Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$

Question

Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$

Proof. So i'm going to prove this by induction. The first case when $n=1$ is trivial since: $$9+16=25,$$ implying $ 5 \mid 25$.

Now we need to show is divisible when $n=k+1$. We will use this later on, but $$9^k+4^{k+1}=5k.$$ The n, $$\begin{align}9^{k+1}+4^{k+2} &= 9 \cdot 9^k+4 \cdot 4^{k+1} \\ &= 9 \cdot 9^k+(9-5) \cdot 4^{k+1} \\ &=9 \cdot 9^k+9 \cdot 4^{k+1} -5 \cdot 4^{k+1} \\ &= 9(9^k + 4^{k+1})-5 \cdot 4^{k+1} \\ &=\underbrace{9(5k)-5 \cdot 4^{k+1}}_{\text{This is where the $5k$ comes in}} \\ &= 5(9k-4^{k+1}),\end{align}$$ thus, the original expression a multiple of $5$.

Is my induction correct?

Edit: I see several answers that took a different approach, all is welcome it really helps me see it in a different way. Thank You!

Answer 1

Yes your proof is correct, as an alternative note that

$$9^n+4^{n+1}\equiv (-1)^n+(-1)^{n+1}\equiv0 \mod 5$$

Answer 2

By the binomial theorem, $9^n+4^{n+1}=(5+4)^n+4^{n+1}=5a+4^n+4^{n+1}=5a+5\cdot4^n$.

Answer 3

Or another way: $$9^n + 4^{n+1} = (10-1)^n +(5-1)^{n+1}=10c+(-1)^n + 5d+(-1)^{n+1}=10c+5d = 5e$$

Edit:

Another similar way: $$9^n + 4^{n+1}= 9^n+4\cdot(9-5)^n = 9^n+4(9^n+5c)= (1+4)9^n + 20c = 5e$$

Answer 4

HINT.-$9^n$ is congruent with $1$ or $9$ modulo $10$ and $4^m$ is congruent with $4$ or $6$ modulo $10$. The integer $9^n+4^{n+1}$always end with $5$.

Answer 5

By Induction:

$n=0$: $9^0+4^1=5$ , divisible by $5$.

Hypothesis:

Assume : $9^n+4^{ n+1}$ is divisible by $5$.

Step:

$9^{n+1}+4^{n+2} =9 \cdot 9^{n} +4 \cdot 4^{n+1}=$

$(5+4)9^n +4 \cdot 4^{n+1} =$

$ 5 \cdot 9^n +4(9^n+4^{n+1})=$

$ 5 \cdot 9^n +4\cdot 5c= 5(9^n+4c)$,

where $c \in \mathbb{Z^+}$, since $(9^n+4^{n+1})$

is by hypothesis divisible by $5$,