Mathematical induction... divisible by 4

Hello I need to proof that the expression $(9^{n}+3)$ is divisible by $4$.

It is true if I calculate it for $n=1$

for $n + 1$ I got stuck in here:

$9 \cdot 9^n + 3$

I don't know how to continue. Can anyone help me please?

Answer

Suppose $9^n + 3$ is a multiple of $4$. Then

$$ (9^{n+1}+3)-(9^n+3)=9^{n+1}-9^n =9^n(9-1)=8*9^n, $$

which is a multiple of four. The difference of $9^{n+1}+3$ and $9^n+3$ is a multiple of 4, and $9^n + 3$ is a multiple of 4, so $9^{n+1}+3$ is a multiple of 4.