<s>[INST] 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.
 [/INST] Observe that

*

*Clearly $ a > 1, b > 1 , c > 1$.

*WLOG $ a < b < c$

*$ 0 \equiv c^2 \equiv c(c+a+b) \equiv (a+b+c)(a+b+c) \pmod{a+b}$ (and similar expressions)

*$\gcd(a+b, b+c ) \mid \gcd( c^2, a^2) = 1$  so $\gcd(a+b, b+c) = 1$. (and similar expressions)

Hence
$$ (a+b)(b+c)(c+a) \mid (a+b+c)^2 \Rightarrow (a+b)(b+c)(c+a) \leq (a+b+c)^2. $$
Thus
$$ 4 (a+b)( b+c)^2 < 4 (a+b)(b+c)(c+a) \leq \left[2(a+b+c)\right]^2 < [3(b+c)]^2$$
which implies that $ (a+b) < \frac{ 9}{4} $, which is a contradiction.
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