Berkeley Math Circle Monthly Contest 4

Solution:
In the factorial form,
$$\frac{n!}{(k-1)!(n-k+1)!} + \frac{n!}{(k+1)!(n-k-1)!} = \frac{2 \cdot n!}{k!(n-k)!}$$
we multiply through by $(k+1)!(n-k+1)!$ to clear the fractions and then divide through by $n!$:
$$k(k+1) + (n-k)(n-k+1) = 2(k+1)(n-k+1)$$
To decrease the number of terms, we let $k+1 = a$ and $n-k+1 = b$:
$$\begin{gathered} (a-1)a + (b-1)b = 2ab \\ a^2 - 2ab + b^2 = a + b \\ (a-b)^2 = a + b \end{gathered}$$
If we let $a-b = c$, then $a + b = c^2$ and we get
$$a = \frac{c^2 + c}{2} \quad \text{and} \quad b = \frac{c^2 - c}{2}$$
Here any integer value of $c$ will yield nonnegative integer values of $a$ and $b$; however, the condition $0 < k < n$ requires that $a = k+1$ and $b = n-k+1$ are each at least $2$. Hence the values $c = -2, -1, 0, 1, 2$ are excluded, while every $c \leq -3$ and every $c \geq 3$ will yield permissible values for
$$k = a - 1 = \frac{c^2 + c - 2}{2}$$
and
$$n = a + b - 2 = c^2 - 2$$
which satisfy the equation.