part a) 1)if $ k = 1$ then $ \left (\begin {array} {c} n \\ j\end{array} \right)$ an arithmetic progression. 2)if $ k = 2$ then $ \left (\begin {array} {c} n \\ j\end{array} \right), \left (\begin {array} {c} n \\ {j + 1}\end{array} \right)$ an arithmetic progression. 3)if $ k \geq 3$ then $ \left (\begin {array} {c} n \\ j\end{array} \right), \left (\begin {array} {c} n \\ {j + 1}\end{array} \right), \left (\begin {array} {c} n \\ {j + 2}\end{array} \right)$ an arithmetic progression so that: $ \left (\begin {array} {c} n \\ j\end{array}\right) + \left (\begin {array} {c} n \\ {j + 2}\end{array}\right) = 2\left (\begin {array} {c} n \\ {j + 1}\end{array}\right)$ so $ (j + 2)(j + 1) + (n - j)(n - j - 1) = 2(j + 2)(n - j)$ so $ 4j^2 + (4n - 8)j + (n^2 - 5n + 2) = 0$ so $ j = \frac {n - 2 + \sqrt {n + 2}}{2}$ or $ j = \frac {n - 2 - \sqrt {n + 2}}{2}$ so $ n$ be a form $ t^2 - 2$ fo $ t \in \mathbb{Z}$ and $ t \geq 2$. so for any $ n$ and $ k \geq 3$ there is not exist $ j$. part b) 1)if $ k =1,2,3$ similary part a case 1,2 for every n there is exist $ j$. 2)if $ k \geq 4$ similary part a case 3,$ n$ be a form $ t^2 - 2,t \in \mathbb{Z},t \geq 2$. so if $ k=1,2,3$ every $ n$, and if $ k = 4$ $ n$ be a form $ t^2 - 2$ for some integer $ t$.