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$4^{n}-1$ is divisible by $3.$

(i) Basis Step: $P(1)$is true because $4^1 -1=3 $ and $3\mid3$.

(ii) Suppose $P(k)$ is true for induction hypothesis $3\mid4^{k}-1$.

Show $(k+1)$ is true $3\mid4^{k+1}$

Now here is where things gets tricky.

$\Rightarrow 4^{k+1} -1 = \left(4^{k}*4 \right)-1 =$

I cannot seem to figure out how one factors this any further. I just cannot seem to make it divisible by 3. Could someone show me how to do this?

Daniel W. Farlow
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Jon
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  • Use \mid instead of the simple | when talking about divisibility. Also, you should specify what your proposition $P(n)$ is. You reference $P(1)$ but never say what $P$ actually is. Also, don't you mean to show that $P(k+1)$ is true due to the assumption that $P(k)$ is true? Not just $(k+1)$ [whatever that might mean]. – Daniel W. Farlow Jul 10 '16 at 01:54
  • In the future, I would strongly recommend that you use this template for writing your induction proofs. Much of it may seem somewhat pointless, but it forces you to write everything out and understand what you are doing. After you have written a number of different induction proofs, then you can relax your expository style, but I imagine it would be helpful at this point for you to flesh everything out as much as possible. – Daniel W. Farlow Jul 10 '16 at 01:59
  • @DanielW.Farlow I will make the template my new desktop wallpaper. If it pleases you. – Jon Jul 10 '16 at 02:04
  • do you know modular arithmetic instead? – Jacob Wakem Jul 10 '16 at 03:13

5 Answers5

3

Inductive assumption: $3 \mid (4^k-1)$, and therefore $(4^k-1) = 3m$ for some $m$.

$4(4^k-1) = 4^{k+1} -4 = 4^{k+1}-3-1$

so $4^{k+1}-1= 4(4^k-1)+3 = 12m+3$ which is divisible by $3$, completing the induction.

Zau
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For $n\geq1$, let $S(n)$ denote the statement $$ S(n) : 3\mid(4^n-1)\Longleftrightarrow4^n-1=3m, m\in\mathbb{Z}. $$ Base case ($n=1$): $S(1)$ says that $3\mid(4^1-1)$, and this is true.

Inductive step: Fix some $k\geq1$, and assume that $S(k)$ is true where $$ S(k) : 3\mid(4^k-1)\Longleftrightarrow4^k-1=3\ell, \ell\in\mathbb{Z}. $$ To be proved is that $S(k+1)$ follows where $$ S(k+1) : 3\mid(4^{k+1}-1)\Longleftrightarrow4^{k+1}-1=3\eta, \eta\in\mathbb{Z}. $$ Beginning with the left-hand side of $S(k+1)$, \begin{align} 4^{k+1}-1&= 4(4^k-1)+3\tag{rearrange}\\[1em] &= 4\cdot3 \ell+3\tag{by $S(k)$}\\[1em] &= 3(4\ell+1)\tag{factor out 3}\\[1em] &= 3\eta,\tag{$\eta=4\ell+1, \eta\in\mathbb{Z}$} \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.

Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq1$. $\blacksquare$

Daniel W. Farlow
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  • If you are wondering where the $m$ and $\ell$ come from and what they are, ask yourself what it means for something to be divisible by 3. – Daniel W. Farlow Jul 10 '16 at 01:51
  • I thought there was a difference between divide $|$ and divisible \ – Jon Jul 10 '16 at 01:52
  • Maybe they were always the same thing I just never thought too deep into it. – Jon Jul 10 '16 at 01:53
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    Saying an integer $\eta$ is divisible by $7$, for instance, means that $\eta$ is a multiple of $7$. And you can write this as follows: $\eta=7\ell$, where $\ell\in\mathbb{Z}$. You could also write $7\mid\eta$, which means "$7$ divides $\eta$." Does that clear things up at all? – Daniel W. Farlow Jul 10 '16 at 01:57
  • Yes now I see the difference. I am still thinking about the m and L though. – Jon Jul 10 '16 at 02:01
  • See the revised proof--it should be much clearer now. – Daniel W. Farlow Jul 10 '16 at 02:09
  • One last question. Is it right to assume. $4^{k+1}-1=-3.$ – Jon Jul 10 '16 at 02:30
  • That is what I think @ZackNi is saying. – Jon Jul 10 '16 at 02:32
  • How could you assume that? Is there anything about my proof that is unclear? If not, then I think you should be on the right track. If so, then please indicate where. – Daniel W. Farlow Jul 10 '16 at 02:33
  • I just thinking where else could the 3 come from. It has to be from the other side. Very thorough proof by the by. – Jon Jul 10 '16 at 02:36
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    $4^{k+1}-1= 4(4^k-1)+3=4\cdot4^k-4\cdot1+3=4^{k+1}-1$ ... you are simply rewriting the expression so that you can use the inductive hypothesis. – Daniel W. Farlow Jul 10 '16 at 02:38
1

$$4^{k+1}-1=4\cdot 4^k-1=4\cdot (4^k-1+1)-1=4(4^k-1)+4(4)-1=4(4^k-1)+15.$$ So $3|(4^k-1)\implies 3|(4^{k+1}-1).$

DanielWainfleet
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Let $S(n)$ be the statement: $4^{n}-1$ is divisible by $3$

Basis step: $S(1)$: $\hspace{5 mm}4^{(1)}-1=3$, which is divisible by $3$

Inductive step:

Assume $S(k)$ is true, i.e. assume that $4^{k}-1$ is divisible by $3$

$\hspace{59 mm}\Rightarrow 4^{k}-1=3A$; $A\in\mathbb{N}$

$\hspace{59 mm}\Rightarrow 4^{k}=3A+1$

$S(k+1)$: $4^{k+1}-1$

$\hspace{12.5 mm}=4\bullet{4^{k}}-1$

$\hspace{12.5 mm}=4\hspace{1 mm}(3A+1)-1$

$\hspace{12.5 mm}=12A+4-1$

$\hspace{12.5 mm}=12A+3$

$\hspace{12.5 mm}=3\hspace{1 mm}(4A+1)$, which is divisible by $3$

So, $S(k+1)$ is true whenever $S(k)$ is true.

Therefore, $4^{n}-1$ is divisible by $3$.

Tazwar
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  • Would this be the same answer as @Daniel W.Farlow? – Jon Jul 13 '16 at 15:40
  • The result is the same, but the methods are slightly different. I directly substituted $3A+1$ in place of $4^{k}$ in the $S(k+1)$ step. – Tazwar Jul 13 '16 at 21:00
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Rewrite statement (*) as $(3+1)^n =3k+1.$ When $n=1$ let $k=1.$

Assume $(3+1)^n =3k+1 $. $(3+1)(3k+1)=3(3k+1)+3k+1=3(3k+1+k)+1 .$

Jon
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Jacob Wakem
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    I hope you realize that it is frowned upon to add multiple answers when one will suffice (unless the answers are wholly different and both quite lengthy, usually you can lump more than answer together in a single post--there is no need for this). – Daniel W. Farlow Jul 10 '16 at 03:12
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    Also, it is high time for you to learn some MathJax. Please use this in the future when asking or answering questions. I believe you would find everyone on MSE to be more receptive. :-) – Daniel W. Farlow Jul 10 '16 at 03:14
  • @DanielW.Farlow I don't want them to be receptive. I want them to think about what I wrote. – Jacob Wakem Dec 29 '16 at 09:31