3

I|m not entirely sure if I'm allowed to ask this, please tell me if I'm not:

As homework our teacher told us to solve that $49^{n} - 2^n$ is divisible by $47$ with $n \in \mathbb N$ using induction. However I'm failing to do this.

If I would be allowed to use modulo calculation this would be proven in 2 lines but induction seems far more complicated to me.

user387499
  • 93
  • Note that the "2 line" modular proof does use induction on $,n,$ (see my answer), so that modular proof certainly counts as such an inductive proof. You can mechanically eliminate the modular arithmetic in my answer and you will obtain the proof in TonyK's answer. I explain how this amounts to compiling the high-level (modular) language into low-level assembly (divisibility) language in this answer. It is essential to learn the high-level modular language in order to master number theory. – Bill Dubuque Nov 13 '16 at 17:58

3 Answers3

2

Using congruences :

Base step : $$47\mid 49^0 - 2^0 = 0$$

Inductive step : Suppose for a certain $n\in \Bbb N$ that : $$47 \mid 49^{n} - 2^n$$ which can, equivalently, be expressed as $$49^{n} \equiv 2^n [47]$$

Then $$49^{n+1} \equiv 2^n 49 \equiv 2^n 2 \equiv 2^{n+1}[47]$$

This ends the induction. Therefore, $$\forall n\in\Bbb N, 47\mid49^n-2^n.$$

amWhy
  • 209,954
Astyx
  • 3,883
  • Isn't that cheating? You could have done the same thing wothout a base step and it wouldn't be called induction right? – user387499 Nov 13 '16 at 14:43
  • 1
    @user387499 What do you mean by cheating ? And why do you think my base step is irrelevant ? – Astyx Nov 13 '16 at 14:50
  • What I meant was that $49^{2n} \equiv 2^n [47]$ already is a true statement and you don't need to suppose it (and so could leave out the rest) if you don't want to work with induction. Or am I missing out onto something? – user387499 Nov 13 '16 at 14:53
  • 1
    @user387499 If you want to prove it you can't use the fact that it is already true, can you ? Or did I missunderstand your claim ? – Astyx Nov 13 '16 at 14:55
  • Actually, user387499, your teacher, according nto your own words, asked you to prove that "$49^{n} - 2^n$ is divisible by $47$ with $n \in \mathbb N$ using induction." Therefore, it cannot be assumed to be true prior to proving it's true. – amWhy Nov 13 '16 at 14:56
  • @Astyx Isn't $49^{2n} \equiv 2^n [47]$ the same as $2 \equiv 2 [47]$? And we know that it is true? – user387499 Nov 13 '16 at 14:57
  • @amWhy What I meant was that given by the properties of modulo it is true, isn't it? – user387499 Nov 13 '16 at 14:58
  • 1
    @MorganRodgers You are right. Anyway I edited my answer to match the question. – Astyx Nov 13 '16 at 15:00
  • A proof using induction on $n$ requires that you first show that the base case (n=1), is true. Only then, do you assume $49^k - 2^k$ is true for $k=n$. But that is merely an assumption, then, given that assumption, if you then prove $49^{k+1} - 2^{k+1}$, you will have proven that $49^n=2^n$ – amWhy Nov 13 '16 at 15:01
  • @amWhy I know that. I didn't try to say that he was doing a mistake. What I tried to point out is that there has to be a way without modulo since if you would be using modulo you would be easier of just leaving out the induction part because you can see that $49^{n} - 2^n \equiv 0[47]$. And since the induction proof given here is inspired by this thought, I don't feel like this is an answer my teacher was looking for why I considered it to be cheating. I don't know if that's clear. – user387499 Nov 13 '16 at 15:04
  • 1
    @user387499 Yes, using modulo properties you can do this avoiding an induction, however this is "cheating" in the sense that you need induction to prove the modulo properties you want to use in the first place. (which is $a\equiv b [c] \implies a^n \equiv b^n [c]$) – Astyx Nov 13 '16 at 15:05
  • 1
    @Astyx Ohh okay. Now I understand it. THANK YOU! – user387499 Nov 13 '16 at 15:08
  • 1
    @user387499 It's my pleasure – Astyx Nov 13 '16 at 15:09
  • Worth explicit mention is that the proof above is a special case of a proof of the general Congruence Power Rule, which proceeds similarly by inductively applying the Congruence Product Rule (e.g. see my answer). To master elementary number theory it helps to know well all of these congruence rules. As here, they often serve to greatly simply inductive proofs. – Bill Dubuque Nov 13 '16 at 17:54
2

$$49^{n+1}-2^{n+1}=49\cdot 49^n - 2\cdot 2^n = 49(49^n-2^n) +47\cdot2^n$$

And the right-hand side is divisible by $47$ if $49^n-2^n$ is.

TonyK
  • 64,559
  • This is exactly the same as the congruence based proofs in the other answers using the Congruence Product Rule, except the language of congruences have been eliminated, i.e. you have inlined the proof of the Product Rule. I explain that e.g. in this answer. – Bill Dubuque Nov 13 '16 at 15:22
  • 2
    @Bill: Yes. But it's exactly what the OP wanted, isn't it? – TonyK Nov 13 '16 at 15:24
  • @TonK I cannot speak for the OP. My point was only to emphasize the these types of proofs are nothing but higher-level languages proofs.compiled into assembly language. – Bill Dubuque Nov 13 '16 at 15:26
1

Hint $\ {\rm mod}\ 47\!:\,\ 49\equiv 2\,\Rightarrow\, 49^n\equiv 2^n\,$ is special case of the Congruence Power Rule (below). You can either prove it as a Lemma, or mimic its inductive proof in this special case.

Congruence Power Rule $\rm\ \ \color{}{A\equiv a}\ \Rightarrow\ \color{#c00}{A^n\equiv a^n}\ \ (mod\ m)\ \ $ for all naturals $\rm\,n.$

Proof $\ $ For $\rm\,n=0\,$ it's $\,1\equiv 1\,$ so true. $\rm\,A\equiv a,\ A^n\equiv a^n \Rightarrow\, \color{#c00}{A^{n+1}\equiv a^{n+1}},\,$ by the Congruence Product Rule, so the result follows by induction on $\rm\,n.$

Community
  • 1
Bill Dubuque
  • 272,048
  • Per the OP: "If I would be allowed to use modulo calculation this would be proven in 2 lines but ..." (OP is asking about "the result (that follows) by induction on $n$.") – amWhy Nov 13 '16 at 15:16
  • @amWhy The OP already accepted an answer that is a special case of this proof, so I wanted to explain how it works in general. See the comments on the other answer. – Bill Dubuque Nov 13 '16 at 15:19
  • Bill D: please don't pester me with your unending self-justifications. I've said what I needed to say regarding the OP's post (and of that, only one comment to your answer .) If you feel you haven't said enough *to* the OP's, then edit your answer rather than pestering me. Convince the asker of the question. Address the asker. – amWhy Nov 13 '16 at 16:33