Bulgaria

By applying the formula $\left(\genfrac{}{}{0ex}{}{a}{b}\right)=\frac{a!}{b!(a-b)!}$ we obtain the equation $$1+\frac{(n-k{)}^2}{(k+1{)}^2}=\frac{(n-k{)}^2(n-k-1{)}^2}{(k+1{)}^2(k+2{)}^2}{\left(\genfrac{}{}{0ex}{}{n}{k+2}\right)}^2.$$ Hence $(k+2{)}^2[(k+1{)}^2+(n-k{)}^2]=(n-k{)}^2(n-k-1{)}^2{\left(\genfrac{}{}{0ex}{}{n}{k+2}\right)}^2$, which implies that $(k+1{)}^2+(n-k{)}^2$ is a perfect square. Let $(k+1{)}^2+(n-k{)}^2=t^2$, where $t\in \mathbb{N}$. We have $$(k+2)t=(n-k)(n-k-1)\left(\genfrac{}{}{0ex}{}{n}{k+2}\right)\ge 2\left(\genfrac{}{}{0ex}{}{n}{k+2}\right).$$ Using that $k+2\le n$ and $t=\sqrt{(k+1{)}^2+(n-k{)}^2}<n+1$ we conclude that $(k+2)t\le n^2$. Let $3\le k+2\le n-3$ (the left hand side of this inequality follows from the condition of the problem). If $n\ge 6$, we have $$2\left(\genfrac{}{}{0ex}{}{n}{k+2}\right)\ge 2\left(\genfrac{}{}{0ex}{}{n}{3}\right)=\frac{n(n-1)(n-2)}{3}>n^2,$$ i.e. the equation has no solution in this case. When $k+2=n-2$ we obtain $t^2=(n-3{)}^2+16$, hence $t=5$, $n=6$, $k=2$. Direct computation shows that $n=6$ and $k=2$ is not a solution. If $k+2=n-1$ we have $t^2=(n-2{)}^2+9$, so $t=5$, $n=6$, $k=3$ and as above we conclude that no solution exists. Therefore positive integers $n$ and $k$ satisfying the equation do not exist.