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I will happy to see how to solve this question I had in my exam today. Prove that this divide by 4 for any n>0 by induction:

$$3^{3n} + 3^{2n}+ 3^n + 1.$$

Lorin Sherry
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4 Answers4

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We have

$$3^{3n}+3^{2n}+3^n+1=(3^{2n}+1)(3^n+1).$$

Now by induction, $3^m$ is odd for any $m$ because $3^0$ is odd and $3^{m+1}=3\cdot 3^m$ is the product of two odd numbers.

Hence $(3^{2n}+1)(3^n+1)$ is a multiple of $4$.

  • there is no induction in this proof – Nathanael Skrepek Jul 22 '19 at 12:55
  • @NathanaelSkrepek: I will add one, but it has no advantage over the direct proof. –  Jul 22 '19 at 12:57
  • i am not saying that it is meaningful to prove it by induction^^ – Nathanael Skrepek Jul 22 '19 at 12:59
  • +1. I think this answer is the right one for the OP. – quasi Jul 22 '19 at 13:29
  • @quasi: I have completely changed my answer. Feel free to undo your upvote. –  Jul 22 '19 at 13:33
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    @Yves Daoust: Your new argument, though not inductive, is super simple (so the upvote stays). – quasi Jul 22 '19 at 13:34
  • @quasi: not inductive ? Check the second sentence. :) –  Jul 22 '19 at 13:35
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    @Yves Daoust: Well, I guess the induction is hidden in the claim that $3^n$ and $3^{2n}$ are odd. Ah yes, I see you said that. – quasi Jul 22 '19 at 13:37
  • @quasi: isn't "Now by induction" followed by an inductive proof explicit enough ? –  Jul 22 '19 at 13:38
  • @Yves Daoust: Yes, I jumped over that line in my first reading. Still, while I think your proof is "optimal", I doubt that the induction in your argument was the one expected of students for the question on the exam. – quasi Jul 22 '19 at 13:41
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    @quasi: I agree with you. I discarded the "assumed expected" proof because I found it ugly. –  Jul 22 '19 at 13:45
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Note that your sequence can be described as $$a_n=27^n+9^n+3^n+1^n$$

so it ought to coincide with a linear recursion, whose characteristic polynomial has roots $1,3,9,27$.

Accordingly, let us make a recursion using the first few values of your sequence as initial conditions and having $$(x-1)(x-3)(x-9)(x-27)$$ as its characteristic polynomial.

By induction, the sequence determined by that recursion will coincide with your sequence.

Indeed, expanding out the polynomial shows that the recursion we want is $$a_n=40 a_{n-1} - 390 a_{n-2} +1080a_{n-3} - 729a_{n-4}$$

with initial conditions $$a_0=4\quad a_1=40\quad a_2=820\quad a_3=20440$$

As each of those initial values is divisible by $4$, so must all the other terms in the recursion and we are done.

Note: there was no need to write out the recursion explicitly. It was clear that there was one (just expand the given polynomial) and that was all we needed to know. The key point is that, by induction, it suffices to check the first four values.

lulu
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    +1. Your argument is surely more sophisticated than the one expected of the students, but nevertheless, it's an elegant way to prove the required result. – quasi Jul 22 '19 at 13:24
  • @quasi My view (no doubt a minority view) is that this sort of argument actually explains why the claim holds. Of course you can manipulate the expression and obtain a more traditional proof by induction, but that always feels like a numerical accident to me. To me the point is that you only need to check $4$ consecutive values to get all subsequent ones. – lulu Jul 22 '19 at 13:30
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Here is a hint you could use for a proof by induction:

$(3^{3(n+1)}+3^{2(n+1)}+3^{n+1}+1) - (3^{3n}+3^{2n}+3^{n}+1)=$

$3^{3n}(3^3-1)+3^{2n}(3^2-1)+3^{n}(3-1)$

and $3^{3n}(26)+3^n(2)=3^n(3^{2n}26+2)$ and $3^{2n}26+2=2(3^{2n}13+1)$ is divisible by $4$

since $3^{2n} $ is odd so $3^{2n}13+1$ is even.

But I find it easier to prove the claim directly:

$3\equiv-1\pmod4,$ so $3^{3n}+3^{2n}+3^{n}+1\equiv(-1)^{3n}+(-1)^{2n}+(-1)^{n}+1$,

which is $-1+1-1+1=0$ if $n$ is odd and $1+1+1+1=4\equiv0\pmod4$ if $n$ is even.

J. W. Tanner
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Try $n=1$:

$3^3 + 3^2 + 3^1 + 1 = 40$, which is divisible by 4.

Now assume it is true for $n=k$ so that:

$3^{3k} + 3^{2k} + 3^{k} + 1$ is divisible by 4. (*)

We want to show that it is true for $n=k+1$, that is:

$3^{3(k+1)} + 3^{2(k+1)} + 3^{(k+1)} + 1$ is divisible by 4. (**)

(**) $= 27\cdot3^{3k} + 9\cdot3^{2k} + 3\cdot3^{k} + 1$

(**) - (*) $= 26\cdot3^{3k} + 8\cdot3^{2k} + 2\cdot3^{k}$

Now any even number multiplied by 2 is a multiple of 4, so it remains to show that (**)/2 is an even number.

(**)/2 $= 13\cdot3^{3k} + 4\cdot3^{2k} + 3^{k}$

$= (odd \times odd) + (even \times odd) + odd$

$= odd + even + odd$

$= even$

and we are done.

battletwink69
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