<s>[INST] 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 $4$.  
 [/INST] 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, $4\mid 3^{2m+1}-3$.
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