1. Determine all natural numbers $n$ such that $9^{n}-7$ can be represented as a product of at least two consecutive natural numbers.
Solution. The product of three consecutive natural numbers is divisible by 3, and $9^{n}-7 \equiv 2(\bmod 3)$, so we conclude that $9^{n}-7$ cannot be written as the product of three consecutive numbers.

Let $9^{n}-7=m(m+1)$ for some natural number $m$. The last equation is equivalent to the equation

$$
4 \cdot 9^{n}-27=(2 m+1)^{2}
$$

i.e., to the equation

$$
4 \cdot 9^{n}-(2 m+1)^{2}=27
$$

Since $\{1,3,9,27\}$ are all the divisors of 27 and

$$
2 \cdot 3^{n}+2 m+1>2 \cdot 3^{n}-2 m-1
$$

there are two possible cases:

1) $\left\{\begin{array}{c}2 \cdot 3^{n}+2 m+1=27 \\ 2 \cdot 3^{n}-2 m-1=1\end{array}\right.$
2) $\left\{\begin{array}{l}2 \cdot 3^{n}+2 m+1=9 \\ 2 \cdot 3^{n}-2 m-1=3\end{array}\right.$.

In case 1) if we add the equations, we get $3^{n}=7$, which is not possible.

In case 2) if we add the equations, we get $3^{n}=3$, from which we find $n=1$. Substituting in the first equation, we get

$$
2 \cdot 3+2 m+1=9, \text { i.e. } m=1
$$

In this case, $9^{1}-7=1 \cdot(1+1)$.