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I was given a problem: Show that $6\mid 7^n - 1$ for $n \in \mathbb{N}$ by induction.

I was then given another problem: Show that $7\mid 8^n - 1$ for $n \in \mathbb{N}$ by induction.

I then speculated that for some $α \in \mathbb{N}, \ α\mid (α + 1)^n - 1$ which I do not have the skills to prove.

I believe that I need to include the binomial theorem to formulate $(α + 1)^n$ but with binomial coefficients included, and with the added use of factorials, I am not familiar with that.

Could somebody please prove/disprove my speculation? Of course I cannot do it by exhaustion because the set of natural numbers $\mathbb{N}$ is infinite. Please do not skip too many steps for I would appreciate an explanation as to why to do this this way instead of that way.

My only attempt I guess for me to try and prove this without using the binomial theorem is proving that for some $x, y \in \mathbb{N}$ we have the following: $$x - y \mid x^n - y^n$$ Though this is just an educated guess, and since I am new to induction, I do not know how to go about this. To let you be aware of my skill level, I can solve expressions that require induction if there is only one variable in it (typically $n, k, t$ and etc).

Thank you in advance :)

Xam
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Mr Pie
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3 Answers3

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Your speculation is correct, for $\alpha\in\mathbb N$ it is true that $\alpha|(\alpha+1)^n-1$, and you can use induction on $n$.

Note that $$(\alpha+1)^{n+1}-1=(\alpha+1)^{n+1}-(\alpha+1)^n+(\alpha+1)^n-1=\alpha(\alpha+1)^n+((\alpha+1)^n-1)$$

and both parts of the last sum are divisible by $\alpha$.

  • I will tick this because it is the only answer that uses an inductive proof and has the less points, so I guess out of generosity, I will give you a tick. You, as for everyone else, will also get an upvote :) $$(+1) \ \ \color{green}{\checkmark}$$ – Mr Pie Dec 11 '17 at 11:28
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Since$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1}),$$it is clear that$$x-y\mid x^n-y^n.$$

José Carlos Santos
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(For a non-inductive proof)

Hint:

Also notice that $$(\alpha+1)^n-1=(\alpha+1)^n -1^n= (\alpha+1-1)(\alpha^{n-1}+…1^{n-1})$$

As a result, what can you conclude?