Prove that this number is divisible by 7

Question

Without using induction, how can it be proved that 7 divides $3^{2n+1}+2^{n+2}$ for each $n\in\mathbb{N}$? I tried to expand it using $\frac{x^{n+1}-1}{x-1}=1+x+..+x^n$ but I had no success. It would be great if more than one proof is provided.

It would be great if even one attempt were provided! What have you tried? — , Aug 18 '20 at 22:50
In plain language, the quantity under consideration is zero modulo 7! — Daniel Akech Thiong, Aug 18 '20 at 23:13
See also: Prove $3^{2n+1} + 2^{n+2}$ is divisible by $7$ for all $n\ge0$, How to prove that $2^{n+2}+3^{2n+1}$ is divisible by 7 using induction?, Find all $n$ such that $3^{2n+1}+2^{n+2}$ is divisible by $7$, Prove by induction: $2^{n+2} +3^{2n+1}$ is divisible by $7$ for all $n \in \mathbb{N}$, Show that $7\mid(3^{2n+1}+2^{n+2})$ for all $n\in\mathbb{N}$. — Martin Sleziak, Aug 19 '20 at 07:01
A general advice is try to search whether the question isn't already answered on the site. For example, you can try Approach0 or SearchOnMath. For more details, see: How to search on this site? — Martin Sleziak, Aug 19 '20 at 07:03

Donald Splutterwit · Accepted Answer · 2020-08-18T23:23:37.823

2

\begin{eqnarray*} \sum_{n=0}^{\infty} (3^{2n+1}+2^{n+2})x^n = \frac{3}{1-9x}+\frac{4}{1-2x} = \frac{ \color{red}{7} (1-6x)}{(1-9x)(1-2x)}. \end{eqnarray*} This function clearly has integer coefficients \begin{eqnarray*} \frac{ (1-6x)}{(1-9x)(1-2x)}=(1-6x) \left( 1 +9x+81x^2+ \cdots \right) \left( 1 +2x+4x^2+ \cdots \right). \end{eqnarray*}

edited Aug 18 '20 at 23:23

answered Aug 18 '20 at 22:58

Donald Splutterwit

37,323

Do not you mean $\frac{2}{1-2x}$ instead of $\frac{4}{1-2x}$? – Aug 18 '20 at 23:08
It is perfect... but how can I deduce that 7 divides each term of the series? – Aug 18 '20 at 23:20
Great solution! – Aug 18 '20 at 23:26
Thanks @Gio $ \ddot \smile$ – Donald Splutterwit Aug 18 '20 at 23:27

score 0 · Answer 2 · answered Aug 18 '20 at 22:54

0

HINT: Simplify $3^{2n+1}+2^{n+2}$ modulo $7$, using the fact that $3^{2n+1}=3\cdot 3^{2n}$ and $3^2\equiv2\pmod7$.

answered Aug 18 '20 at 22:54

Brian M. Scott

631,399

1

Was just about to write this too – arunoruto Aug 18 '20 at 22:55
for this method, would you have to do case work? – C Squared Aug 19 '20 at 00:36
@CSquared: No; it’s a one-line proof. – Brian M. Scott Aug 19 '20 at 00:46

score 0 · Answer 3 · edited Aug 18 '20 at 23:34

0

$3^{2n + 1} + 2^{n+2} = 3\cdot 3^{2n} + 2^2\cdot 2^n = 3\cdot(9)^n + 4\cdot s2^n\equiv 3\cdot(2)^n + 4\times 2^n = 7\cdot 2^n\equiv 0\pmod 7$.

edited Aug 18 '20 at 23:34

amWhy

210,739

answered Aug 18 '20 at 23:00

fleablood

130,341

score 0 · Answer 4 · answered Aug 19 '20 at 02:02

0

$3^{2n+1}+2^{n+2}=3\times9^n+4\times2^n=7\times2^n+3\times(9^n-2^n)$

$=7\times2^n+3\times(9-2)(9^{n-1}+\cdots+2^{n-1})=\color{red}7\times2^n+3\times\color{red}7(9^{n-1}+\cdots+2^{n-1})$

answered Aug 19 '20 at 02:02

J. W. Tanner

63,683
4
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Prove that this number is divisible by 7

4 Answers4

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