Problems of Ukrainian Authors

Let $p$ be some prime divisor of the number $2^{2^n} + 1$. Consider numbers $x$ of the form $2^k$. By the statement of the problem we need $x^x - 1 = 2^{2^s} - 1 \equiv 0 \pmod{p}$. If $s \ge n+1$ then
$$2^{2^{2^k}} - 1 = (2^{2^{n+1}})^{2^{2^k-(n+1)}} - 1 : 2^{2^{n+1}} - 1 = (2^{2^k} - 1)(2^{2^k} + 1),$$
where $a: p$ denotes "a is divisible by $p", i.e.$p|a$. Thus$2^{2^k} - 1$is divisible by$p$. So if all the numbers$$2^{2^{n+1}}, 2^{2^{n+2}}, \dots, 2^{2^{2008}}$$are less than$p, the problem is solved. The solution of the problem follows from the lemma, stated below, for A = 2^{2008}$.
Lemma 1. For any positive integer number $A$ there exists such $n_0$ that all numbers of a kind $2^{2^n} + 1$, where $n \ge n_0$, have prime divisor greater than $A \cdot 2^{2^n}$.
To prove lemma 1, we need an additional lemma.
Lemma 2. If the prime number $p$ is a divisor of the number $2^{2^n} + 1$, then $2^{n+1} \le p-1$.
Proof (of lemma 2). Let $\Delta$ be the least positive integer such that $p|2^\Delta - 1$. We show that $\Delta | p-1$. It follows from Fermat's (little) theorem that $2^{p-1} \equiv 1 \pmod{p}$. Let $\Delta_1$ be the remainder in division of $p-1$ by $\Delta$. Then from $2^\Delta \equiv 1 \pmod{p}$ and $2^{p-1} \equiv 1 \pmod{p}$ it follows that $2^{\Delta_1} \equiv 1 \pmod{p}$. From the supposition of minimality of $\Delta$ it follows that $\Delta_1 = 0$, i.e. $\Delta | p-1$.

$2^{2^{n+1}} - 1 = (2^{2^n} - 1)(2^{2^n} + 1)$, so $p \mid 2^{2^{n+1}}$. Reasoning by analogy with the above, one gets $\Delta \mid 2^{n+1}$, that is $\Delta = 2^k$, $k \le n+1$. If $k \le n$ then
$2^{2^n} + 1 = (2^{2^k})^{2^n-2^k} + 1 \equiv 1^{2^n-2^k} + 1 \equiv 2 \pmod p$, which is impossible due to $p \mid 2^{2^{n+1}}$. Hence $\Delta = 2^{n+1}$, which (with $\Delta \mid p-1$ proven above) finishes the proof of the lemma.
Proof (of lemma 1). It follows from lemma 2 that all prime divisors of the number $2^{2^n} + 1$ can be expressed as $2^{n+1} \beta_j + 1$. Let
$$2^{2^n} + 1 = p_1^{\alpha_1} \cdots p_s^{\alpha_s}, \text{ where } p_j = 2^{n+1} \beta_j + 1.$$
Then
$$p_j^{\alpha_j} = (2^{n+1} \beta_j + 1)^{\alpha_j} = 1 + \binom{\alpha_j}{1} 2^{n+1} \beta_j + \binom{\alpha_j}{2} 2^{2(n+1)} \beta_j^2 + \dots \equiv 1 + \alpha_j 2^{n+1} \beta_j \pmod{2^{2(n+1)}}.$$
Thus,
$$p_1^{\alpha_1} \cdots p_s^{\alpha_s} \equiv \prod_{j=1}^s (1 + \alpha_j 2^{n+1} \beta_j) \equiv 1 + 2^{n+1} \sum_{j=1}^s \alpha_j \beta_j \pmod{2^{2(n+1)}}. \quad (1)$$
If $2^{2n+2} < 2^{2^n}$ (which is true for $n > 2$), then $2^{2^n} + 1 \equiv 1 \pmod{2^{2(n+1)}}$, hence we can get from (1) $2^{n+1} \sum_{j=1}^s \alpha_j \beta_j \equiv 0 \pmod{2^{2(n+1)}}$, therefore $\sum_{j=1}^s \alpha_j \beta_j \equiv 0 \pmod{2^{n+1}}$. Note that $\alpha_j > 0$ and $\beta_j > 0$, so $\sum_{j=1}^s \alpha_j \beta_j \ge 2^{n+1}$.
Suppose that all $\beta_j < A$ (otherwise, when one of the $\beta_j \ge A$, let it be $\beta_{j_0}$, the number $2^{2^n} + 1$ has a prime divisor $2^{n+1} \beta_{j_0} + 1 > 2^{n+1} A$ and the lemma is proved). Then
$$2^{n+1} \le \sum_{j=1}^s \alpha_j \beta_j < A \sum_{j=1}^s \alpha_j \le A s \alpha_k, \text{ where } \alpha_k = \max_{1 \le j \le s} \alpha_j.$$
All $p_j$ are distinct numbers, thus all $\beta_j$ are also distinct numbers, and the number of $\beta_j$ is $s$. By supposition all $\beta_j < A$, whence we get $s < A$. So $2^{n+1} < A^2 \alpha_k$, that is $\alpha_k > \frac{2^{n+1}}{A^2}$, but then $p_k^{\alpha_k} = (2^{n+1} \beta_k + 1)^{\alpha_k} > (2^{n+1})^{\alpha_k} > 2^{(n+1)2^{n+1}/A^2}$. Since $p_k^{\alpha_k} \le 2^{2^n} + 1$, then $2^{(n+1)2^{n+1}/A^2} < 2^{2^n} + 1$. (2)
There exists such $n_0$ that for $n \ge n_0$ we have $\frac{n+1}{A^2} > 1$. Then it follows from (2) that
$$2^{2^n} + 1 \ge 2^{(n+1)2^{n-1}/A^2} > 2^{2^{n+1}} > 2^{2^n} + 1.$$
a contradiction, which proves the lemma.