I think the only solution is $(1,1,1)$. I used FMID. Not sure of it though

there are infinite solutions (of one parameter)

Let $p|(a,b)$ where $p$ is prime. Then $p|n^2 \to p|n \to p| n \cdot lcm(b, c) + n^2 \to p|c$ If we divide $a,b,c,$ by $p$ then we get same system So let $(a,b)=(b,c)=(c,a)=1$ Then $lcm(a,b)=ab$ and so we have system $a^2 + b^2 = n\cdot ab + n^2$ $b^2 + c^2 = n \cdot bc + n^2$ $c^2 + a^2 = n \cdot ca + n^2$ Let $a \geq b,a \geq c$ $b^2-c^2=na(b-c)$ $(b-c)(b+c-na)=0$ Case 1: $b=c$ $2b^2=nb^2+n^2$ $b^2=\frac{n^2}{2-n} \to n=1,b=c=1, a=1$ Case 2: $b+c=na \to n=2,a=b=c \to $ contradiction So solution is $a=b=c=n$