To solve the problem of finding the number of positive integer solutions ((x, y)) to (x^2 + y^2 = 2023), we use modular arithmetic analysis:

Key Observations:

• Squares modulo 4 are either (0) or (1):

  • Even number squared: ((2k)^2 = 4k^2 \equiv 0 \mod 4).
  • Odd number squared: ((2k+1)^2 = 4k^2 + 4k + 1 \equiv 1 \mod 4).

• Compute (2023 \mod 4):
  (2023 = 4 \times 505 + 3 \implies 2023 \equiv 3 \mod 4).

Sum of Two Squares Modulo 4:

The possible sums of two squares mod 4 are:
(0+0=0), (0+1=1), (1+0=1), (1+1=2).

None of these equal (3 \mod 4). Thus, there are no positive integer solutions ((x,y)) to (x^2 + y^2 = 2023).

Answer: (\boxed{0})

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