The square of every odd integer is of the form

Question 9Answer

a.

8k+1,k∈Z

8

𝑘

+

1

,

𝑘

∈

𝑍


b.

4k+1,k∈Z

4

𝑘

+

1

,

𝑘

∈

𝑍



c.

3k+1,k∈Z

3

𝑘

+

1

,

𝑘

∈

𝑍



d.

4k+3,k∈Z

To determine the correct form of the square of every odd integer, let's consider the properties of odd integers.

Let \( n \) be any odd integer. An odd integer can be expressed in the form:

\[ n = 2k + 1 \]

where \( k \) is an integer.

Now, let's square this expression:

\[ n^2 = (2k + 1)^2 \]

Expanding the square:

\[ n^2 = (2k + 1)(2k + 1) = 4k^2 + 4k + 1 \]

We can factor out the 4 from the first two terms:

\[ n^2 = 4(k^2 + k) + 1 \]

Notice that \( k^2 + k \) is an integer (let's call it \( m \)), so:

\[ n^2 = 4m + 1 \]

This shows that the square of any odd integer is always of the form:

\[ n^2 = 4k + 1 \]

where \( k \) is an integer.

Therefore, the correct answer is:

**Option b. \( \boxed{4k + 1, \, k \in \mathbb{Z}} \)**

Would you like more details or have any other questions?

Here are five related questions for further exploration:

1. What is the form of the square of an even integer?
2. How would you prove that the square of any integer is congruent to either 0 or 1 modulo 4?
3. Can you determine the form of the cube of an odd integer?
4. What can be said about the square of a number in relation to its remainder when divided by 4?
5. How can this result be used to prove that a number is not a perfect square?

**Tip:** Understanding the algebraic manipulation of expressions like \( (2k + 1)^2 \) can be useful in a variety of number theory problems.

Explore the algebraic form of the square of every odd integer, defined as 4k + 1 where k is an integer. Understand the properties of odd integers and their squares with detailed explanations and examples. Suitable for high school students.

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