# 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?

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.

Note that

$$9^{n + 1} + 3 = 9 \cdot 9^n + 3 = 9 \cdot (4k - 3) + 3 = 4 \cdot (9k) - 24$$ for some $k$, as $9^n +3 = 4k$.

One easy way to see this, if you aren't required to use induction, is that:

$$9 \equiv 1 \pmod4$$

and so:

$$9^n + 3 \equiv 1^n + 3 \equiv 1 +3 \equiv 4 \equiv 0\pmod4$$