Vietnamese Mathematical Olympiad

a) It is easy to find the general formula of $(a_n)$, which is
$$a_n = 2^n + 3^n, \forall n \in \mathbb{Z}^+.$$
Suppose that there exists $n \ge 1$ that $a_n, a_{n+1}$ have common prime divisor $p$. Clearly, $\gcd(p, 6) = 1$. We have
$$\begin{cases} p|2^n + 3^n, \\ p|2^{n+1} + 3^{n+1}, \end{cases} \quad \text{so} \quad \begin{cases} p|3 \cdot 2^n + 3^{n+1}, \\ p|2 \cdot 2^n + 3^{n+1}, \end{cases}$$
implies that $p | 2^n$, which is a contradiction since $\gcd(p, 6) = 1$.

b) Let $p$ be the prime divisor of $2^{2^k} + 3^{2^k}$. Clearly, $2^{2^k} \equiv -3^{2^k} \pmod p$ so $2^{2^{k+1}} \equiv 3^{2^{k+1}} \pmod p$. By Fermat's little theorem,
$$2^{p-1} \equiv 3^{p-1} \equiv 1 \pmod p.$$
Let $h$ be the smallest positive integer that $2^h \equiv 3^h \pmod p$. It is well-known that for all $h' \ge h$ satisfying this condition, $h|h'$. Now, we obtain that $h' = 2^{k+1}$ satisfying that condition then $h|2^{k+1}$, thus $h = 2^x$ with $0 \le x \le k+1$. Suppose that $x \le k$ then
$$2^x \equiv 3^x \pmod p \text{ so } p|2^x - 3^x|2^{2k} - 3^{2^k},$$
which is a contradiction since $p|2^{2^k} + 3^{2^k}$. Therefore, we must get $x = k+1$, which implies $2^{k+1} | p-1$. $\square$