4. Let $\left\{x_{n}\right\}_{n>1}$ be the sequence of positive integers defined by $x_{1}=2$ and $x_{n+1}=$ $2 x_{n}^{3}+x_{n}$ for all $n \geq 1$. Determine the highest power of 5 that divides the number $x_{2014}^{2}+1$.
Solution. For $n=1$, we have $x_{1}=2$ and $x_{1}^{2}+1=2^{2}+1=5$ which is a multiple of 5 but not of $5^{2}$. For $n=2$, it is $x_{2}=2 \cdot 2^{3}+2=18$ and $x_{2}^{2}+1=18^{2}+1=325$ which is a multiple of $5^{2}$ but not of $5^{3}$. This suggests we conjecture that $x_{n}^{2}+1$ is a multiple of $5^{n}$ but not of $5^{n+1}$. We will prove this by induction. The cases $n=1$ and $n=2$ are already proven. Suppose that $n \geq 3$ and that $x_{n}^{2}+1=k 5^{n}$ with $(k, 5)=1$. Then,

$$
\begin{aligned}
x_{n+1}^{2}+1 & =\left(2 x_{n}^{3}+x_{n}\right)^{2}+1=x_{n}^{2}\left(2 x_{n}^{2}+1\right)^{2}+1=\left(k 5^{n}-1\right)\left(2 k 5^{n}-1\right)^{2}+1 \\
& =\left(k 5^{n}-1\right)\left(4 k^{2} 5^{2 n}-4 k 5^{n}+1\right)+1=4 k^{3} 5^{3 n}-8 k^{2} 5^{2 n}+k 5^{n+1} \\
& =k 5^{n+1}(\underbrace{4 k^{2} 5^{2 n-1}-8 k 5^{n-1}}_{\dot{5}}+1)
\end{aligned}
$$

which is a multiple of $5^{n+1}$ but not a multiple of $5^{n+2}$ since for $n \geq 3$ it is $2 n-1 \geq 1$ and $n-1 \geq 1$ respectively and the term in parentheses is not a multiple of 5. This proves the conjecture and, therefore, the highest power of 5 that divides $x_{2014}^{2}+1$ is $5^{2014}$.