Suppose for contradiction that this is not true. Then n^1/2 = p/q for integers p, q where gcd(p,q) = 1. n is not a perfect square.

n = p² / q² , n * q² = p^2.

Now consider the prime factorizations of these numbers, specifically n:

n = (p_0)^j_0 * (p_1)^j_1 * (p_2)^j_2 * ... (p*k)^j_k

Since n is not a perfect square, there exists at least one prime in the factorization of n, call it r, such that r is raised to an odd power (else n would be a perfect square). That is, the maximal power of r that divides n is odd. For some positive integer k, r^k divides n, and r^k+1 does not. And k is odd.

Applying this same logic to p² and q^2, if the prime number r which divides n divides p, then it must be raised to an even power. If r divides p, then it divides p² and the maximal power of r that divides p² is even since the prime factorization of p² is the same as the prime factorization of p, with every term duplicated. Same for q and q^2. If r does not divide these, then r⁰ still works (and is even).

Then n * q² = p² , r raised to some even power divides p² , r raised to some even power divides q² , and r raised to some odd power divides n. Then the maximal power of r which divides p² is both odd and even which is a contradiction.