1. Find all natural numbers $a, b$, and $c$ that satisfy the equation

$$
\frac{a+b}{a+c}=\frac{b+c}{b+a}
$$

and for which $a b+a c+b c$ is a prime number.
II/1. Let's denote the prime number $a b+b c+c a$ by $p$. From the relation, we get that $(a+b)^{2}=c^{2}+p$ or $p=(a+b-c)(a+b+c)$. Since $p$ is a prime number and $0<a+b-c<a+b+c$, it must be that $a+b-c=1$ and $a+b+c=p$ or $c=\frac{p-1}{2}$ and $a+b=\frac{p+1}{2}$. On the other hand, $a b=p-b c-c a=p-c(a+b)=p-\frac{p-1}{2} \cdot \frac{p+1}{2}=-\frac{p^{2}-4 p-1}{4}$ must be a positive number. The roots of the quadratic function $f(x)=x^{2}-4 x-1$, which has a positive leading coefficient, are $2-\sqrt{5}$ and $2+\sqrt{5}<5$, so $p$ can only be 2 or 3. Since 2 cannot be, as $c=\frac{p-1}{2}$ would not be an integer, $p=3$. Thus, $c=1, a+b=2$ and since $a$ and $b$ are natural numbers, the only solution is $a=b=c=1$. The triplet $(1,1,1)$ clearly satisfies all the conditions of the problem.