Let $c$ be a positive integer. The sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ is defined by $a_{1}=c$, and $a_{n+1}=a_{n}^{2}+a_{n}+c^{3}$, for every positive integer $n$. Find all values of $c$ for which there exist some integers $k \geq 1$ and $m \geq 2$, such that $a_{k}^{2}+c^{3}$ is the $m^{\text {th }}$ power of some positive integer.
First, notice:

$$
a_{n+1}^{2}+c^{3}=\left(a_{n}^{2}+a_{n}+c^{3}\right)^{2}+c^{3}=\left(a_{n}^{2}+c^{3}\right)\left(a_{n}^{2}+2 a_{n}+1+c^{3}\right)
$$

We first prove that $a_{n}^{2}+c^{3}$ and $a_{n}^{2}+2 a_{n}+1+c^{3}$ are coprime.
We prove by induction that $4 c^{3}+1$ is coprime with $2 a_{n}+1$, for every $n \geq 1$.
Let $n=1$ and $p$ be a prime divisor of $4 c^{3}+1$ and $2 a_{1}+1=2 c+1$. Then $p$ divides $2\left(4 c^{3}+1\right)=(2 c+1)\left(4 c^{2}-2 c+1\right)+1$, hence $p$ divides 1, a contradiction. Assume now that $\left(4 c^{3}+1,2 a_{n}+1\right)=1$ for some $n \geq 1$ and the prime $p$ divides $4 c^{3}+1$ and $2 a_{n+1}+1$. Then $p$ divides $4 a_{n+1}+2=\left(2 a_{n}+1\right)^{2}+4 c^{3}+1$, which gives a contradiction.

Assume that for some $n \geq 1$ the number

$$
a_{n+1}^{2}+c^{3}=\left(a_{n}^{2}+a_{n}+c^{3}\right)^{2}+c^{3}=\left(a_{n}^{2}+c^{3}\right)\left(a_{n}^{2}+2 a_{n}+1+c^{3}\right)
$$

is a power. Since $a_{n}^{2}+c^{3}$ and $a_{n}^{2}+2 a_{n}+1+c^{3}$ are coprime, then $a_{n}^{2}+c^{3}$ is a power as well.
The same argument can be further applied giving that $a_{1}^{2}+c^{3}=c^{2}+c^{3}=c^{2}(c+1)$ is a power.
If $a^{2}(a+1)=t^{m}$ with odd $m \geq 3$, then $a=t_{1}^{m}$ and $a+1=t_{2}^{m}$, which is impossible. If $a^{2}(a+1)=t^{2 m_{1}}$ with $m_{1} \geq 2$, then $a=t_{1}^{m_{1}}$ and $a+1=t_{2}^{m_{1}}$, which is impossible.

Therefore $a^{2}(a+1)=t^{2}$ whence we obtain the solutions $a=s^{2}-1, s \geq 2, s \in \mathbb{N}$.