To determine the number of integer solutions ((x, y, z)) to the equation (x^2 + y^2 + z^2 = 2023), we use modular arithmetic analysis:

Step 1: Analyze squares modulo 4

Squares modulo 4 are either (0) (even numbers: ((2k)^2 = 4k^2 \equiv 0 \mod 4)) or (1) (odd numbers: ((2k+1)^2 = 4k^2 +4k +1 \equiv 1 \mod 4)).

(2023 \mod 4 = 3) (since (2023 = 4 \times 505 + 3)).

The sum of three squares modulo 4 equals 3 only if all three squares are (1 \mod 4) (since (1+1+1=3)). Thus, (x, y, z) must all be odd.

Step 2: Substitute odd variables

Let (x=2a+1), (y=2b+1), (z=2c+1) (where (a,b,c) are integers). Expanding:
[ (2a+1)^2 + (2b+1)^2 + (2c+1)^2 = 2023 \
4(a^2+a + b^2+b + c^2+c) + 3 = 2023 \
4(...) = 2020 \implies a^2+a + b^2+b + c^2+c = 505.
]

But (a^2+a = a(a+1)) is even (product of consecutive integers). The sum of three even terms is even, but 505 is odd. This is a contradiction.

Conclusion: No integer solutions exist.

Answer: (\boxed{0})

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