Using induction, prove that $(-7)^n -9^n$ is divisible by $16$

Question

First of all, I think the problem should be $(-7)^n -9^n$ is divisible by $-16$ because if I test the basis by letting $n=1$, I have $-16$ instead of $16$.

Edit: Alright ... I sort of understand why it really is divisible by $16$. Even though I had $-16$ as the answer for the basis, if I divide $-16$ by $16$ then I get $-1$ , so the book didn't make an error, but the fact that I was dealing with a negative number for the basis made me freak out a little bit.

$(-7)^1 -9^1 = -7-9 = - 16$

Now for the induction...this is my attempt.

$P(k) = (-7)^k -9^k $

For $P(k+1)$

$(-7)^{k+1} - 9^{k+1}$

$(-7)^k * (-7)^1 + [9^k * (-9)^1]$

$(-7)^k -9^k = -16m$

$(-7)^k = -16m +9^k$

$(-16m +9^k )* (-7) + [9^k * (-9)]$

$(-16m * -7) +(9^k * -7) + (9^k * -9)$

$(-16m * -7) +(9^k) * (-7-9)$

$(-16m * -7) +(9^k) * (-16)$

Btw, the start is "even more clear" if on starts with $n=0$ instead of $n=1$ — Hagen von Eitzen, Feb 08 '14 at 09:57
if I start with $N=0$ that would mean that $(-7)^0-9^0$ and it's $1-1 = 0$ — usukidoll, Feb 08 '14 at 09:58
Note that $-16=-1\times 16$ is divisible by $16$. It is not quite clear what your question is. — Mark Bennet, Feb 08 '14 at 09:58
You finished the entire proof. finally, it is coming: $(-16)*[(-7m)+(9^k)]$. So what is the problem? — Richard J, Feb 08 '14 at 09:58
I just wanted to see if it's correct... should've put the proof-verification tag — usukidoll, Feb 08 '14 at 09:59
@MarkBennet $-16$ is divisible by $16$? So I guess the book didn't make a mistake, but still it freaked me out a bit because when I did the basis and let $n=1$ I got $-7-9 = -16$ — usukidoll, Feb 08 '14 at 10:04

score 1 · Answer 1 · answered Feb 08 '14 at 10:16

1

At $n=0$, we have $16\mid(-7)^0-9^0$.

Then, suppose that $16\mid(-7)^n-9^n$, $$ \begin{align} (-7)^{n+1}-9^{n+1} &=(-7)(-7)^n-9\cdot9^n\\ &=-\color{#C00000}{16}(-7)^n+9(\color{#C00000}{(-7)^n-9^n}) \end{align} $$

answered Feb 08 '14 at 10:16

robjohn

345,667

But a base case of 1 is clearer. – Tom Collinge Feb 08 '14 at 15:48
1

@TomCollinge: I really don't find a base case of $1$ any clearer, and a base case of $0$ extends your result to one more integer. – robjohn Feb 08 '14 at 16:38

score 0 · Answer 2 · answered Feb 08 '14 at 10:29

If $P(k)=(-7)^k-9^k$

$P(k+1)-9P(k)=(-7)^{k+1}-9^{k+1}-9[(-7)^k-9^k]=(-7)^k(-7-9)=-16(-7)^k$

Or $P(k+1)-(-7)P(k)=(-7)^{k+1}-9^{k+1}-(-7)[(-7)^k-9^k]=-9^k(9+7)=-16\cdot9^k$

So in either way, $P(k+1)$ will be divisible by $16,$ if $P(k)$

Now establish the base case i.e., $k=1$

Using induction, prove that $(-7)^n -9^n$ is divisible by $16$

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