5. Given a natural number $x=5^{n}-1$, 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: 3124.

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 remainders of the powers of 5 when divided by 11 are $5, 3, 4, 9, 1$ and then repeat cyclically. Thus, the divisibility of $x$ by 11 means that $n$ is a multiple of 5, that is, $x = 5^{5m} - 1$. From here, using the formula for the sum of a geometric progression,

$$
x = p \cdot q, \quad \text{where } p = 5^m - 1, \quad q = 1 + 5^m + 5^{2m} + 5^{3m} + 5^{4m}
$$

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

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

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

We will prove that the number $p$ is a power of two only when $m = 1$. Indeed, let $m > 1$. If $m = 2k$, then write $p = (5^k - 1)(5^k + 1)$. On the right side, we have the product of two consecutive even numbers. At least one of them is greater than 4 and does not divide by 4, so it is not a power of 2. Now let $m = 2k + 1$. Then

$$
p = 5^m - 5 + 4 = 5 \cdot (5^k - 1)(5^k + 1) + 4.
$$

The right side is not divisible by 8, since $(5^k - 1)(5^k + 1)$ is divisible by 8 as the product of two consecutive even numbers. Since $p > 4$, the number $p$ cannot be a power of two.

When $m = 1$, we get $x = 3124 = 4 \cdot 11 \cdot 71$, which fits. We will show that there are no solutions when $m > 1$. One of the numbers $p$ and $q$ is divisible by 11. 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 71. 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.
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 $5^m$ when divided by 8 is 5 for odd $m$ and 1 for even $m$. If $m$ is odd, the remainder of $q$ when divided by 8 is the same as that of $1 + 5 + 1 + 5 + 1 = 13$, which is 5. For even $m$, the remainder of $q$ when divided by 8 is also 5. But the remainders of $11^s$ when divided by 8 are only 3 and 1, so $q$ cannot be a power of 11.