Find all positive integers n such that n!+2023 is a perfect square.

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The video illustrates step-by-step how to find all positive integers n such that n!+2023 is a perfect square. Use proof by contradiction to prove that n!+2023 can't be a perfect square when n is greater than 3.

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For additional problem:

(even number)^2 can be divided by 4. After (odd number)^2 is divided by 4, the remainder is 1.
From the condition that m is an even number, (m^2)*(n^2) can be divided by 4. By the way, 9999÷4=2499・・・3
∴　If (m^2)*(n^2)+9999 is divided by 4, the remainder is 3. (If it were perfect square, the remainder should be 0 or 1.)
Therefore, (m^2)*(n^2)+9999 is NOT a perfect square number.

Is this proof OK?

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