Determine with proof, all the positive integers n for which i n is not the square of any integer, and i i n 3 divides n 2 , where ( x denotes the largest integer that is less than or equal to x . )

Since, n is not the square of any integer, hence let n = k , k ∈ Z . We know that if x = I , then I ≤ x < I + 1 . But, n ≠ k 2 , hence, we have k 2 < n < k + 1 2 ⇒ k 2 < n < k 2 + 2 k + 1 . . . i Also, since k 3 divides n 2 , we have that k 2 divides n 2 and hence k divides n . Thus, from i the only possibilities for n are n = k 2 + k and n = k 2 + 2 k . Let n = k 2 + k . Then k 3 n 2 ⇒ k 3 k 2 + k 2 ⇒ k 3 k 4 + 2 k 3 + k 2 Since k 4 + 2 k 3 is divisible by k 3 hence, for the number k 4 + 2 k 3 + k 2 to be divisible by k 3 , we must have ⇒ k 3 k 2 ⇒ k = 1 , i.e., n = k 2 + k = 1 + 1 = 2 . Again, let n = k 2 + 2 k . Then k 3 n 2 ⇒ k 3 k 2 + 2 k 2 ⇒ k 3 k 4 + 4 k 3 + 4 k 2 Since k 4 + 4 k 3 is divisible by k 3 hence, for the number k 4 + 4 k 3 + 4 k 2 to be divisible by k 3 , we must have k 3 4 k 2 ⇒ k 4 . Therefore, k = 1 , 2 or 4 . And, n = k 2 + 2 k = 1 2 + 2 × 1 = 3 , 2 2 + 2 × 2 = 8 and 4 2 + 2 × 4 = 24 . Thus n = 2 , 3 , 8 and 24 are the positive integers satisfying the given conditions.