Square numbers and cube numbers

Since $\begin{aligned} (n+2)n+1=n^2+2n+1=(n+1)^2 \\ \ \\ (n^2+3n+3)n+1=n^2+3n^2+3n+1=(n+1)^3 \end{aligned}$ the $n+2$ . item of the sequence is always a square number, and the $n^2+3n+3$ . item of the sequence is always a cube number.

Therefore the answer is $\text{Both of A and B}$ .

The general term of the sequence is $kn+1$ , where $k \in \mathbb N$ . Assuming statement A is true that is a perfect square $m^2$ term exists in the sequence. Then we have:

$\begin{aligned} kn + 1 & = m^2 \\ kn & = m^2 - 1 \\ kn & = (m-1)(m+1) \end{aligned}$

Let $n=m-1$ and $k=m+1$ , $\implies k = n+2$ , then $kn+1 = (n+2)n+1 = n^2+2n+1=(n+1)^2$ . Therefore, statement A is true.

Similarly for statement B :

$\begin{aligned} kn + 1 & = m^3 \\ kn & = m^3 - 1 \\ kn & = (m-1)(m^2+m+1) \end{aligned}$

Let $n=m-1$ and $k = m^2+m+1$ , $\implies k = n^2+3n+3$ , $kn+1 = n^3+3n^2+3n + 1 = (n+1)^3$ . Therefore, statement B is true.

Therefore, the answer is both A and B .