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The problem asks to prove that the square of any odd integer \( n \) has the form \( 8m + 1 \) for some integer \( m \). Let's go through the steps shown in the image and fill in the blanks as needed.

### Step 1: Division Algorithm
For any integer \( n \), it can be expressed as:
\[
n = 4q + r \quad \text{with} \quad 0 \leq r < 4
\]

### Step 2: Possible Values of \( r \)
- **(a) If \( r = 0 \)**, then \( n = 4q \).
- **(b) If \( r = 1 \)**, then \( n = 4q + 1 \).
- **(c) If \( r = 2 \)**, then \( n = 4q + 2 \).
- **(d) If \( r = 3 \)**, then \( n = 4q + 3 \).

### Step 3: Ignoring Even \( n \) Cases
Since we are concerned with odd integers \( n \), we can ignore cases (a) and (c):
- **Case (a)**: \( n = 4q = 2(2q) \), which is even.
- **Case (c)**: \( n = 4q + 2 = 2(2q + 1) \), which is also even.

This means \( n \) is even in both cases, so they don't apply to our situation.

### Step 4: Consider Odd Cases
Now consider the cases where \( n \) is odd, i.e., cases (b) and (d).

- **Case (b)**: 
  \[
  n = 4q + 1 
  \]
  Then:
  \[
  n^2 = (4q + 1)^2 = 16q^2 + 8q + 1 = 8(2q^2 + q) + 1
  \]
  Here, \( m = 2q^2 + q \), so \( n^2 = 8m + 1 \).

- **Case (d)**:
  \[
  n = 4q + 3
  \]
  Then:
  \[
  n^2 = (4q + 3)^2 = 16q^2 + 24q + 9 = 8(2q^2 + 3q + 1) + 1
  \]
  Here, \( m = 2q^2 + 3q + 1 \), so \( n^2 = 8m + 1 \).

### Conclusion
In both cases where \( n \) is odd, \( n^2 \) has the form \( 8m + 1 \). Thus, the claim is proved.

---

Would you like more details on any part of this explanation? Here are five related questions you could consider:

1. How can we use this method to prove properties of other forms of integers?
2. What happens if we consider different modulo conditions instead of \( n = 4q + r \)?
3. Can you show how this proof method works for other cases, like \( n^2 = 4m + 1 \) for even numbers?
4. Why is the choice of \( r \) limited to 0, 1, 2, and 3 in this context?
5. How does this proof generalize to different mathematical structures, like polynomials?

**Tip:** When proving properties about squares of integers, consider the division algorithm and modular arithmetic for simplification.

Prove the square of any odd integer n has the form 8m + 1 for some integer m.

This page provides a step-by-step proof of why the square of any odd integer n has the form 8m + 1 for some integer m. Using the division algorithm and modular arithmetic, we explore the cases for odd integers and prove the statement.

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