Proof:

Let the odd integer be represented as $n$. Since $n$ is odd, it can be written in the form:

$n = 2k + 1$

where $k$ is some integer.

Now, we are asked to prove that the difference between the square of this odd integer and the integer itself is always an even integer. This means we need to show that the following expression is even:

$n^2 - n$

Step 1: Substitute $n = 2k + 1$

We substitute $n$ with $2k + 1$ in the expression $n^2 - n$:

$n^2 - n = (2k + 1)^2 - (2k + 1)$

Step 2: Expand the expression

Now, expand $(2k + 1)^2$:

$(2k + 1)^2 = (2k)^2 + 2(2k)(1) + 1^2 = 4k^2 + 4k + 1$

Substitute this back into the expression for $n^2 - n$:

$n^2 - n = (4k^2 + 4k + 1) - (2k + 1)$

Step 3: Simplify the expression

Now, subtract $(2k + 1)$ from $(4k^2 + 4k + 1)$:

$n^2 - n = 4k^2 + 4k + 1 - 2k - 1$

Simplify the terms:

$n^2 - n = 4k^2 + 2k$

Step 4: Factor out the common factor

Notice that both terms, $4k^2$ and $2k$, have a common factor of 2. So, we can factor out 2:

$n^2 - n = 2(2k^2 + k)$

Step 5: Conclusion

Since $n^2 - n = 2(2k^2 + k)$, the expression is clearly a multiple of 2. Therefore, $n^2 - n$ is always an even integer.

Thus, we have proven that the difference between the square of any odd integer and the integer itself is always even.

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Do you need any further clarification on this proof?

Here are five related questions to explore:

1. How can we prove the same for even integers?
2. What happens if we apply this reasoning to a negative odd integer?
3. Can this concept be generalized to powers greater than 2?
4. How does this relate to the fact that $n^2 - n = n(n-1)$?
5. How can we prove that the sum of an odd number and an even number is always odd?

Tip: The key to recognizing whether a number is even or odd in algebraic proofs is using the form $2k$ for even and $2k+1$ for odd, where $k$ is an integer.