Prove that $4$ divides $3^{2m+1} - 3$ 
Prove that $4$ divides $3^{2m+1} - 3$.

By plugging in numbers I can see this is true, but I can't figure out a way to prove this, I was thinking maybe proving first that it is divisible by $2$, and concluding its divisible by $4$.  
 A: There's actually the stronger result 
$$3^{2m +1} - 3 = 3(9^m - 1) = 24\sum_{i=1}^m 9^{m-i}$$
A: There are generally several ways to approach these types of divisibility problems. I am showing two of them.
By Mathematical Induction: Putting $m=0$ we have $3^{2m+1}-3=0$, which is obviously divisible by $4$. Now let $4\mid 3^{2m+1}-3$ for some $m\in N_0$. Let $3^{2m+1}-3=4k$, for some $k\in N_0$. Now $3^{2(m+1)+1}-3=9(3^{2m+1}-3)+24=9\cdot 4k+24=4(9k+6)$, which is also divisible by $4$. So we see that $4\mid 3^{2m+1}-3$ for all $m\in N_0$. 
By Modular Arithmetic: We see that $3^2\equiv1\pmod{4}$. So for any $m\in N_0$, $3^{2m}\equiv1\pmod{4}$ and $3^{2m+1}\equiv3\pmod{4}$. That is, $4\mid 3^{2m+1}-3$.
A: $$3^{2m+1}-3=(4-1)^{2m+1}-3=\sum\limits_{i=0}^{2m+1}{\left( \begin{matrix}
   2m+1  \\
   i  \\
\end{matrix} \right)}{{4}^{2m+1-i}}{{(-1)}^{i}}\,\,\,-3$$
$$3^{2m+1}-3=\underbrace{\sum\limits_{i=0}^{2m}{\left( \begin{matrix}
   2m+1  \\
   i  \\
\end{matrix} \right)}{{4}^{2m+1-i}}{{(-1)}^{i}}}_{4q}\,\,\,-4=4(q-1)$$
A: Here is a different approach.
Note that $3^{2m+1}-3 = 3(3^{2m}-1)$, so it enough to show that $4$ divides $3^{2m}-1$.
To do this, write $3^{2m}-1$ in base $3$ as $\underbrace{22\dots2_3}_{2m \text{ copies of 2}}$ and note that $4_{10} = 11_3$.  Thus, we have : $$\frac{3^{2m}-1}{4}=\frac{22\dots2_3}{11_3} = {2020\dots202}_{3}$$
A: $3^{2m+1}-3=3(3^m-1)(3^m+1)$, and both of the factors $3^m \pm 1$ are even, so their product is divisible by 4.  
A: Proof by induction:
First, show that this is true for $m=1$:
$3^{2+1}-3=24$
Second, assume that this is true for $m$:
$3^{2m+1}-3=4k$
Third, prove that this is true for $m+1$:
$3^{2(m+1)+1}-3=$
$3^{2m+3}-3=$
$9\cdot(\color\red{3^{2m+1}-3})+24=$
$9\cdot\color\red{4k}+24=$
$4\cdot(9k+6)$
Please note that the assumption is used only in the part marked red.
A: Let $S(m)$ be the statement: $3^{2m+1}-3$ is divisible by 4
Basis step: $S(1)$:
$\Rightarrow 3^{2(1)+1}-3=3^{2+1}-3$
$\hspace{26 mm}=3^{3}-3$
$\hspace{26 mm}=27-3$
$\hspace{26 mm}=24$, which is divisible by $4$
Inductive step:
Assume $S(k)$ is true, i.e. assume that $3^{2k+1}-3$ is divisible by 4
$\hspace{59 mm}\Rightarrow 3^{2k+1}-3=4A$; $A\in\mathbb{N}$
$\hspace{59 mm}\Rightarrow 3^{2k+1}=4A+3$
$S(k+1)$: $3^{2(k+1)+1}-3$
$\hspace{12.5 mm}=3^{2k+2+1}-3$
$\hspace{12.5 mm}=3^{2k+3}-3$
$\hspace{12.5 mm}=3^{2}\bullet{3^{2k+1}}-3$
$\hspace{12.5 mm}=9(4A+3)-3$
$\hspace{12.5 mm}=36A+27-3$
$\hspace{12.5 mm}=36A+24$
$\hspace{12.5 mm}=4(9A+6)$, which is divisible by $4$
So, $S(k+1)$ is true whenever $S(k)$ is true.
Therefore, $3^{2m+1}-3$ is divisible by 4.
A: $3=-1(\mod 4)$, $3^{2m}=1(\mod 4)$ and $3^{2m+1}=3(\mod 4) \implies 3^{2m+1}-3=0(\mod 4)$
