Divisiblity proof with coprimes of 6 Can someone please tell me if this proof is okay? Define "okay" as acceptable and rigorous. I wanted to avoid being to wordy so I did not prove a few trivial aspects. I feel the proof is quite verbose but I think as concise as I could get it without skipping too many details.
Suppose $(6,a) = (6,b) = 1$. We wish to prove $24 \mid (a^2-b^2)$. We will use the division algorithm for this proof. If when $a$ is divided by $6$ there is a remainder of $2$, then a is some multiple of $6$ plus $2$, then $2 \mid a$. The same applies for a remainder of $4$. If when a is divided by $6$ there is a remainder of $3$, then $a$ is a multiple of $6$ plus $3$, so a would be divisible by $3$. $a$ is not divisible by $2$ or $3$, therefore the only possible remainders when $a$ is divided by $6$ are $1$ and $5$. The same applies to $b$. so $a = 6p+1$ or $a = 6k+5$ and $b = 6m + 1$ or $b = 6n + 5$ for $p,k,m,n \in \mathbb{Z}$. Note that $a^2 - b^2 = (a+b)(a-b)$. We will prove for all 4 cases that $24 \mid (a+b)(a-b)$. Case 1: $(a+b)(a-b) = (6p + 1 + 6m + 1)(6p-6m) = 12(3(p+m) + 1)(p-m)$. Suppose $p$ and $m$ are both even or odd, then $p-m$ is even. Suppose one of $p, m$ is even and the other is odd. Then $p+m$ is odd, so $p+m = 2x + 1$ for some $x \in \mathbb{Z}$. So $3(p+m)+1 = 3(2x+1)+1 = 2(3x+2)$, so $3(p+m)+1$ is even. So $24 \mid (a^2-b^2)$. Case 2: $(a+b)(a-b) = (6p+1 +6n+5)(6p+1 - 6n - 5) = 12(p + n + 1)(3(p - n) -2)$. Suppose both $p$, $n$ are odd or even, then $p-n$ is even so $p-n = 2x$ for some $x \in \mathbb{Z}$. So $3(p-n)-2 = 2(3x - 1)$ which is even. Suppose one of $p,m$ are even and the other odd, then $p+n$ is odd so $p+n+1$ is even. So $24 \mid a^2-b^2$. Case 3: $(a+b)(a-b) = (6k+5 + 6m + 1)(6k+5 - 6m - 1) = 12(k + m + 1)(3k-3m+2)$. The proof of case 3 is identical to case 2, so $24 \mid a^2-b^2$. Case 4: $(a+b)(a-b) = (6k+5 + 6m +5)(6k-6m) = 12(3(k+m) + 5)(k-m)$. Suppose both k, m are even or odd. Then $k-m$ is even. Suppose one of $k,m$ is even and the other is odd. Then $k+m$ is odd. So $k+m = 2x+1$ for some $x \in \mathbb{Z}$. Then $3(k+m) + 5 = 2(3x+4)$ which is even. So $24 \mid a^2-b^2$. In all cases of a,b, we have proven that $24 \mid a^2 - b^2$. $\square$
 A: May I suggest a shorter proof? (With $a \equiv_n \pm 1$ is meant: $a \equiv 1 \mod n$ or $a \equiv -1 \mod n$ etc.)
$$3\not | a,b \Rightarrow a\equiv_3 \pm 1\mbox{ and } b\equiv_3 \pm 1 \Rightarrow a^2\equiv_3 b^2 \equiv_3 1 \Rightarrow a^2-b^2 \equiv_3 0$$
Hence, $\boxed{3|a^2-b^2}$
$$2 \not | a,b \Rightarrow a\equiv_4 \pm 1\mbox{ and } b\equiv_4 \pm 1 \Rightarrow a^2\equiv_4 b^2 \equiv_4 1 \Rightarrow a^2-b^2 \equiv_4 0$$
$$a^2-b^2 \equiv_4 (a-b)(a+b) \equiv 0 \Rightarrow a-b \equiv_4 0 \mbox{ or } a+b \equiv_4 0 $$
$$2 \not | a,b \Rightarrow a+b\equiv_2 0 \mbox{ and } a-b \equiv_2 0$$
From the last two lines follows: $4 | (a+b) \mbox{ and } 2|(a-b)$ or $2 | (a+b) \mbox{ and } 4|(a-b) \Rightarrow \boxed{8|a^2-b^2}$
A: The logic behind your argument is solid.  Nice job!  So I'll critique the style a bit.
As you gain experience and confidence, your proofs will get shorter, as you get a feel for what is and isn't going to be trivial to your readers.  In that sense, if you looked back on the first paragraph in a few years, you will find the first paragraph to be longer than it needs to be.  So, without seeing them, let me thank you for all the things you did choose to leave out of that paragraph.
You chose to go with four variables there, and I'd have made do with two.  For instance, you can say that $a=6m\pm1$ and $b=6n\pm1$ for some integers $m,n$.  Then you'd have had fewer variables and more regularity in your four cases, which is easier to write and read.
The one yellow flag I'm going to throw is that using $p$ as a variable in a number theory proof is something you should be careful of.  Historically, $p$ and $q$ will represent prime numbers, so there will be a little head-scratching and backtracking to realize that you're using it for an arbitrary integer.  Pairing $k$ with $j$ would have been a better choice.
All in all, though, nice proof for a beginner!
