I'm working on a homework problem for my discrete math class, and I'm stuck. (Note: I made a post about this earlier, but I read the problem incorrectly, thus the work was wrong, so I deleted the post.)
Prove by mathematical induction that for every integer n $\ge$ 0, $12\mid8^{2n+1}+2^{4n+2}$
I start out by proving the base case, $F(0)$, to be true:
$$ F(0)=8^1+2^2=12\quad \text{Obviously, 12 is divisible by 12} $$
I then move on to the induction step to prove the $F(n+1)$ is true:
I assume that $8^{2n+1}+2^{4n+2}$ is divisible by 12, and then plug in $(n+1)$:
$$ F(n+1)=8^{2(n+1)+1}+2^{4(n+1)+2}=8^{2n+3}+2^{4n+6} $$
I then do $F(n+1)-F(n)$:
$$ F(n+1)-F(n)=(8^{2n+3}+2^{4n+6})-(8^{2n+1}+2^{4n+2}) $$
$$ =8^{2n+3}-8^{2n+1}+2^{4n+6}-2^{4n+2} $$
I then factor out the terms used in $F(n)$:
$$ 8^{2n+1}(8^2-1)+2^{4n+2}(2^4-1)=8^{2n+1}(63)+2^{4n+2}(15) $$
I can re-write the result as:
$$ 8^{2n+1}(63)+2^{4n+2}(12+3) $$
This is where I'm stuck. I broke up the $15$ into $12+3$ since I need to prove that there is a multiple of 12, but I don't know what to do with the 63, since (I think) you're supposed to have the terms in $F(n)$ multiplied by 3 after you distribute so that you can factor out the 3 and have $F(n)$ in the equation, which is proven to be divisible by 12.
I tried splitting the $63$ into $(21*3)$
$$ 8^{2n+1}(21*3)+2^{4n+2}(12+3) $$
But I'm not sure what to do next. Any ideas?
If you already have $F(n)=12k$ for some $k\in\mathbb{N}$ and $$ F(n+1)=8^{2(n+1)+1}+2^{4(n+1)+2}=8^{2n+3}+2^{4n+6}, $$ then \begin{align} 8^{2n+3}+2^{4n+6} &=64\cdot8^{2n+1}+16\cdot2^{4n+2}\\ &=12(5\cdot8^{2n+1}+2^{4n+2})+4(8^{2n+1}+2^{4n+2})\\ &=12(5\cdot8^{2n+1}+2^{4n+2}+4k), \end{align} which is divisible by 12.