Step 1: Represent X using a parameter.
Let the four consecutive integers be k, k+1, k+2, k+3 (with k ∈ ℤ>0). Then:
X = k(k+1)(k+2)(k+3), n = X + 1.
Step 2: Prove the perfect-square identity.
Observe:
k(k+1)(k+2)(k+3) + 1 = (k² + 3k + 1)².
Hence n is always a perfect square ⇒ Statement 3 is true.
Step 3: Check parity of n.
Among four consecutive integers there are two even numbers, so X is even. Therefore n = X+1 is odd ⇒ Statement 1 is true.
Step 4: Is n prime?
Since n = (k² + 3k + 1)² and k ≥ 1, we have n > 1 and n is a non-trivial square, hence composite. Statement 2 is false.
Answer: Statements 1 and 3 only are true.