International Mathematical Arhimede Contest

To prove the first part, we begin observing that $2013 = 3 \cdot 11 \cdot 61 = 33 \cdot 61$. Since $9^5 = 81 \cdot 81 \cdot 9 = (61+20)(61+20) \cdot 9$ and $20 \cdot 20 \cdot 9 = 3600 = 61 \cdot 59 + 1$, then we have $9^5 \equiv 1 \pmod{61}$. Since $4^6 = 4096 = 61 \cdot 67 + 9 \equiv 9 \pmod{61}$ and $6^{n-1} = (5+1)^{n-1} = 5m + 1$, $m \in \mathbb{N}$, then for all $n \ge 1$, holds
$$4^{6^n} = (4^6)^{6^{n-1}} = (4^6)^{5m+1} \equiv 9^{5m} \cdot 9 \pmod{61} \equiv 9 \pmod{61}$$
So, $a_n = 4^{6^n} + 1943 \equiv 1952 \pmod{61} \equiv 0 \pmod{61}$ and $61|a_n$, for all $n \ge 1$.

On the other hand, $a_n = 4^{6^n} - 4 + 1947 = 4(4^{6^n-1} - 1) + 33 \cdot 59$. Since $6^n - 1 \equiv 0 \pmod{5}$, then $6^n - 1 = 5p, p \in \mathbb{N}$. Then, we have
$$\begin{aligned} 4^{6^n-1} - 1 &= 4^{5p} - 1 = (4^5)^p - 1 = 1024^p - 1 \\ &= (1024 - 1)(1024^{p-1} + \dots + 1) = 1023 \cdot q = 33 \cdot 31 \cdot q \end{aligned}$$

and $33|a_n$ the jointly with the preceding yields $2013 = 33 \cdot 61|a_n$ for all $n \ge 1$.

To solve the second part of the statement, we observe that $a_n - 207 = 4^{6^n} + 1736$ is an even integer, say $2x$ with $x \in \mathbb{N}$. From $4^{6^n} + 1736 = (2x)^3$ follows $2^{2 \cdot 6^{n-3}} + 217 = x^3$ or $2^{3(4 \cdot 6^{n-1}-1)} + 217 = x^3$. Putting $2^{4 \cdot 6^{n-1}-1} = y$ in the last equation yields
$$x^3 - y^3 = 217 \Leftrightarrow (x - y)(x^2 + xy + y^2) = 217 = 7 \cdot 31$$
Since $x - y < x^2 + xy + y^2$, then we have two possibilities
$$\begin{cases} x - y = 1, \\ x^2 + xy + y^2 = 217. \end{cases} \quad \text{or} \quad \begin{cases} x - y = 7, \\ x^2 + xy + y^2 = 31. \end{cases}$$
The solutions of the first system are $(9, 8)$, $(-8, -9)$ and the solutions of the second one $(6, -1)$ and $(1, -6)$. Finally, since $y = 2^{4 \cdot 6^{n-1}-1}$ is a positive integer, then $y = 2^{4 \cdot 6^{n-1}-1} = 8 = 2^3$ from which follows $n = 1$, and we are done. $\square$