5

Question:

Let $a$ be an integer different than $1$. Prove by induction, that for any $n \in\Bbb N$, $ a^n -1$ is divisible by $a-1$.

My attempt:

Base Case is trivial.

I.H: Assume $ a^k -1$ is divisible by $a-1$, $k \in N$

Then,

$$ a^{k+1} -1 = a\cdot a^k -1 + a^k -1 - a^k + 1 = a\cdot a^k - a^k + a^k -1 = (a^k)(a-1) + (a^k -1)$$which is divisible by $ a-1$.

Hence, proven by induction.

Is this the correct way to prove it using induction? Is there a more efficient way to prove it using induction?

Bill Dubuque
  • 272,048
  • 4
    This is clever and correct. You should probably note explicitly where you use the inductive hypothesis. That said I think it's terrible to ask you to prove this by induction when there's an easy well known one line argument. – Ethan Bolker Jul 05 '17 at 00:37
  • 4
    Looks pretty efficient to me (of course there are non-inductive ways to prove it which are more intuitively satisfying). – lulu Jul 05 '17 at 00:37
  • 1
    You are basically translating a proof that of geometric series, that $\sum^k a^j = \frac{a^{k+1} - 1}{a - 1}$, into an inductive proof. – DanielV Jul 05 '17 at 08:54

2 Answers2

2

Hint $\ $ The inductive step follows by telescopy. $ $ Let $f(k)= a^k-1$

Then $\,\color{#0a0}{f(k\!+\!1)\!-\!f(k)} = a^{k+1}-a^k = (a\!-\!1)a^k$ is divisible by $\,a\!-\!1$

so if $\,\color{#c00}{f(k)}$ is divisible by $\,a\!-\!1\,$ then so too is $\,f(k\!+\!1) = \color{#0a0}{f(k+1)\!-\!f(k)} + \color{#c00}{f(k)}$

Remark $\, $ Summing the above makes the telescopic cancellation explicit

$$\begin{align}a^{k+1}-1 =\, &\,\ \ \color{#c00}{a^{k+1}\!-a^k}+\color{#0a0}{a^k\!-a^{k-1}} + \cdots+ a^1\!-a^0\\[.3em] =\, &\,(a-1)\, (\color{#c00}{a^k} + \color{#0a0}{a^{k-1}}+\cdots+1),\ \ \text{or, more formally}\\ f(k\!+\!1)-f(0) =\,& \sum_{i=0}^k (f(i\!+\!1)\!-\!f(i)) = \sum_{i=0}^ka^i(a\!-\!1) = (a\!-\!1)\sum_{i=0}^k a^i\end{align} $$

The first equality above, expressing $f(k\!+\!1)-f(0)$ as the sum of its first differences, has a trivial (and obvious) proof by induction. Once you prove this general form inductively, you can use it as a lemma to inductively prove many special cases - which are ubiquitous. You can find many examples and much further discussion of telescopic induction in various prior posts.

Bill Dubuque
  • 272,048
  • 1
    Nice solution, Bill! This is the method that I had originally thought of but the OP wanted a proof by induction; thus, I wrote the above. One could also write out the telescopic cancellation explicitly and be done with the proof after this line (since $a-1$ divides the RHS, $a-1$ divides the LHS as well). – Mee Seong Im Jul 05 '17 at 03:37
  • @MeeSeongIm The first proof hinted above is most certainly a standard proof by induction - but structured in a way that highlights the role played by telescopy. The remark shows how the first proof is related to telescopic cancellation. Telescopic induction greatly simplifies many induction proofs, e.g. follow the link I gave for many examples. – Bill Dubuque Jul 05 '17 at 12:54
  • @MeeSeongIm I expanded the remark to clarify how using telescopy (as a lemma) does count as doing the proof by induction. – Bill Dubuque Jul 05 '17 at 14:05
0

Let $a>1$ be an integer. Prove by induction that for any $n \in \mathbb{N}$, $ a^n -1$ is divisible by $a-1$.

$\textbf{Proof}.$ Base case: let $n=1$. Then it is clear that $a-1|a-1$.

Induction hypothesis: assume that $a-1|a^k-1$ for some integer $k$. Consider $a^{k+1}-1$, which could be rewritten as $$ \begin{align*} a^{k+1}-1 &= a\cdot a^k + a^k-a^k-1 \\ &= a^k(a-1)+ (a^k-1). \\ \end{align*} $$ Since $a-1|a^k(a-1)$ and $a-1|a^k-1$ by the induction hypothesis, $a-1|a^{k+1}-1$.

Thus $a-1|a^n-1$.

Mee Seong Im
  • 3,190
  • 1
  • 15
  • 22