5. Given a natural number $x=9^{n}-1$, where $n-$ is an odd natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 61. 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 61. The remainders of the powers of 9 when divided by 61 are $9, 20, 58, 34, 1$ and then repeat cyclically. Thus, the divisibility of $x$ by 61 means that $n$ is a multiple of 5, i.e., $x = 9^{5m} - 1$, where $m$ is odd. By the formula for the sum of a geometric progression,

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

We will 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 $r$. Hence, $r$ is a divisor of 5. But $p$ is not divisible by 5 when $m$ is odd, so $r = 1$.

We will prove that the number $p$ is a power of two exactly when $m = 1$. Indeed, for $m > 1$, we can write $p = (3^m - 1)(3^m + 1)$. In the right-hand side, we have the product of two consecutive even numbers, both greater than 2. Therefore, at least one of them is not divisible by 4 and, hence, is not a power of 2.

If $m = 1$, we get $x = 9^5 - 1 = 59048 = 2^3 \cdot 11^2 \cdot 61$, which fits our criteria. We will show that there are no solutions for $m > 1$. Consider two cases.

1) $p$ is divisible by 61. Then $m \vdots 5$, 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, neither of which is a power of 2. Therefore, $x$ has at least three different odd prime divisors, which is impossible.
2) $q$ is divisible by 61. Note that $p$ has an odd divisor, and $q$ is odd and coprime with $p$. Then $q$ must be a power of 61, i.e., $q = 61^s$. Note that $s \vdots 3$, since $q - 1$ is divisible by 9, and the remainders of $61^s$ when divided by 9 are $7, 4, 1$ and then repeat cyclically. Additionally, when divided by 8, the number $q$ gives a remainder of 5, while the remainder of $61^s$ is 5 for odd $s$ and 1 for even $s$. Therefore, $s$ is odd, so $s$ has the form $6k + 3$ for some natural number $k$.

Now let's compare the remainders of $q$ and $61^s$ when divided by 7. The remainders of $61^s$ are $5, 4, 6, 2, 3, 1$ and then repeat cyclically. For $s = 6k + 3$, all of them coincide with 6. The remainders of $9^m$ are $2, 4, 1$ and then repeat cyclically. Therefore, it is sufficient to find the remainders of $q$ for $m = 1, 2, 3$. They are

$$
\begin{aligned}
& (1 + 2 + 4 + 1 + 2) \bmod 7 = 3, \quad \text{if } m = 1, \\
& (1 + 4 + 1 + 2 + 4) \bmod 7 = 5, \quad \text{if } m = 2, \\
& (1 + 1 + 2 + 4 + 1) \bmod 7 = 2, \quad \text{if } m = 3.
\end{aligned}
$$

Since none of these numbers coincide with 6, $q$ cannot be a power of 61.