Using mathematical induction prove that for $$ {n \ge 1} \in \mathbb{Z} $$
$$ 3^{2k}-1 $$ is divisible by 8.

Question

anonymous · Answer

show it for certain number n ... show it hods for n+1

phoenixfire · Answer

Yeah I got the base step.
n=1
$$3^2-1=8k$$ where k is some Integer.
$$8=8k$$ so it's divisible.

phoenixfire · Answer

Then the inductive hypothesis is: Assume $$3^{2k}-1$$ is divisible by 8 when k=n
Now to prove it hold for k=n+1, I get confused and don't know what to do.

anonymous · Answer

this is difficult.
let's try without induction first.
you have to show that there is 3 even factors.

anonymous · Answer

3^(2k) - 1 = (3^k)^2 - 1^2 = (3^k - 1)(3^k + 1) <-- one of them is divisible by 4. and both is divisible by 2 ... hence it is divisible by 8

phoenixfire · Answer

Okay, I understand that. Now what?

anonymous · Answer

i don't see short sweet method.

anonymous · Answer

Prove ... if (3^k - 1) is divisible by 2 prove (3^k + 1) is always divisible by 4 using induction.
and vice-versa ...

anonymous · Answer

you want to show that  give $3^{2k}-1$ is divisible by $8$ then $3^{2(k+1)}-1$ is as well

anonymous · Answer

that is, if it is true for $n=k$ then it is true for $k+1$
now some algebra slight of hand

anonymous · Answer

3^(2(k+1)) - 1 =  9(3^(2k) - 1) + 1) - 1 = 9(x+1) - 1 = 9*8x + 8 which is divisible by 8

anonymous · Answer

and the goal is to get $3^{2k}-1$ somehow out of the exression $3^{2(k+1)}-1$ so that you can use the "induction hypothesis" of the term $3^{2k}-1$

anonymous · Answer

Ley 8x = 3^(2k) - 1
3^(2(k+1)) - 1  = 9 ( (3^(2k) - 1) + 1) - 1 = 9( 8x + 1 ) - 1 = 9*8x + 8 <-- hence it is divisible by 8

mimi_x3 · Answer

$$k = n$$
$$3^{2n}  - 1 = 8Q (where ~Q~ is~ any~ integer) => 3^{2n}  = 8Q+1$$
$$k= n+1 $$
$$3^{2(n+1)}  - 1 => 3^{2n+1}  - 1 => 3^{2n}  *3 - 1 => \left(8Q+1\right)  *3 -1$$

anonymous · Answer

attachment

anonymous · Answer

$$3^{2(k+1)}-1=3^{2k+2}-1=3^{2k}3^2-1=9\times 3^{2k}-1$$
$$=(3^{2k}-1)\times 9+8$$

anonymous · Answer

by induction, the terms $3^{2k}-1$ is divisible by $8$ and clearly $9\times 8$ is divisible by $8$ so the sum is divisible by $8$

phoenixfire · Answer

Okay, I now understand, however I have one question... Where the heck did the +8 come from?
You have 
$$(9)3^{2k}-1$$ 
how did that become $$9(3^{2k}-1)+8$$

Actually, I understand @RaphaelFilgueiras 
Is it much different? @satellite73 or am I just missing something in the working.

anonymous · Answer

3²-1

phoenixfire · Answer

Thanks everyone!
I've got it now.

anonymous · Answer

3^(2k) = (3^(2k) - 1) + 1 <--- 8 comes from here

anonymous · Answer

it is a trick of algebra, and the reason to use the trick is to make sure you can get $3^{2k}-1$ out of $3^{2k+2}-1$ so you can use the induction hypothesis

that is how i arrived at it to being with
i just wrote $3^{2k}\times 9-1$ and forced $3^{2k}-1$ out of it