To solve the problem of finding the number of positive integers (n) ((1 \leq n \leq 1000)) such that (x^2 + y^2 = n) has exactly 12 solutions, we use the following key insights:

Key Observations:

The number of integer solutions to (x^2 + y^2 = n) is (r_2(n) = 4 \times \prod (b_i + 1)), where (b_i) are exponents of primes (\equiv 1 \mod 4) in the factorization of (n). For (r_2(n)=12), (\prod (b_i + 1)=3), so exactly one prime (\equiv 1 \mod 4) with exponent 2, and primes (\equiv 3 \mod 4) have even exponents.

Steps:

1. Primes (\equiv 1 \mod 4): ({5,13,17,29}) (since (p^2 \leq 1000)).
2. Count valid (n) for each (p):
   • (p=5) ((p^2=25)): (25 \times 2^a \times m^2 \leq 1000) → 9 values.
   • (p=13) ((p^2=169)): (169 \times 2^a \times m^2 \leq 1000) → 3 values.
   • (p=17) ((p^2=289)): (289 \times 2^a \times m^2 \leq 1000) → 2 values.
   • (p=29) ((p^2=841)): (841 \times 2^a \times m^2 \leq 1000) → 1 value.

Total:](#total)

Summing these gives (9+3+2+1=15).

Answer: (\boxed{15})

作者声明:本文包含人工智能生成内容。