Determine all pairs $(n, k)$ of integers such that $0<k<n$ and

$$
\left(\begin{array}{c}
n \\
k-1
\end{array}\right)+\left(\begin{array}{c}
n \\
k+1
\end{array}\right)=2\left(\begin{array}{l}
n \\
k
\end{array}\right)
$$
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 $n-k+1=b$ :

$$
\begin{gathered}
(a-1) a+(b-1) b=2 a b \\
a^{2}-2 a b+b^{2}=a+b \\
(a-b)^{2}=a+b
\end{gathered}
$$

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.