To solve the problem, let's analyze the square of an odd integer and see which of the given forms it matches.

Step 1: Express an Odd Integer

An odd integer can be expressed as: $n = 2m + 1$ where $m$ is an integer.

Step 2: Square the Odd Integer

Let's square $n$: $n^2 = (2m + 1)^2 = 4m^2 + 4m + 1 = 4(m^2 + m) + 1$ Notice that $m^2 + m$ is an integer, so let's set $k = m^2 + m$. Therefore, we can write: $n^2 = 4k + 1$ where $k$ is an integer.

Conclusion:

The square of every odd integer is of the form $4k + 1$, where $k$ is an integer.

Thus, the correct answer is:

b. $4k + 1, \ k \in \mathbb{Z}$

Would you like any further explanation on this topic or have any questions?

Here are five related questions for further practice:

1. What is the general form for the square of an even integer?
2. Prove that the square of an odd integer is always congruent to 1 modulo 4.
3. If $n = 5$, what is the value of $n^2$ and how does it fit into the form $4k + 1$?
4. Is the statement "The square of any integer is of the form $3k$, $k \in \mathbb{Z}$" true or false?
5. What is the square of the odd integer $7$, and into what form does it fit from the options provided?

Tip: Whenever you're given a question involving odd or even integers, remember to express them as $2m + 1$ or $2m$ to simplify the problem.