suppose n is an odd natural number. prove that n^2 can have only remainder 1 or 3 when divided by 6. 

To prove that \( n^2 \) can have only a remainder of 1 or 3 when \( n \) is an odd natural number and divided by 6, we can proceed with a number-theoretic approach using modular arithmetic.

### Step-by-Step Proof

1. **Representing Odd Natural Numbers**:
   Since \( n \) is an odd natural number, we can represent it in the form:
   \[
   n = 2k + 1
   \]
   where \( k \) is an integer. This representation covers all odd numbers (e.g., 1, 3, 5, ...).

2. **Finding \( n^2 \) Modulo 6**:
   Let's calculate \( n^2 \):
   \[
   n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1
   \]
   Since \( 2(2k^2 + 2k) \) is clearly divisible by 2, it is an even number, and \( n^2 \) is of the form \( 2m + 1 \) where \( m = 2k^2 + 2k \). Therefore, \( n^2 \) is odd.

   To find the remainder when \( n^2 \) is divided by 6, we look at possible values of \( n \mod 6 \).

3. **Possible Odd Residues Modulo 6**:
   Odd numbers can have two possible residues modulo 6: 1 and 5. Other odd residues (e.g., 3) aren't considered because we are focusing on odd numbers, and other numbers like 0, 2, 4 are even. Therefore, we check for:

   - Case 1: \( n \equiv 1 \pmod{6} \)
   - Case 2: \( n \equiv 5 \pmod{6} \)

4. **Calculating \( n^2 \) in Each Case**:

   - **Case 1: \( n \equiv 1 \pmod{6} \)**
     \[
     n^2 \equiv 1^2 \equiv 1 \pmod{6}
     \]

   - **Case 2: \( n \equiv 5 \pmod{6} \)**
     \[
     n^2 \equiv 5^2 \equiv 25 \equiv 1 \pmod{6} \quad (\text{since } 25 = 6 \times 4 + 1)
     \]

5. **Conclusion**:
   In both cases, we observe that \( n^2 \equiv 1 \pmod{6} \). Since these are the only possible cases for odd numbers \( n \) modulo 6, we conclude that:

   \[
   n^2 \text{ has a remainder of } 1 \text{ when divided by 6.}
   \]

Therefore, \( n^2 \) can only have a remainder of 1 when \( n \) is an odd natural number and divided by 6.

### Summary
- For any odd \( n \), \( n \equiv 1 \) or \( n \equiv 5 \pmod{6} \).
- \( n^2 \equiv 1 \pmod{6} \) in both cases.

Thus, we've shown that \( n^2 \) will only have a remainder of 1 when \( n \) is an odd natural number divided by 6.

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Would you like more details or have any questions? Here are some related questions for further exploration:

1. What is the remainder of \( n^2 \) when \( n \) is even and divided by 6?
2. Can the proof be extended to show that \( n^2 \) for all integers \( n \) has specific remainders when divided by 6?
3. How can we generalize this proof to other modulo bases, such as 8 or 12?
4. What happens if we consider the cube of an odd number, \( n^3 \), modulo 6?
5. How does this result connect to quadratic residues in number theory?

**Tip:** Understanding modular arithmetic is crucial in number theory and can help solve a variety of problems related to divisibility and remainders.

Explore the proof that n^2 can only have remainders of 1 or 3 when divided by 6 for odd natural numbers. This mathematical problem utilizes modular arithmetic and number theory, crucial for understanding number properties and remainders in advanced high school mathematics.

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