# Prove that $6^{n+1} + 4$ is divisible by 4

I hope you can understand me, English isn't my main language.

I have a superior algebra problem that I can't solve.

Prove that for every n in Natural numbers ($N$) $$6^{n+1} + 4$$ is divisible by 4

-First I prove that for 1 is True. Correct.

-Then I create a group of numbers that make that True $$M = \{ r \in N / 6^{r+1} + 4 \to ~\text{is divisible by 4} \}$$

-The I have to prove that $$6^{(m+1)+1} + 4$$ is divisible by 4

This is where I'm stuck, none of my ideas have worked.

Hope you can help me.

This actually doesn't require an inductive proof; you can note that for $n\geq1$, $$6^{n+1}+4=36\cdot 6^{n-1}+4=4(9\cdot 6^{n-1}+1)$$ and is therefore divisible by 4.
If you WANT to use an inductive proof, you have to figure out how the assumption that $6^{n+1}+4$ is a multiple of 4 helps to show that $6^{n+2}+4$ is a multiple of four.
You could start by writing $6^{n+1}+4=4k$ for some $k$, and noting $$6^{n+2}+4=6(6^n+4-4)+4=6(4k-4)+4=4(6(k-1)+1))$$ and is therefore divisible by $4$ as well.
Note that $6=2*3$, you see that $6^{n+1}=2^{n+1}3^{n+1}$, and it follows immediately.
Note that the part +4 is useless, so you can investigate only $6^{n+1}$. But $6^{n+1} = 2^{n+1} * 3^{n+1}$,
So for any $n >= 1$ it is true.