Let's determine the integer solutions of the equation $\sqrt{a}+\sqrt{b}=\sqrt{c}$.
It is clear that when $c=0$, only $a=b=0$ is possible. In the following, we can assume that $c>0$. Then we can divide the initial equality by $\sqrt{c}$, and thus we arrive at the equality $\sqrt{u}+\sqrt{v}=1$, where $u=\frac{a}{c}$ and $v=\frac{b}{c}$ are rational numbers. However, from $(\sqrt{u}+\sqrt{v})(\sqrt{u}-\sqrt{v})=u-v$, it follows that $\sqrt{u}-\sqrt{v}=u-v$ is also a rational number. From this, we get that $\sqrt{u}=\frac{1+u-v}{2}$ and $\sqrt{v}=\frac{1-u+v}{2}$ are also rational, meaning $u=p^{2}$ and $v=q^{2}$, with suitable rational numbers $p$ and $q$. In the end, the only possible solutions are $a=c \cdot p^{2}, b=c \cdot q^{2}$. Of course, the initial equality can only hold if $p+q=1$. Additionally, $c$ must be chosen such that both $a=c \cdot p^{2}$ and $b=c \cdot q^{2}$ are integers; when indeed the equality holds.