5. Given a natural number $x=7^{n}+1$, where $n$ is an odd natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 11. Find $x$.
Answer: 16808.

Solution. Since the number $x$ is even, one of its prime divisors is 2. Therefore, we need to find all $x$ that have exactly two odd prime divisors, one of which is 11. The divisibility of $x$ by 11 is equivalent to $n$ being an odd multiple of 5, that is, $x=7^{5m}+1$ for some odd $m$. From here, using the formula for the sum of a geometric progression,

$$
x=p \cdot q, \quad \text{where } p=7^{m}+1, q=1-7^{m}+7^{2m}-7^{3m}+7^{4m}
$$

Note that $p$ and $q$ are coprime. Indeed, let the number $r$ divide both $p$ and $q$. Then

$$
5=q-(7^{m}+1)-(7^{2m}-1)-(7^{3m}+1)-(7^{4m}-1)
$$

The expressions in parentheses on the right-hand side are divisible by $p$ and, therefore, by $r$. Thus, $r$ is a divisor of 5. But the remainders of the powers of 7 when divided by 5 are 2, 4, 3, 1, and then they cyclically repeat. Therefore, $p$ is not divisible by 5 for odd $m$, hence $r=1$.

We will prove that the number $p$ is a power of two only when $m=1$. Indeed, let $m>1$ and $7^{m}+1=2^{s}$. Then $s \geqslant 4$ and the number $7^{m}+1$ is divisible by 16. But this is impossible, since the remainders of $7^{m}$ when divided by 16 are only 7 and 1.

If $m=1$, we get $x=16808=2^{3} \cdot 11 \cdot 191$, which fits our criteria. We will show that there are no solutions for $m \geqslant 3$. Consider two cases.

1) $p$ is divisible by 11. Then $m \vdots 5$, that is, $m=5k$. If $k=1$, then $p$ is divisible by 11 and 191. For $k>1$, we can apply the same reasoning to $p$ as to $x$. In both cases, $p$ will decompose into two coprime factors that are not powers of two. Therefore, $x$ will have at least three different odd prime divisors, which is impossible.
2) $q$ is divisible by 11. Note that $p$ has an odd divisor, and $q$ is odd and coprime with $p$. Then $q$ must be a power of 11. The remainder of $7^{m}$ when divided by 8 is -1 due to the oddness of $m$. Therefore, the number $q$ gives a remainder of 5 when divided by 8. But the remainders of $11^{s}$ when divided by 8 are only 3 and 1, so $q$ cannot be a power of 11.