Final answer:
The claim is incorrect; if n is a perfect square, then n^2 is also a perfect square, because squaring a square simply raises the base integer to the fourth power, which can be expressed as the square of a square.
Step-by-step explanation:
The claim that n is a perfect square, then n2 is not a perfect square, is mathematically inconsistent. In fact, if n is a perfect square, this simply means that there exists an integer m such that n = m2. When you take the square of n (which is already a square), you get n2 = m4, which is still a perfect square because it can be expressed as the square of m2
To visualize this, consider that squaring a number is like doubling the exponent in the prime factorization. If n = m2, then all the primes in m are raised to an even power. Squaring n doubles these exponents, resulting in an even power across all primes, which by definition means n2 is a perfect square.
The student's statement might arise from confusion or a typo in the problem statement. It's important to revisit the original question to ensure it's correctly understood. Mathematical operations like squaring or finding the square root are reversible, and applying them doesn't change the 'perfect square' property of a number.