Problem 403

We note that $n=p^{2} q^{2}=(p q)^{2}$. Since $n<1000$, then $(p q)^{2}<1000$ and so $p q<\sqrt{1000} \approx 31.6$. Finding the number of possible values of $n$ is thus equivalent to finding the number of positive integers $m$ with $1 \leq m \leq 31<\sqrt{1000}$ that are the product of two prime numbers. The prime numbers that are at most 31 are $2,3,5,7,11,13,17,19,23,29,31$. The distinct products of pairs of these that are at most 31 are: $2 \times 3=6, 2 \times 5=10, 2 \times 7=14, 2 \times 11=22, 2 \times 13=26, 3 \times 5=15, 3 \times 7=21$. Any other product either duplicates one that we have counted already, or is larger than 31. Therefore, there are 7 such values of $n$.