When:

$$\frac{a^2}{b+c}＋\frac{b^2}{a+c}＋\frac{c^2}{a+b}\in\mathbb Z$$ and $\{a,b,c\}$ are coprime natural numbers and $a,b,c>1$, is there a solution except $\{183,77,13\}$?

Given $a,b,c\in \Bbb{N}$ such that $\{a,b,c\}$ are coprime natural numbers and $a,b,c>1$. When $$\frac{a^2}{b+c}＋\frac{b^2}{c+a}＋\frac{c^2}{a+b}\in\mathbb Z\,?$$ I know the solution $\{183,77,13\}$. Is there any other solution?

When:

$$\frac{a^2}{b+c}＋\frac{b^2}{a+c}＋\frac{c^2}{a+b}\in\mathbb Z$$ and $\{a,b,c\}$ are coprime natural numbers, is there a solution except $\{183,77,13\}$?

When:

$$\frac{a^2}{b+c}＋\frac{b^2}{a+c}＋\frac{c^2}{a+b}\in\mathbb Z$$ and $\{a,b,c\}$ are coprime natural numbers and $a,b,c>1$, is there a solution except $\{183,77,13\}$?