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.