Solution:

For

Therefore, k > 4. Let us prove that the number of divisors is not 10. If n² - 4 has exactly 10 divisors, then n² - 4 = p⁹ or n² - 4 = p ^. q⁴ for some prime numbers p and q. In the first case, (n - 2)(n + 2) = p⁹; if n = 3 contradiction, if n- 2 > 1, then p divides both n- 2 and n + 2, and hence p divides n + 2 - (n - 2) = 4. But the only possibility p = 2 fails. In the second case, (n- 2)(n + 2) = pq⁴, as in the first case, if q divides both n- 2 and n + 2, then q = 2, therefore n = 2m. By substitution n = 2m we get m² - 1 = (m - 1)(m + 1) = 4p, hence m is odd. But if m is odd, then m² - 1 = 0 (mod 8) and 4p = 0 (mod 8), and p is even, contradiction. Therefore, q must divide only one of two numbers n - 2 and n + 2, words, one of this number is q⁴, other one p. We have two cases:

1. n + 2 = p, n - 2 = q⁴ and p = 4 + q⁴. q is not 5 (4 + 5⁴ = 629 is not prime). Therefore, q⁴ = 1 (mod 5) and p = 4 + q⁴ = 0 (mod 5), and p = 5. Contradiction.

2. n + 2 = q⁴, n- 2 = p and p = q⁴ - 4 = (q² - 2)(q² + 2). Since q > 1, p is not prime. Contradiction.