Solution:
For
k = 1, | n = 7 | n^{2} - k = 48 = 3 ^{.} 2^{4}, |
k = 2, | n = 235 | n^{2} - k = 55227 = 23 ^{.} 7^{4}, |
k = 3, | n = 4936 | n^{2} - k = 24364093 = 17 ^{.} 37^{4}. |
Therefore, k > 4. Let us prove that the number of divisors is not 10. If n^{2} - 4 has exactly 10 divisors, then n^{2} - 4 = p^{9} or n^{2} - 4 = p ^{.} q^{4} for some prime numbers p and q. In the first case, (n - 2)(n + 2) = p^{9}; 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) = pq^{4}, 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 m^{2} - 1 = (m - 1)(m + 1) = 4p, hence m is odd. But if m is odd, then m^{2} - 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 q^{4}, other one p. We have two cases:
1. n + 2 = p, n - 2 = q^{4} and p = 4 + q^{4}. q is not 5 (4 + 5^{4} = 629 is not prime). Therefore, q^{4} = 1 (mod 5) and p = 4 + q^{4} = 0 (mod 5), and p = 5. Contradiction.
2. n + 2 = q^{4}, n- 2 = p and p = q^{4} - 4 = (q^{2} - 2)(q^{2} + 2). Since q > 1, p is not prime. Contradiction.