# Show that $7\mid(3^{2n+1}+2^{n+2})$ for all $n\in\mathbb{N}$ [duplicate]

Prove that the following is true for every $$n∈ℕ$$:

$$7\mid(3^{2n+1}+2^{n+2}).$$

I've noticed $$3^{2n+1}+2^{n+2} =3^{2n} \cdot 3+2^{n} \cdot 4.$$

Any suggestions how to continue from there to get something like $$7k$$ for $$k\in\mathbb{N}$$.

• What are your thoughts so far? :) May 26 '14 at 15:25
• Two options: Induction, or use $3^{2n}=9^n$ and modular arithmetic. May 26 '14 at 15:28
• Hint: if $a \equiv b (\text{mod } c)$, then we'll have $a^n \equiv b^n (\text{mod } c)$, and $9 \equiv 2 (\text{mod } 7)$. Or you can try proof by induction, it should work too. :) May 26 '14 at 15:30
• @ThomasAndrews There is also a method using recurrence relations which can be used to find the sequence $k_n$ May 26 '14 at 15:30
• @MarkBennet What is the difference between recurrence and induction? May 26 '14 at 15:45

Hint $\ \ 7\mid\color{#c00}{3^2}\!\color{#c00}-\!\color{#c00}2,\,\ 7\mid\overbrace{\color{#0a0}{3^{2k+1}}\!\color{#0a0}+\! \color{#0a0}{2^{k+2}}}^{\large P(k)} \Rightarrow\, 7\mid 2(\color{#0a0}{3^{2k+1}\!\!+\!2^{k+2}}) + (\color{#c00}{3^2\!-2})3^{2k+1}\! =\, \overbrace{3^{2k+3}\!+2^{k+3}}^{\large P(k+1)}$

Or, notice $\ 2^2\!+2+1 = 7\$ so we can apply

Lemma $\ \ a^2\!+a+1\mid a^{n+2}+(a+1)^{2n+1}\! =: b$

Proof $\, \ {\rm mod}\,\ a^2\!+a+1\!:\ \color{#0a0}{a(a+1)\equiv -1}$ and $\,\color{#c00}{a^3\equiv 1}\$ by $\,0\equiv (a\!-\!1)(a^2\!+a+1) = a^3\!-1,\,$ so

$\qquad\quad\! a^{2n+1}b = a^{3n+3} + (\color{#0c0}{a(a+1)})^{2n+1} \equiv\, (\color{#c00}{a^3})^{n+1}\!-1\equiv 0\$ so $\ b\equiv 0\ \$ QED

Remark $\$ Below I explain how the first explicit inductive proof is a special case of the latter congruence arithmetic proof, which boils down to $\color{#0a0}{(-1)^{2n+1}\equiv -1}\,$ and $\color{#c00}{1^{n+1}\equiv 1},\,$ both of which have trivial inductive proofs (a special case of the Congruence Power Rule inductive proof).

Here is the inductive step $\,P(k)\,\Rightarrow\,P(k\!+\!1)\,$ written in intuitive congruence arithmetical form

$$\begin{eqnarray} {\rm mod}\ 7\!:\quad\ \ \color{#c00}{3^{\large 2}} &\equiv& \color{#c00}2\\ \color{#0a0}{3^{\large 2k+1}}&\equiv& \color{#0a0}{-2^{\large k+2}},\quad\ {\rm i.e.}\ \ \ P(k)\\ \Rightarrow\ \ 3^{\large 2(k+1)+1} &\equiv& \color{#c00}{3^{\large 2}}\: \color{#0a0}{3^{\large 2k+1}}\\ &\equiv& \color{#c00}2 (\color{#0a0}{- 2^{\large k+2}})\\ &\equiv& {-}\!2^{\large k+3},\quad {\rm i.e.}\ \ \ P(k+1) \end{eqnarray}\qquad$$

by the Congruence Product Rule $\ A\equiv a,\ B\equiv b\,\Rightarrow\, AB\equiv ab.\,$ If congruence arithmetic is unfamiliar it can be eliminated by unwinding the proof of the Product Rule, yielding

$\quad \begin{eqnarray} 0\,\equiv\, \color{#c00}{A}&\color{#c00}-&\color{#c00}a, &&\ \color{#0a0}{B}&\color{#0a0}-& \color{#0a0}b &\Rightarrow& \qquad AB\ -\ ab &=& a\ \ (\color{#0a0}{\ B\ \ -\ \ b}\ ) &+& (\color{#c00}{A-a})B\,\equiv\, 0\\ 7\,\mid\, \color{#c00}{3^2}&\color{#c00}-&\color{#c00}2, && \color{#0a0}{3^{2k+1}}\!&\color{#0a0}+&\! \color{#0a0}{2^{k+2}} &\Rightarrow& 7\mid 3^{2k+3}\!+2^{k+3}\! &=& 2(\color{#0a0}{3^{2k+1}\!+2^{k+2}}) &+& (\color{#c00}{3^2\!-2})3^{2k+1}\phantom{I^{I^I}} \\ \end{eqnarray}$

The prior is precisely the standard inductive proof that is usually pulled out of hat, like magic, without any intuitive motivation. We can employ further congruence arithmetic to make it even more obvious than above. Note $\,P(k)\,$ is $\,3\cdot 9^k\equiv -4\cdot 2^k\equiv 3\cdot 2^k\$ so $\,P(k\!+\!1)$ arises simply by multiplying by $\,9\equiv 2\$ to get $\, P(k\!+\!1)\!:\ 3\cdot 9^{k+1}\equiv 3\cdot 2^{k+1}.\,$ Even more clearly, by dividing, we see that $\,P(k)\,$ is equivalent to $\,(9/2)^k\equiv 1.\,$ But $\,9\equiv 2\,$ so $\,9/2\equiv 1,\,$ so the induction boils down to the trivial induction that $\,1^k\equiv 1,\,$ which is a simple special case of the inductive proof of the Congruence Power Rule.

Similarly, many inductions can be transformed into such standard or trivial inductions. Hence it is well-worth the effort to spend some time looking for such innate structure. This is especially true for divisibility problems, since transforming to congruence form allow us to exploit our well-honed arithmetical intuition, which is much stronger than our intuition on divisibility relation caculus.

Note that $2 \equiv 3^2 \pmod 7$, therefore $$3^{2n+1} + 2^{n+2} \equiv 3^{2n+1} + 3^{2n+4} \equiv 3^{2n+1} \cdot 28 \pmod 7,$$ and that is of course divisible by $7$.

\begin{align}3^{2n+1}+2^{n+2}=3\cdot 9^n + 4\cdot 2^n&=7\cdot 9^n -4(9^n-2^n)\\ &=7\cdot 9^n-4(9-2)(9^{n-1}+9^{n-2}\cdot2+\cdots+2^{n-1})\\ &=7[ 9^n-4(9^{n-1}+9^{n-2}\cdot 2\cdots+2^{n-1})]\end{align}

Which is divisible by $7$

Here is a solution based on congruences,

$$3\cdot 9^n +4\cdot 2^n \equiv 3\cdot 2^n +4\cdot 2^n \equiv 7\cdot 2^n \equiv 0\text{(mod 7)}$$

• Thank you. I was closer to the first method. I tried rewriting $3$ to $(7-4)$ and $4$ to $(7-3)$, whereas I should only rewrite $3$ to $(7-4)$ and leave alone the number $4$. I wonder when I would figure that out on my own. May 26 '14 at 15:42

We have

$$3^{2n+1}+2^{n+2}=3\times\color{red}9^n+4\times2^n\equiv3\times\color{red}2^n+4\times2^n=7\times2^n\equiv0\mod7$$

Note that a sequence with form $u_n=a\cdot 9^n+b\cdot 2^n$ satisfies a linear recurrence with auxiliary equation $$(x-9)(x-2)=0$$ so that $$u_{n+1}=11u_n-18u_{n-1}$$

Hence if $7|u_n \& 7|u_{n-1}$ then $7|u_{n+1}$. We have $u_0=7, u_1=35$ both divisible by $7$, so with a base case established we are done by induction.

If $u_n=7k_n$, then $k_n$ satisfies the same recurrence with $k_0=1, k_1=5$

Just surprised that the very crude induction method with $u_n=3\cdot 9^n+4\cdot 2^n$ hasn't been posted yet - it goes $$u_{n+1} = 3\cdot 9^{n+1}+4\cdot 2^{n+1}=7\cdot3\cdot 9^n+2\cdot 3\cdot 9^n+2\cdot 4\cdot 2^n=7\cdot 3\cdot 9^n+2\cdot u_n$$ which is divisible by $7$ if $u_n$ is divisible by $7$. Base case is $u_0=7$. It depends on the fact that $9-2$ is divisible by $7$.