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.