## Task A-1.4.

Only one divisor of the number $3^{12}-1$ is greater than 70 and less than 80. Determine it.
## First solution.

Let's factorize the given number:

$$
\begin{aligned}
3^{12}-1 & =\left(3^{6}-1\right)\left(3^{6}+1\right) & & 1 \text { point } \\
& =\left(3^{3}-1\right)\left(3^{3}+1\right)\left(3^{6}+1\right) & & 1 \text { point } \\
& =(3-1)\left(3^{3}+3+1\right)(3+1)\left(3^{2}-3+1\right)\left(3^{2}+1\right)\left(3^{4}-3^{2}+1\right) & & \\
& =2 \cdot 13 \cdot 4 \cdot 7 \cdot 10 \cdot 73 . & & 3 \text { points }
\end{aligned}
$$

The desired divisor is the number 73.

Note: It is not required for the student to show that there are no other divisors between 70 and 80. However, when the prime factorization of \(3^{12}-1=2^{4} \cdot 5 \cdot 7 \cdot 13 \cdot 73\) is known, this can be easily verified.