Find all natural numbers $a>1$, with the property that every prime divisor of $a^6-1$ divides also at least one of the numbers $a^3-1$, $a^2-1$.

[i]K. Dochev[/i]
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 \( \boxed{ a = 2 } \).