Proof by Cases involving divisibility

Question

To prove that if $n$ is odd, then $8 \mid n^2−1$, observe that as $n$ is odd either $n = 4k+1$ for some integer $k$ or $n=4k+3$ for some integer $k$. In the first case $n^2 -1=8(2k^2+k)$ and in the second case $n^2-1 = 8(2k^2 + 3k + 1)$.

Given an odd integer $a$, establish that $a^2 +(a+2)^2 +(a+4)^2 + 1$ is divisible by $12$. Solve by consideration of six cases.

Answer 1

Hint $\,f(n\!+\!2)-f(n) = (a\!+\!6)^2\!-a^2 = \color{#c00}{12}(n\!+\!3),\ $ so $\ {\rm mod}\ \color{#c00}{12}\!:\ f(n\!+\!2)\equiv f(n)\ $ therefore by induction $\ 0\equiv f(1)\equiv f(3)\equiv f(5)\equiv \cdots$

Answer 2

HINT:

We don't really need to handle cases separately as

$$(2m+1)^2=8\cdot\dfrac{m(m+1)}2+1$$

Now, $$(a-2)^2+a^2+(a+2)^2+1=3a^2+9=3(a^2-1)+12$$