b,c)=1$).
As divisors of pairwise coprime numbers, $a+b$, $a+c$, $b+c$ are also pairwise coprime. 
Now, since $(a+b)\mid c^2$, $(a+c)\mid b^2$, $(b+c)\mid c^2$, each of $a+b$, $a+c$, $b+c$  divides
$$D = (a+b)^2 + (a+c)^2 + (b+c)^2 - a^2 - b^2 - c^2.$$
Then their product $(a+b)(a+c)(b+c)$  must also divide $D$.
Without loss of generality, assume that $a\leq b\leq c$ and so $a+b\leq a+c\leq b+c$. Then
$(a+b)(a+c)(b+c) \leq D < 3(b+c)^2$ and hence 
$$(a+b)c < (a+b)(a+c) < 3(b+c) \leq 6c,$$ 
implying that $a+b < 6$.
So, there is a finite number of cases to consider. It is easy to check that none of them gives a solution. Namely, for any fixed $a,b$, possible $c$ must belong to the finite set:
$$\{ d-b\ :\ d\mid a^2\} \cap \{ d-a\ :\ d\mid b^2\}.$$
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