15.13. Find all solutions in natural numbers of the equation $C_{n}^{m-1}=C_{n-1}^{m}$, where $C_{n}^{m}=\frac{n!}{m!(n-m)!}$ is the binomial coefficient.
15.13. The considered equation is equivalent to the equation $m n=(n-m)(n-m+1)$. Let $d=$ GCD $(m, n), n=d a$ and $m=d b$. After reducing by $d$, we get the equation $a b d=(a-b)((a-b) d+1)$. The number $d$ is coprime with $(a-b) d+1$, and the number $a b$ is coprime with $a-b$, so $a b=(a-b) d+1$ and $d=a-b$. Substituting the expression $a=b+d$ into the first equation and replacing $a-b$ with $d$, we obtain $b^{2}+b d=d^{2}+1$, i.e., $b^{2}+b d-d^{2}=1$. In problem 15.12, a solution to a slightly more general equation is given, where the right-hand side is $\pm 1$. Discarding the solutions that correspond to -1, we get $d=F_{2 k}, b=F_{2 k-1}$ and $a=b+d=F_{2 k+1}$. Thus, $n=F_{2 k} F_{2 k+1}$ and $m=F_{2 k} F_{2 k-1}$.

Remark. The equality $C_{n}^{m-1}=C_{n-1}^{m}$ is equivalent to the equality $C_{n-1}^{m-1}+C_{n-1}^{m-2}=C_{n-1}^{m}$.