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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.

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  • $\begingroup$ Not up to 1024, at any rate. The following mindless exhaustive search took 3 minutes on gp; one can surely do better by searching over $b+c$ and $a^2$ first since only a tiny fraction of candidates will survive this initial test. $\endgroup$ – Noam D. Elkies Mar 12 '17 at 0:31
  • $\begingroup$ H = 1024; for(a=1,H,for(b=a,H,for(c=b,H,if(content([a,b,c])==1,if(a^2%(b+c)+b^2%(c+a)+c^2%(a+b),,print([a,b,c])))))) $\endgroup$ – Noam D. Elkies Mar 12 '17 at 0:32
  • $\begingroup$ I suspect there are no positive integer solutions. No proof to back it up with. $\endgroup$ – T. Amdeberhan Mar 12 '17 at 0:44
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    $\begingroup$ A quite similar question appeared in a German math contest for high school students "Bundeswettbewerb Mathematik", see mathe-wettbewerbe.de/bwm/aufgaben (2008, round 2). $\endgroup$ – Jens Reinhold Mar 12 '17 at 7:11
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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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