How to prove that $4 \mid 3^{2n} - 1$ Prove that $3^{2n} - 1$ is divisible by $4$ for any positive integer $n$.
My work:
Let $p(n)$ be true, and let $p(n+1)$ be true. So, $3^{2(n+1)}-1$.
Can anyone help to solve this?
 A: $3^{2n} - 1 = (3^n)^2 - 1 = (3^n - 1)(3^n + 1)$.
$3$ is odd.  So $3^n$ is odd for all positive $n$.  So $3^n\pm 1$ is even.
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You are clearly Trying to do a proof by induction but have no idea what a proof by induction is.
So this is how you do a proof by induction:
0) Clearly identify the statement you are trying to prove in terms of a specific variable $n$.
In this case:  ** $3^{2n} - 1$ is divisible by $4$**.  
Now some texts will have you write that is statement notation as $P(n):$$3^{2n} - 1$ is divisible by $4$, but I don't see that as helpful if a student isn't going to know what that means.  Instead we get students parroting meaningless statements such as "Let $p(n)$ be true" that the clearly don't understand but thing is ritual incantation.  (What is $p(n)$?  You can't let something be true; either it is or it is not, etc.)
1) Base step:  You prove the statement is true for $n= 1$.
So:  Prove $3^{2*1} -1$ is divisible by $4$.
The base step is often just a matter of doing a single calculation.
In this case:  $3^{2*1} -1 = 3^2 - 1 = 9-1 = 8 = 4*2$ is divisible by $4$.
2) Induction step: You assume that there is an number $k$ for which the statement is true.  You know there is a number $k$ for which it is true because you just showed it is true for $1$.  So it is true when $n=k=1$.  Then show that IF it is true for $k$ then the statement is also true for $k + 1$.
(So because you know it is true for $k=1$ then if you proof this induction step you will know it is true for $n = k + 1 =2$.  Then because you know this is true for $k = 2$ that it will be true for $n = k + 1 = $.  Then because you know this is true for $k = 3$ then it will be true for ....)
If $3^{2k} - 1$ is divisible by $4$.  That is if $3^{2k}-1 = 4M$ for some integer $M$, your task is to prove $3^{2(k+1)} -1$ is divisible by $4$, probably by using the fact that $3^{2k} - 1$ is divisible by $4$.
So you need to prove if $3^{2k} - 1 = 4M$ then $3^{2(k+1)} - 1 = 4N$ for some integer $N$.
.....
And then we are done:
We have proven that for a) $n= 1$ then $3^{2n} -1$ is divisible by $4$.  b) that if for $n=k$ that if $3^{2k} -1$ is divisible by $4$ then $3^{2(k+1)} - 1$ is divisible by $4$.  So it is true for $n = 1+1=2$ and for $n=2+1 =3$, anad for $n=3+1 =4 $ and so on for all possible natural numbers $n$.
A: Hint: $9 \equiv 1 \pmod 4$, and $3^{2n} = {(3^2)}^n =9^n$.
A: Hint for the induction step:
Suppose $3^{2n}-1$ is divisible by $4$. Then $$3^{2(n+1)}=9\times 3^{2n}-1=9(3^{2n}-1)+9-1=9(3^{2n}-1)+8$$
must be divisible by $4$.
A: Note that 
$$3\equiv -1 \mod 4 \implies 3^{2n}-1\equiv (-1)^{2n}-1\equiv 0 \mod 4$$
A: Without using induction, observe that
$$
3^{2n}-1=9^n-1=(9-1)(9^{n-1}+9^{n-2}+\dotsb+1)
$$
for $n\geq 1$ by the formula for a finite geometric series. Thus $8\mid 3^{2n} -1$ and in particular $4\mid3^{2n}-1$.
A: $3^{2n}-1=(4-1)^{2n}= 4M +1-1= 4M$
