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I want to prove by induction that for all $n$, 8 is a factor of $$7^{2n+1}+1$$

I have proved it true for the base case and assumed it true for $n=k$, but when I cannot figure out when to go towards the end of proving it true for $n=k+1$ assuming it is true for $n=k$.

I let $$7^{2k+1}+1 = 8m$$

Then I work with $$7^{2(k+1)+1}+1$$ to get eventually $$7^2(8m)$$

I am not sure if this is correct or if it is I am not sure how to prove that this too is dividable by $8$.

I would appreciate any help.

Thanks

Bill Dubuque
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7 Answers7

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Below I explain how nonmodular inductive proofs (e.g. other answers) are really just applications of modular arithmetic product rules. Let's examine closely the inductive step (which is usually presented unmotivated - pulled out of a hat like magic). Below we present the inductive proof in a way that makes it crystal clear how it is a special case of the following general product rules, viz. specialize: $\,\rm X=7^{2n+1},\,\color{#c00}{x=-1},\ \color{#0a0}{Y=7^2},\,y=1\,$ below.

$\qquad\qquad\quad\! 8\mid \overbrace{(7^{\large 2n+1}\!+1)}^{\large\rm P(n)},\quad\ 8\mid (7^{\large 2} - 1)$

$\qquad\: \Longrightarrow\ \ 8\mid (7^{\large 2n+1}\!+\color{#c00}1)\,\color{#0a0}{7^{\large 2}}\color{#c00}{-1}\,(\color{#0a0}{7^{\large 2}}-1)\, =\, \smash{\overbrace{7^{\large 2n+3}-1}^{\large\rm P(n+1)}}\ \ $ is a special case of Proof below

$\begin{eqnarray} \rm {\bf Lemma}\ \ &\rm m\ \ |&\rm\ \, X\!-\!x\quad\ \& &&\rm\! m\ |\: Y\!-\!y \ \Rightarrow\ m\:|\!\!&&\rm XY - \: xy\ \ \ \, {\bf [Divisibility\ Product\ Rule]} \\[.3em] \rm {\bf Proof}\ \ \ \ \ &\rm m\ \ |&\rm (X\!-\!\color{#c00}x)\:\color{#0a0}Y\ \,+ &&\rm\, \color{#c00}x\ (\color{#0a0}Y\!-\!y)\ \ \ \ = &&\rm XY - \: xy \\[.5em] \rm {\bf Lemma}\ \ & &\rm\ \, X\equiv x\quad\ \ \& &&\rm\quad\ Y\equiv y \ \ \ \ \Rightarrow\ &&\rm XY\equiv xy \ \ \ \, {\bf [Congruence\ Product\ Rule]}\\[.3em] \rm {\bf Proof}\ \ \ \ \ &0\equiv& \rm (X\!-\!\color{#c00}x)\:\color{#0a0}Y\,\ + &&\rm\, \color{#C00}x\ (\color{#0a0}Y\!-\!y)\ \ \ \ \equiv &&\rm XY - \: xy \\ \end{eqnarray}$

If congruences are known, the inductive step is obvious by applying said Congruence Product Rule

$$\begin{align} \bmod 8\!:\ \ \ \ \ \color{#c00}{7^{\large 2n+1}}&\equiv\color{#c00}{-1}\\[.2em] \color{#0a0}{7^{\large 2}}&\equiv\ \color{#0a0}1\\[.2em] \Longrightarrow\, 7^{\large 2n+3}= \color{#c00}{7^{\large 2n+1}}\color{#0a0}{ 7^{\large 2}}&\equiv\ \color{#0a0}{1}\:\!\color{#c00}{(-1)}\end{align}\qquad\qquad$$

If congruences are not known then we can still preserve this conceptual structure by rewriting the congruence product rule into lower-level divisibility language, cf. above divisibility product rule.

In either case - to preserve the conceptual structure - we should invoke these rules by name, since instead invoking them by value (i.e. repeating the product rule proofs in this special case) greatly obfuscates the crucial innate arithmetical product structure at the heart of the inductive step.

Thus, with the help of modular language, we see that the induction boils down to the trivial induction that $\, a\equiv 1\,\Rightarrow\, a^n\equiv 1\:\!$ (here $\,\color{#0a0}{a = 7^2}).\,$ Or, $\,7\equiv -1\,\Rightarrow\, 7^n\equiv (-1)^n\equiv -1$ for odd $\,n\,$ by the Congruence Power Rule, which abstracts such iterated applications of the Product Rule.

So the nonmodular proofs can be viewed as the result of compiling the higher-level (congruence) language proofs into lower-level (divisibility) assembly language. We can do such compilation mechanically for any such congruence proof. Just as for software, the low-level assembly language is far less comprehensible since it eliminates higher level conceptual structure - which often leads to great simplification, e.g $\,a\equiv 1\,\Rightarrow\,a^n\equiv 1\,$ above.

Also worth mention is that this proof can be discovered mechanically, i.e. without any required insight, be using the method of multiplicative telescopy.

Bill Dubuque
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    Who could doubt that this is the best answer to the proposed problem? Again the bad habit of the beginners of Stack Exchange to mark as best answer the first that appears to them. In view of this, this mark should be changed to "satisfactory" and not "the best". – Ataulfo Nov 13 '16 at 16:22
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    Justifying downvotes should be a prevailing rule in MSE. It is my conviction, except when the error is blatant and very evident. – Ataulfo Mar 12 '20 at 02:30
  • See this answer for a similar example and further elaboration. $\ \ $ – Bill Dubuque Jul 15 '25 at 22:37
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Hint:

$$7^{2(n+1)+1}+1=49\cdot\left(7^{2n+1}+1\right)-48\implies\ldots$$

DonAntonio
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  • @user2352274 If you wish to understand the arithmetical essence of the inductive step then see my answer, which shows that one does not need to pull this inference out of a hat. – Bill Dubuque Mar 01 '14 at 22:02
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Inductive step

$$7^{(2(n+1)+1)}+1=7^{(2n+1+2)}+1=49\times\left(7^{(2n+1)}+1\right)-8\times 6$$

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Your approach was right until you considered $7^{2k+1}+1=8m$

$49 \times 7^{2k+1} + 1 = 49 \times(8m-1) +1 = 8 \times 49m -48$ which is divisible by 8.

Priyatham
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use modules...

$7 \equiv -1 \pmod 8$, so $$7^{2n + 1} + 1 \equiv (-1)^{2n + 1} + 1 \equiv -1 + 1 \equiv 0 \pmod 8$$ so it is divisble by $8$

(Note that $2n + 1$ is odd, so $(-1)^{2n+1} = -1$)

Ant
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    He did say "by induction." This sort of problem is very often posed before the class has covered modular arithmetic. – Thomas Andrews Mar 01 '14 at 21:05
  • @Thomas Whether explicit or not, the proofs all essentially employ modular product rules - see my answer. – Bill Dubuque Mar 01 '14 at 22:00
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    Of course they do, @BillDubuque, doesn't negate my comment. – Thomas Andrews Mar 01 '14 at 23:16
  • @Thomas The point of the comment was to make that relationship explicit, not to negate anything. It was really targeted at the author (and readers) rather than you. But I pinged you as a courtesy in case you might have something to add (the author of the answer is pinged automatically). – Bill Dubuque Mar 01 '14 at 23:22
  • To be honest, I didn't read induction was required. On the other hand,
    1. It can't hurt to know a second way of doing things, and
    2. My application of modulat arithmetics is really, really trivial. Is something anyone would learn in up to 5 minutes, so it is useful to whoever read the answer. Then, as Bill points out, every proof employ basically the same reasoning, only in an implicit way
     – Ant Mar 02 '14 at 01:26
  • @Ant Agreed (but apparently not everyone agrees, presuming that's what the downvote on my answer signifies). – Bill Dubuque Mar 04 '14 at 20:27
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Hint.-$$7=8-1\Rightarrow 7^{2n+1}=8M-1\Rightarrow7^{2n+1}+1=8M$$

Ataulfo
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$7^{2(0)+1}+1=8: $ Get on a step of the ladder.

$7^{(2k+1)}+1=8m$ : Given you are on a step on the ladder...

$7^{(2(k+1)+1)}+49=49\cdot 8m$: after multiplying by $7^2=49$

$7^{(2(k+1)+1)}+1=49\cdot 8m-48=8(49m-6)$ the property holds for the next step of the ladder.

You added a 1 to your exponent and not your $k$ value in your proof.

TurlocTheRed
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  • This repeats the standard congruence language proof using the Congruence Product Rule (cf. my answer), but eliminates all congruence language by replacing $,a\equiv b\pmod{!8},$ by its definition $,a=b+8m,$ for some $,m\in\Bbb Z.,$ As explained at length in my $11$ year-old answer, doing this compilation into "assembly language" greatly obfuscates the conceptual arithmetical structure at the heart of the proof - which I explain there is simply $,\color{#c00}{7^K\equiv -1},,7^2\equiv 1\Rightarrow 7^{K+2}\equiv \color{#c00}{7^K}7^2\equiv (\color{#c00}{-1})1\equiv -1.\ \ $ – Bill Dubuque Jul 14 '25 at 21:03
  • I don't think OP knew congruences at the time of writing the question. Notice the language in which he asks his question. Mathematically, it might be better to engage the higher concepts. Pedagogically, I have doubts. – TurlocTheRed Jul 14 '25 at 21:06
  • Non-congruence proofs are already given in $4$ eleven year old answers. This answer duplicates those. Please don't duplicate prior answers (unless you have something significantly novel to add). $\ \ $ – Bill Dubuque Jul 14 '25 at 21:10