If p is a prime number greater than 3, then `(p^2-1)` is always divisible by (a) 6 but not 12 (b) 12 but not 24 (c) 24 (d) None of these

If n is a whole number greater than 1, then n^(2)(n^(2)-1) is always divisible by 8 (b) 10 (c) 12 (d) 16

n being any odd number greater than 1n^(65)-n backslash is always divisible by 5 (b) 13 (c) 24 (d) None of these

If n be any natural number then by which largest number (n^3-n) is always divisible? 3 (b) 6 (c) 12 (d) 18

If n be any natural number then by which largest number (n^(3)-n) is always divisible? 3 (b) 6(c)12 (d) 18

The number 6n^(2)+6n for natural number n is always divisible by 6 only (b) 6 and 12(c)12 only (d) 18 only

The least number divisible by 15, 20, 24, 32 and 36 is (a) 1440 (b) 1660 (c) 2880 (d) None of these

If n is any odd number greater than 1, then n backslash(n^(2)-1) is divisible by 24 always (b) divisible by 48 always (c) divisible by 96 always (d) None of these

Prove that if p is a prime number greater than 2, then the difference [(2+sqrt(5))^p]-2^(p+1) is divisible by p, where [.] denotes greatest integer.

Prove that if p is a prime number greater than 2, then the difference [(2+sqrt(5))^p]-2^(p+1) is divisible by p, where [.] denotes greatest integer.

Prove that if p is a prime number greater than 2, then the difference [(2+sqrt(5))^(p)]-2^(p+1) is divisible by p,where [.] denotes greatest integer.