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To solve the problem of finding the number of non-negative integer solutions ((x,y)) to the equation (2x^2 + y^2 = 2020), we proceed step by step:

Step 1: Analyze parity of (y)

The equation rearranges to (y^2 = 2(1010 - x^2)). Since the right-hand side is even, (y^2) must be even → (y) is even. Let (y=2k) ( (k) non-negative integer). Substitute into the equation:
[2x^2 + (2k)^2 = 2020 \implies x^2 + 2k^2 = 1010]

Step 2: Analyze parity of (x)

(x^2 = 1010 - 2k^2) → even, so (x) is even. Let (x=2m) ( (m) non-negative integer):
[ (2m)^2 + 2k^2 = 1010 \implies 2m^2 + k^2 = 505]

Step 3: Check possible values of (k)

(k^2 = 505 - 2m^2). For (k^2) to be non-negative: (2m^2 \leq 505 \implies m \leq \sqrt{252.5} \approx 15).

Additionally, mod 4 analysis:

• (505 \mod 4 = 1), (2m^2 \mod 4) is 0 or 2.
• (k^2 \mod 4) can only be 0 or 1. Thus, (2m^2 \mod 4 = 0) → (m) even.

Check even (m) values (0,2,...,14) and corresponding (k^2):
None of these yield integer (k) (e.g., (m=0) → (k^2=505) not square; (m=2) → (k^2=497) not square, etc.).

Step 4: Verify all possible (y)

All even (y) (0 to 44) are checked, and none give integer (x).

Answer: (\boxed{0})

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