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Prove that $3^n+8^n$ is not divisible by 5.

I know that this can be proved by using congruence and I am providing the proof by congruence below. But is there any way to Prove It By Induction.

The proof by congruence goes like this:

$3\equiv 3\pmod 5 \\ 3^2 \equiv 4\pmod 5 \\ 3^3\equiv 7\pmod 5 \\ 3^4\equiv 1\pmod 5 \\ 3^5\equiv 3\pmod 5$

Also,

$8\equiv 3\pmod 5 \\ 8^2 \equiv 4\pmod 5 \\ 8^3\equiv 7\pmod 5 \\ 8^4\equiv 1\pmod 5 \\ 8^5\equiv 3\pmod 5$

Adding the congruence up (since the same cycle repeats after the 4th power) none of them are divisible by 5 or equal to 0.

But I need a proof by Induction.

Any help will be appreciated.

John Omielan
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rockin numbers
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    You can use this question and add $2\cdot 3^n$, which is not divisible by $5$. Hence $3^n+8^n$ is not divisible by $5$. – Dietrich Burde Jul 22 '19 at 15:04
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    Does it have to be by induction? It seems to be much faster just to observe that $$3^n+8^n \equiv 3^n+3^n = 2\cdot 3^n \pmod 5$$ which certainly can never have any factor of $5$. If you want to shoehorn in an induction step, prove by induction that $3^n\equiv 8^n\pmod 5$. – hmakholm left over Monica Jul 22 '19 at 15:05
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    It would help if you explained why you need a proof by induction, otherwise your potential answerers who are perfectly happy with the congruence proof have little motivation to help you. – Lee Mosher Jul 22 '19 at 15:07
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    @HenningMakholm Today all these questions are with induction, whether reasonably or not. – Dietrich Burde Jul 22 '19 at 15:07
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    This is not by induction, but if you multiply through by $2^n$ (not divisible by $5$) you obtain (modulo $5$) $6^n+16^n\equiv 1^n+1^n= 2$ - just for the record, and to illustrate another approach. – Mark Bennet Jul 22 '19 at 15:20
  • $$\large \begin{align}{\bf Hint}\ \ &f(n!+!1) \ =\ 3,f(n) + 5\cdot 8^n\[.4em] \Rightarrow\ \ 5,\mid &,f(n!+!1)!\iff! 5\mid f(n)\end{align}\qquad\qquad$$ – Bill Dubuque Jul 22 '19 at 15:31
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    The proof you gave !!*IS*!! a proof by induction! – fleablood Jul 22 '19 at 15:33
  • A proof that isn't by induction would be that $8\equiv 3 \pmod 5$ so $3^n + 8^n\equiv 23^n\pmod 5$ and $5,2,3$ being prime $5\not\mid 23^n$. But the proof you did that $3^{n+4k} + 8^{n+4k}\equiv 3^{n} + 8^n$ and that is not divisible by $5$ if $n = 1,2,3,4,5$ IS a proof by induction. – fleablood Jul 22 '19 at 15:38
  • Please don't be misled by the false claims that the congruence argument you gave already is a proof by induction. That is not true. What is true is that it is possible to use that idea to give a rigorous inductive proof, e.g. by inductively proving $,3^{\large n} = 3^{\large 4k +r}\equiv 3^{\large r},$ using $,3^{\large 4}\equiv 1,,$ and similarly for $,8^{\large n}.\ \ $ – Bill Dubuque Jul 22 '19 at 16:38
  • If one is going to insist this is not a proof by induction, then I think one should say the reason is because it is not a proof at all. But it is an argument by induction. It is not the case that this is a non-inductive proof. It very much is an argument by induction. The question is whether this argument by induction is a proof or not. Bill Dubuque seems pretty insistent it is not. I don't really care but do agree it can use some tightening up. Particularly "the cycle repeats" should be stated more explicitly as to how that implies it's true for all values. – fleablood Jul 22 '19 at 18:01

7 Answers7

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Yes, you can do it by induction as well. Note that:

$3^{n+1}+8^{n+1}=3^n\times 3+8^n\times (3+5)=(3^n+8^n)\times 3+5\times 8^n$

By induction hypothesis the first term is not divisible by $5$ while the second term is obviously divisible by $5$. The required result follows.

Mark
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  • "you can do it by induction as well". BUt what the OP had done, was induction. This is doing it by induction differently. – fleablood Jul 22 '19 at 15:45
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You can go like this. For the base case, let $n=1$. Then $$3^n+8^n = 3+8 = 11,$$ which is not divisible by $5$.

For the induction step, let $n\geq 1$ be arbitrary and assume that $3^n+8^n$ is not divisible by $5$. Now

\begin{align*} 3^{n+1} + 8^{n+1} &= 3\cdot 3^n + 8\cdot 8^n\\ &= 3\cdot 3^n + (3+5)\cdot 8^n\\ &= 3\cdot 3^n + 3\cdot 8^n + 5\cdot 8^n\\ &= 3(3^n + 8^n) + 5\cdot 8^n. \end{align*} The first term is not divisible by $5$ by the induction hypothesis, and because $3$ and $5$ are relatively prime, so since $5$ divides the second term, it does not divide their sum.

Mankind
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Just to be a bit different:

note that $3^{n+1}+8^{n+1}=8(3^{n}+8^n)-5(3^n)$

J. W. Tanner
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=====Answer 3:======

It's important that one realizes that whenever they use an argument that a pattern repeats or an observation will recur indefinitely, they are fundamentally relying upon and using the Principal of induction. To wit:

They are showing something is true for a few base cases; They are (hopefully-- sometimes this step is weak---) that if it is true for some cases is will follow through for the next cases; and they make it clear that this will repeat and be true for an infinite or indefinite number of iterations.

So your argument is an argument of induction.

You've shown for base cases: $n = 1,2,3,4,5$ That $3^n + 8^n $ are none divisible by $4$.

You state that the cycle repeats. (You actually need to give a reason why the cycle repeats. That is why if $3^n + 8^n\equiv K \pmod 5$ why $3^{n+4} + 8^{n+4} $ is also $\equiv K \pmod 5$. You just noticed that $3^{5} \equiv 3^{1}$ and $8^{5} \equiv 8^{1}$ and assumed that means it is true for all $n$ and $n + 4$. You have to justify this.)

And therefore you concluded it is true for all $n$.

It is a principal if induction that allows you to conclude this.

So if you can give a reason why $3^{n+4}\equiv 3^{n}$ and $8^{n+4}\equiv 8^n$ you would be done.

(Hint: $3^{n+4} = 3^{n-1}3^5\equiv 3^{n-1}3^1 \equiv 3^n\pmod 5$. That is, after all, the reason you assumed the cycle repeated, isn't it?)

===== Answer 2: =======

You DID a proof by induction!

Notice the key phrase in you proof and the phrase that assures that you are done is:

since the same cycle repeats after the 4th power

This means that if it is true for $3^n + 8^n$ it will be true for $3^{n+4} + 8^{n+4}$ and so by induction:

As you showed a Base case that it is true for $n = 1,2,3, 4$ (as well as $n=5$ and an induction case that if it is true for $n$, we can conclude it it true for all $n = 1+4k, 2+4k, 3+4k, 4+4k$. WHich means it is true for all $n$.

That IS a prove by induction.

....

But another proof by induction follows.

===== Answer 1: ======

Well, follow the rules of a proof by induction.

Base case: $n=1$

$3^1 + 8^1 =11 $ which is not divisible by $5$.

Base case done:

Inductive case:

Assume that $3^n + 8^n$ is not divisible $5$.

Now we need to prove that thereform $3^{n+1} + 8^{n+1}$ is not divisible by $5$.

Now my advice is that when you need to prove something about $P(n+1)$ is to put it into terms of $P(n)$ and use what you know about $P(n)$.

$3^{n+1} + 8^{n+1} = 3*3^n + 8*8^n = 3*3^n + 3*8^n + 5*8^n= 3*(3^n + 8^n) + 5*8^n$ and....

$5$ is prime. $5\not \mid 3$ and $5\not \mid (3^n + 8^n)$ and $5|5*8^n$ so $5 \not \mid 3*(3^n+8^n) + 5*8^n$.

Induction step done.

Principal of induction declares we are done. Base case: $3^n + 8^n$ is not divisible by $5$ for $n = 1$. Induction case: If $3^n+8^n$ is not divisible by $5$ for a value of $n$ then in will not be divisible by $5$ for the next value of $n$. Therefore: As we can get to all values of $n$ by starting at $1$ and then going the next, and the next after that, and so on.... it must be true that $3^n +8^n$ is not divisible by $5$ for any natural $n$.

By the way...

fleablood
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  • Au contraire: there is no "proof by induction" in the OP. OP has an idea for an inductive proof, but does not have a rigorous proof. It is absolutely essential to distinguish between the two. Handwaving is not proof. – Bill Dubuque Jul 22 '19 at 15:37
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    Hmmmm.... point taken. But if one thinks it is "absolutely essential" that his proof by formal, then I think our efforts need to be to point out that his proof (which the OP does think is complete) is not; and although the OP thinks his idea is not induction, that we should point out that it most certainly is, and our advice should not be to find a different proof but to help him formally complete the induction idea he began. – fleablood Jul 22 '19 at 15:44
  • The OP's claim that "the cycle repeats" is an idea for an inductive proof. The point of teaching induction is to show how to make rigorous such pre-inductive intuition. It is pedagogically damaging to claim that such handwaving suffices for proof. As such it would be much better to rephrase your claim "That IS a prove(sic) by induction" – Bill Dubuque Jul 22 '19 at 16:48
  • Thanks for your help. I have understood the parts I have left out in my proof. – rockin numbers Jul 22 '19 at 17:48
  • Maybe... but that misses the point. The OP should learn to recognize inductive thinking and "the cycle repeats" is inductive thinking. I think I am going to disagree with you about what is and is not pedagogically damaging. I think expecting proofs to have magic formulas that if you say the right magic words they work, is pedagogically damaging. And again, if this is not an acceptable proof it is because it needs, formalizing; NOT because it isn't induction. – fleablood Jul 22 '19 at 17:51
  • Ok, then I'm afraid I will have to downvote. "The cycle repeats" is a claim that needs to be both precisely formulated then rigorously proved by induction. Your claim that such vague handwaving suffices as proof is very misleading and damaging to those who wish to learn induction. OP did NOT do a proof by induction. – Bill Dubuque Jul 22 '19 at 20:33
  • I never said such "handwaving" suffices as a formal proof. I said such handwaving was an induction argument. I don't care and I don't think any question was asked about formality or rigor. But a question was asked about induction. I think it is vital that a student is taught that any "the cycle repeats" and "and so on statement" rely upon and imply induction and induction is essential to such arguments. To NOT do so is to be misleading and damaging. Any criticism about handwaving or rigor is just arguing over price. – fleablood Jul 22 '19 at 21:32
  • But we don't seem to disagree about inductive intuition. Rather, we disagree about what constitutes an inductive proof (you twice wrongly claimed that the OP did a proof by induction). If you replaced "proof" by "idea" or "intuition" then I would not object. In my experience teaching induction I have seen many students struggle greatly while attempting to convert their intuition about similar "cycling" into a rigorous inductive proof. You may be misjudging this difficulty given that you already know well such techniques. – Bill Dubuque Jul 23 '19 at 00:55
  • In particular it will not be clear to many readers precisely which "principal(sic) of induction" you implicitly invoke in "Answer 3" (it is not even clear to me, since I can see various ways). It is essential to be precise, otherwise anyone can claim that some handwaving about "cycling" is a "proof by induction". But it is not. – Bill Dubuque Jul 23 '19 at 01:04
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Alternatively,

$$3^n + 8^n \equiv 3^n + 3^n \equiv 2*3^n \not\equiv 0 \bmod 5$$

because $2*3^n$ does not contain a factor $5$.

J. De Ro
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I think that

$$3^{n+1}+8^{n+1}=3(3^n+8^n)+(8-3)8^n=3P+5Q$$

with P not a multiple of 5, according to the recurrence, and Q integer, will do.

Guillaume Bell
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COMMENT.- Another way very easy: Take modulo $10$

If $n=4k,\space 4k+1,\space 4k+2,\space 4k+3$ we have respectively

$3^n+8^n\equiv7,\space 1,\space 3,\space 9\pmod{10}$ It is never congruent to $5$ because all multiple of $5$ ends in $5$ or $0$.

Piquito
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