Problem 1287

1. We start with the given expression $a^6 - 1$:
$$a^6 - 1 = (a^3 - 1)(a^3 + 1)$$
Further factorizing $a^3 + 1$:
$$a^3 + 1 = (a + 1)(a^2 - a + 1)$$
Therefore:
$$a^6 - 1 = (a^3 - 1)(a + 1)(a^2 - a + 1)$$

2. We need to consider the prime divisors of $a^6 - 1$. Every prime divisor of $a^3 - 1$ is already a prime divisor of $a^3 - 1$, so we focus on the prime divisors of $a + 1$ and $a^2 - a + 1$.

3. Consider the prime divisors of $a + 1$. Any prime divisor $p$ of $a + 1$ must also divide:
$$(a + 1)(a - 1) = a^2 - 1$$
Hence, $p$ divides $a^2 - 1$.

4. Now, consider the prime divisors of $a^2 - a + 1$. Let $p$ be a prime divisor of $a^2 - a + 1$. We need to check if $p$ divides $a^3 - 1$ or $a^2 - 1$.

5. Suppose $p$ divides $a^3 - 1$. Then:
$$p \mid a(a^2 - a + 1) - (a^3 - 1) = a^2 - a - 1$$
Therefore:
$$p \mid (a^2 - a + 1) - (a^2 - a - 1) = 2$$
This implies $p = 2$. However, if $2 \mid a^2 - a + 1$, then $2 \mid a(a - 1)$, which means $2 \mid 1$, a contradiction. Thus, $p \neq 2$.

6. Next, suppose $p$ divides $a^2 - 1$. Then:
$$p \mid (a^2 - a + 1) - (a^2 - 1) = a - 2$$
Therefore:
$$p \mid a(a - 2) - (a^2 - 1) = 2a - 1$$
This implies:
$$p \mid (2a - 1) - 2(a - 2) = 3$$
Hence, $p = 3$.

7. We conclude that the only prime divisor of $a^2 - a + 1$ is $3$. Let:
$$a^2 - a + 1 = 3^\beta$$
If $\beta \geq 2$, then $a^2 - a + 1 \equiv 0 \pmod{9}$. Checking all residues modulo 9 for $a$, we find this is impossible. Therefore, $\beta = 1$, and:
$$a^2 - a + 1 = 3$$
Solving this quadratic equation:
$$a^2 - a - 2 = 0$$
Factoring:
$$(a - 2)(a + 1) = 0$$
Thus, $a = 2$ (since $a > 1$).

The final answer is $\boxed{ a = 2 }$.