求救…召喚數學師

同上一條有啲似不過我唔想用binomial 廢事寫
不過的確可以試下， 都係一個approach 💀🙏

0^2 = 0, 0 mod 4 = 0
1^2 = 1, 1 mod 4 = 1
2^2 = 4, 4 mod 4 = 0
3^2 = 9, 9 mod 4 = 1

我直接寫咁o5ok：

Since numbers larger than 4 simply repeat the same behavior modulo 4, we don’t need to check numbers like 5, 6, 7, 8, etc. They behave exactly the same as 1, 2, 3, and 0 (respectively) modulo 4.

because 5 and 1 are congruent modulo 4 (5 ≡ 1 mod 4).
because 6 and 2 are congruent modulo 4 (6 ≡ 2 mod 4).
because 7 and 3 are congruent modulo 4 (7 ≡ 3 mod 4).
And so on …

Sure, let's prove why squares of integers modulo 4 can only be 0 or 1.

### Proof:

Consider any integer \( n \). When we take its square \( n^2 \), we need to look at all possible residues (remainders) of \( n \) modulo 4.

An integer \( n \) can be congruent to 0, 1, 2, or 3 modulo 4. Let's analyze each case:

1. **Case 1: \( n \equiv 0 \pmod{4} \)**
\[ n = 4k \, \text{for some integer} \, k \]
\[ n^2 = (4k)^2 = 16k^2 \equiv 0 \pmod{4} \]

2. **Case 2: \( n \equiv 1 \pmod{4} \)**
\[ n = 4k + 1 \]
\[ n^2 = (4k + 1)^2 = 16k^2 + 8k + 1 \equiv 1 \pmod{4} \]

3. **Case 3: \( n \equiv 2 \pmod{4} \)**
\[ n = 4k + 2 \]
\[ n^2 = (4k + 2)^2 = 16k^2 + 16k + 4 \equiv 0 \pmod{4} \]

4. **Case 4: \( n \equiv 3 \pmod{4} \)**
\[ n = 4k + 3 \]
\[ n^2 = (4k + 3)^2 = 16k^2 + 24k + 9 \equiv 1 \pmod{4} \]

### Summary:
\[
\begin{aligned}
&\text{If} \; n \equiv 0 \pmod{4}, \; n^2 \equiv 0 \pmod{4} \\
&\text{If} \; n \equiv 1 \pmod{4}, \; n^2 \equiv 1 \pmod{4} \\
&\text{If} \; n \equiv 2 \pmod{4}, \; n^2 \equiv 0 \pmod{4} \\
&\text{If} \; n \equiv 3 \pmod{4}, \; n^2 \equiv 1 \pmod{4} \\
\end{aligned}
\]

As we see, the possible results for \( n^2 \) modulo 4 are always 0 or 1, proving the statement.

If you have more questions or need further clarification, let's dive in! 😊

…叫佢gen 啲 readable text 好啲.. 咁好難睇

姐係寫呢個包埋n > 4 個d cases? 算係唔用MI，用分case 做咁，哦4k +1 包埋5， 4k+2 包埋6 咁 ， for some integer k, okok 又學撚到

Btw 呢個咩AI嚟 兄弟💀🙏， gpt 4-o?

咪所有數除4餘數係0,1,2,3 ， 呢個寫法好常用

copilot