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 $24 \mid a^2 - b^2$. $\square$

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 $2 | (a+b) \mbox{ and } 4|(a-b) \Rightarrow \boxed{8|a^2-b^2}$
