Prove that $3^{2n-1} + 2^{n+1}$ is always a multiple of $7$.

Question

I'm trying to prove the following statement: $P(n) = 3^{2n-1} + 2^{n+1}$ is always a multiple of $7$ $\forall n\geq1$. I want to use induction, so the base case is $P(1) = 7$ so that's okay.

Now I need to prove that if $P(n)$ is true then $P(n+1)$ is true. So there exists a $d \in \mathbb{N}$ such that $$ 3^{2n-1} + 2^{n+1} = 7d $$ From this I need to say that there exists a $k \in \mathbb{N}$ such that: $$ 3^{2n+1} + 2^{n+2} = 7k $$ With a little algebraic manipulation, I have managed to say: $$ 2 \cdot 3^{2n+1} + 9 \cdot 2^{n+2} = 7\cdot(18d) $$ But now I am stuck. How should I keep going?

Answer 1

If

$$2 \cdot 3^{2n+1} + 9 \cdot 2^{n+2} = 7\cdot 18d$$

Then

$$2 \cdot 3^{2n+1} + 2 \cdot 2^{n+2} = 7\cdot 18d - 7 \cdot 2^{n+2}$$

And we conclude

$$3^{2n+1} + 2^{n+2} = \frac{7(18d - 2^{n+2})}{2} = 7(9d - 2^{n+1})$$

Answer 2

$3^{2n-1}+2^{n+1}=9^n3^{-1}+2^n2\equiv 2^n3^{-1}+2^n2=5\times2^n+2\times 2^n=7\times2^n\equiv0 \pmod7 $ since $3^{-1}\equiv5 \pmod7$

Answer 3

$$2\equiv 9$$ $$2\equiv 3^2$$

$$2^{n+1}\equiv 3^{2n+2} $$

$$3^{2n-1}+3^{2n+2}=3^{2n-1}(1+27)$$

$$=4.7.3^{2n-1} $$

Done

Answer 4

The idea is to extract the $n$-th case from $(n+1)$-th case. To show it for $n+1$: $$\begin{align}&3^{2(n+1)-1}+2^{(n+1)+1}= \\ &3^{2n+1}+2^{n+2}= \\ &9\cdot 3^{2n-1}+2\cdot 2^{n+1}= \\ &2(3^{2n-1}+2^{n+1})+7\cdot 3^{2n-1}.\end{align}$$

Answer 5

We can proof it by induction; it is clear that $P(n)$ hold for $n=1$ and we Assume that it hold for $n$ so $$2^{n+1}=7k - 3^{2n-1}$$ Show it for $n+1$ $$ 3^{2(n+1)-1} + 2^{(n+1)+1} = 3^{2n+1} + 2.2^{n+1}= 3^{2n+1}+2(7k - 3^{2n-1})= 3^{2n+1} + 2.3^{2n-1}+14k = (9-2).3^{2n-1}+14k= 7(3^{2n-1}+2k)$$

Answer 6

Alternatively, suppose $3^{2n-1}+2^{n+1}= 7x$.

Then, $x=\frac17\left(3^{2n-1}+2^{n+1}\right)=\frac{1}{21}\left(3^{2n}+6(2^{n})\right)=\frac{1}{21}\left(9^{n}+6(2^{n})\right)\equiv 0 \pmod3$

Now, $9\equiv 2 \pmod7\Rightarrow9^n\equiv 2^n \pmod3$, $6\equiv -1 \pmod7$ and hence, $6\cdot2^n\equiv (-1)\cdot2^n\pmod7$

Therefore, $9^{n}+6(2^{n})\equiv 0 \pmod7$, because $9^{n}\equiv 2^n \pmod7$ and $6\cdot2^{n}\equiv -2^n \pmod7$

Therefore, since $9^{n}+6(2^{n})\equiv 0 \pmod3$, $9^{n}+6(2^{n})\equiv 0 \pmod{21}$, hence $\frac{1}{21}\left(9^{n}+6(2^{n})\right)=\frac{1}{7}\left(3^{2n-1}+2^{n+1}\right)$ is an integer.