5. Given a natural number $x=9^{n}-1$, where $n$ is a natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 11. Find $x$.
Answer: 59048.

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. First, let's prove that the number $3^{s}+1$ cannot be a power of two when $s>1$, and the number $3^{s}-1$ cannot be a power of two when $s>2$. Indeed, for $s>1$, the number $3^{s}+1$ is greater than 8 and is not divisible by 8, as its remainder when divided by 8 is 4 or 2. Now let $s>2$. If $s$ is odd, the number $3^{s}-1$ is greater than 8 and is not divisible by 8, as its remainder when divided by 8 is 2. If $s=2k$, then $3^{s}-1$ is divisible by the number $3^{k}+1$, which is not a power of two. From the above, it follows, in particular, that $9^{s}-1$ will be a power of two only when $s=1$.

The remainders of the powers of 9 when divided by 11 are $9, 4, 3, 5, 1$ and then repeat cyclically. Therefore, the divisibility of $x$ by 11 means that $n$ is a multiple of 5, i.e., $x=9^{5m}-1$. If $m=1$, we get $x=9^{5}-1=59048=2^{3} \cdot 11^{2} \cdot 61$, which fits our criteria. Let's show that there are no solutions for $m>1$. Consider two cases.

1) $m$ is odd. By the formula for the sum of a geometric progression,

$$
x=p \cdot q, \quad \text{where } p=9^{m}-1, \quad q=1+9^{m}+9^{2m}+9^{3m}+9^{4m}
$$

Let's prove that the numbers $p$ and $q$ are coprime. Indeed, let the number $r$ divide both $p$ and $q$. Then

$$
5=q-(9^{m}-1)-(9^{2m}-1)-(9^{3m}-1)-(9^{4m}-1)
$$

Differences of the form $9^{km}-1$ are divisible by $p$ and, therefore, by $r$. Thus, $r$ is a divisor of 5. But $p$ is not divisible by 5 when $m$ is odd, so $r=1$.

Suppose $p$ is divisible by 11. Then $m \vdots \leq 5$, i.e., $m=5k$. If $k=1$, then $p$ is divisible by 11 and 61. 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, $x$ has at least three different odd prime divisors, which is impossible.

Now suppose $q$ 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, i.e., $q=11^{s}$. But the remainder of $11^{s}$ when divided by 8 is 3 for odd $s$ and 1 for even $s$, while the remainder of $q$ when divided by 8 is 5. Therefore, the number $q$ cannot be a power of 11.
2) $m$ is even. Then $n=2k$ for $k \geq 5$, and

$$
x=(9^{k}-1)(9^{k}+1)=(3^{k}-1)(3^{k}+1)(9^{k}+1)
$$

The factors on the right side are coprime, since the pairs of numbers $3^{k} \pm 1$ and $9^{k} \pm 1$ are odd and differ by 2. Moreover, none of the factors is a power of two. Indeed, we have already proven this for the first two, and $9^{k}+1$ is greater than 8 and gives a remainder of 2 when divided by 8. Thus, $x$ has at least three different odd prime divisors, which is impossible.