你是陈浩南 ？

Sure! Let’s prove that the square of an odd number is always 1 more than a multiple of 4.

Represent an odd number: Any odd number can be written in the form (2n + 1), where (n) is an integer.
Square the odd number: [ (2n + 1)^2 = 4n^2 + 4n + 1 ]
Factor out 4: [ 4n^2 + 4n + 1 = 4(n^2 + n) + 1 ]
Analyze the expression: The term (4(n^2 + n)) is clearly a multiple of 4 because it is 4 times an integer. Therefore, (4(n^2 + n) + 1) is 1 more than a multiple of 4.
Thus, the square of any odd number is always 1 more than a multiple of 4.

Feel free to ask if you have any more questions!

真实

谢谢 已经获得miss neko3的DLC了

2个人加我已经有了

真实