Do there exist any three relatively prime natural numbers so that the square of each of them is divisible by the sum of the two remaining numbers? $\textbf{Question:}$Do there exist any three relatively prime natural numbers so that the square of each of them is divisible by the sum of the two remaining numbers?
that is pairwise  relatively prime $a,b,c \in \mathbb N  $($0$ not included) such that $a+b \mid c^2,b+c \mid a^2,c+a \mid b^2$
$\textbf{Stuffs I have tried:}$
I have tried a bunch of constructions like :$(p,\frac{p+1}{2},\frac{p-1}{2})$,one of them are the sum of other two,tried small values.But was unable to find any valid triple.So,I began trying to prove such triple can't exist.I tried to bound the numbers,tried to use modular arithmatic but failed again.
Any kind of hint or solution will both be appreciated.

Well, let's see. So we have
$$\begin{cases}
a^2 = x(b+c) \\
b^2 = y(a+c) \\
c^2 = z(a+b)
\end{cases}$$
where all quantities are integers. What does this mean?
For one thing, each of $a,b,c$ is strictly greater than 1.
For another, if $a,b,c$ are all pairwise coprime (as per the problem statement), then so are $b+c, a+c, a+b$.
Now let's make one peculiar combination out of these numbers, for no reason at all:
$$(a+b+c)^2 = a^2+2ab+2ac+(b+c)^2=x(b+c)+2a(b+c)+(b+c)^2$$
So it is divisible by $b+c$, and also (following similar reasoning) by $a+b$ and $a+c$. If these three are all mutually coprime, then it is divisible by their product.
Assume that $a<b<c$. Then $b+c>{2\over3}(a+b+c)$ and $a+c>{1\over3}(a+b+c)$. So...
$$(a+b)(b+c)(a+c)\;|\;(a+b+c)^2\\
(a+b)(b+c)(a+c)\leqslant(a+b+c)^2\\
(a+b)\cdot{1\over3}(a+b+c)\cdot{2\over3}(a+b+c)<(a+b+c)^2\\
a+b < {9\over2}\\
a+b \leqslant 4$$
With $a,b>1$ this only leaves us the option of $a=b=2$, which are not coprime.
Finally, there are no such coprime triples. Q.e.d.
