Verify my proof: "for all odd integers $a$ and $b$, $b^2-a^2 \neq 4$" I'm learning math without a math professor. I need some feedback from community regarding my proof.
The book is : "Discrete Mathematics with Applications" by Susanna S. Epp, 5th edition.
Exercise 16 from page 226. Please verify my proof.

Prove the following statement: "for all odd integers $a$ and $b$, $b^2-a^2 \neq 4$"

Proof. Let's prove by contradiction. Let say there are two integers $a$ and $b$ and $b^2-a^2 = 4$. Because $a$ and $b$ are odd integers, then $a=2n+1$ and $b=2k+1$, where $n \in Z$ and $k \in Z$. Now let's perform the following computations:
$b^2 - a^2 = (2k+1)^2 - (2n+1)^2 = 4k^2 + 4k + 1 - 4n^2 - 4n - 1 = 4(k^2 + k - n^2 - n) = 4(k^2 - n^2 + k - n) = 4[(k+n)(k-n)+(k-n)] = 4(k-n)(k+n+1)  $
So, in order for the expression $b^2 - a^2 = 4(k-n)(k+n+1) $ to be true, $(k-n)(k+n+1)$ must be equal to 1 or -1. Then $k-n = 1$ and $k+n+1 = 1$ or $k-n = -1$ and $k+n+1 = -1$, because both of expressions should be equal to one (or minus one), we have :
$k-n = k+n+1 \implies -n = n + 1 \implies -n-n = 1 \implies -2n = 1 \implies n = -\frac{1}{2} $, so $n$ is not an integer $\blacksquare$
 A: I'm merely critiquing your presentation.


Prove the following statement: "for all odd integers $a$ and $b$, $b^2-a^2 \neq 4$"

Proof. Let's prove by contradiction. Let say there are two integers $a$ and $b$ and $b^2-a^2 = 4$. Because $a$ and $b$ are odd integers, then $a=2n+1$ and $b=2k+1$, where $n \in Z$ and $k \in Z$.

I suggest replacing the "where $n\in\mathbb Z$" with "for some $n\in\mathbb Z$" (as opposed to "for all $n\in\mathbb Z$"), so that it is immediately clear that you are not referring to an arbitrary $n$ (consequently, an arbitrary odd integer).

Now let's perform the following computations:
$b^2 - a^2 = (2k+1)^2 - (2n+1)^2 = 4k^2 + 4k + 1 - 4n^2 - 4n - 1 = 4(k^2 + k - n^2 - n) = 4(k^2 - n^2 + k - n) = 4[(k+n)(k-n)+(k-n)] = 4(k-n)(k+n+1)  $


So, in order for the expression $b^2 - a^2 = 4(k-n)(k+n+1) $ to be true,

This sentence is confusing: you've already, by assumption, shown it to be true. What you mean is instead something like "Equating both expressions for $b^2-a^2,$ we have that $4(k-n)(k+n+1)=4.$"

$(k-n)(k+n+1)$ must be equal to 1 or -1. Then $k-n = 1$ and $k+n+1 = 1$ or $k-n = -1$ and $k+n+1 = -1$, because both of expressions should be equal to one (or minus one), we have :


$k-n = k+n+1 \implies -n = n + 1 \implies -n-n = 1 \implies -2n = 1 \implies n = -\frac{1}{2} $, so $n$ is not an integer. $\blacksquare$

This is slightly confusing. If I'm merely skimming or if your text is dense, then it is not immediately clear that you are even asserting that $k-n = k+n+1$ (the leftmost antecedent) is true; in other words, it is not immediately clear whether you have actually derived that conclusion $n=-\frac12$ or whether it is provisional on that leftmost antecedent being true. I suggest using words like "so", "thus", "therefore" and "hence" instead, and reserving ⟹ for when you are merely asserting a conditional/implication.
