## Task A-4.2.

Determine all ordered pairs $(p, n)$, where $p$ is a prime and $n$ is a natural number, for which

$$
1+p+p^{2}+p^{3}+\cdots+p^{n}=2801
$$
## First Solution.

Subtract $1$ from both sides of the equation $1+p+p^{2}+p^{3}+\cdots+p^{n}=2801$. Since $p+p^{2}+p^{3}+\cdots+p^{n}=p \cdot \frac{p^{n}-1}{p-1}$, we obtain the equation

$$
p \cdot \frac{p^{n}-1}{p-1}=2800
$$

Notice that the left side is divisible by $p$, so $p$ must be one of the prime factors of 2800. Therefore, $p$ must be 2, 5, or 7.

If $p=2$, the equation becomes $2^{n}-1=1400$. This equation has no solutions in natural numbers, so this case is not possible.

If $p=5$, the equation becomes $\frac{5^{n}-1}{4}=560$, or $5^{n}=2241$. This equation has no solutions in natural numbers, so this case is not possible either.

If $p=7$, the equation becomes $\frac{7^{n}-1}{6}=400$, or $7^{n}=2401$. From this, we find that $n=4$. Therefore, the only solution to the original equation is $(p, n)=(7,4)$.