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The following are two exercises from "Guide to Abstract Algebra" by Carol Whitehead, 1988 first edition.

  1. Prove by induction that the integer $u_n = 4^n + 6n -1$ is divisible by 9, for every $n \in \mathbb{Z}^+$.

  2. Prove by induction that the integer $u_n = 7^n - 4^n$ is divisible by 3, for every $n \in \mathbb{Z}^+$.

I can do exercise 6, as follows:

  • We show the proposition is true for $n=1$. That is, $u_1 = 7^1 - 4^1 = 3$, which is divisible by 3.

  • Next we show that if the proposition is true for $n$, it is also true for $n+1$. We can do this by showing that $u_{n+1} - u_n$ is a multiple of 3.

  • $u_{n+1} - u_n = 7^{n+1} - 4^{n+1} - 7^n + 4^n = 7^n\cdot(7-1) - 4^n\cdot(4-1) = \textbf{3}\cdot2\cdot7^n - \textbf{3}\cdot4^n$, which is clearly divisible by 3.

Trying the same approach for the earlier exercise 5:

  • We show the proposition is true for $n=1$. That is, $u_1 = 4^1 + 6\cdot1 -1 = 9$, which is divisible by 9.

  • Next we show that if the proposition is true for $n$, it is also true for $n+1$. We can do this by showing that $u_{n+1} - u_n$ is a multiple of 9.

  • $u_{n+1} - u_n = 4^{n+1}+6(n+1)-1 - 4^n - 6 n + 1 = 3\cdot4^n + 6$.

Question: I can't see how $3\cdot4^n + 6$ is divisible by 9.

There are many questions on this site that ask for proofs of this same proposal but none actually take the solution in this direction, and don't answer the specific question asked above.

Penelope
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  • Try to show that $u_{n+1}+5u_{n}$ is divisible by $9$ – Student Mar 26 '23 at 01:43
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    Without induction: $;4^n+6n-1=\color{blue}{(3+1)^n}+6n-1=\color{blue}{3^2\cdot(\dots) + 3n + 1} + 6n -1,$. – dxiv Mar 26 '23 at 01:57
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    Again induction... – Bob Dobbs Mar 26 '23 at 02:11
  • See the linked dupe (and its links) for more proofs than you can count. – Bill Dubuque Mar 26 '23 at 02:21
  • @dxiv No, that does use induction (on $n)$ - it simply hides the induction in a Lemma, viz. the proof of the Binomial Theorem (or its first two terms). – Bill Dubuque Mar 26 '23 at 03:09
  • @BillDubuque Right, though difference is whether said induction must be included as part of the proof, or is part of "previous art" that can be used without proving. It doesn't have to be the binomial theorem, it follows just as easily from Vieta's relations. – dxiv Mar 26 '23 at 03:22
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    @BillDubuque Thanks but the other solutions on this site take the solution in a different direction. They don't answer the specific question here. – Penelope Mar 26 '23 at 03:59

1 Answers1

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For exercise 5, with the assumption that $u_n$ is divisible by $9$, to continue on your thoughts,

$$\begin{align*} u_{n+1} - u_n &= 3\cdot 4^{n} + 6\\ &= 3(4^n +6n-1) - 3(6n-1) + 6\\ &= 3u_n - 18n + 3+6\\ &= 3u_n + 9(-2n + 1) \end{align*}$$

which is divisible by $9$.

Alternatively,

$$\begin{align*} u_{n+1} &= 4^{n+1} + 6(n+1) - 1\\ &= 4(4^n + 6n -1) - 4(6n -1) + 6(n+1) - 1\\ &= 4u_n - 24n + 4 + 6n +6-1\\ &= 4u_n +9(-2n +1) \end{align*}$$

which is divisible by $9$.

peterwhy
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