Proof by induction: Prove that $6$ divides $9^n - 3^n$

Question

Induction: prove that $6| 9^n - 3^n$, where $n$ is a positive integer

inductive step: trying to prove $6| 9^{k+1} - 3^{k+1}$,

$= 9^k \cdot 9 - 3^k \cdot 3$

$= 6(\frac3 2 \cdot 9^k - \frac1 2 \cdot 3^k)$

maybe it's not going in a good direction...

Answer 1

Hint: Observe \begin{align} 9^{k+1}-3^{k+1}&=9^{k+1}-9^k\cdot3+9^k\cdot 3-3^{k+1}\\ &=(9-3)9^k+3(9^k-3^k)\\ &=6\cdot 9^k+3(9^k-3^k) \end{align}

Answer 2

Assume $9^k-3^k=6c$

$9^k\cdot 9 - 3^k \cdot 3=(6c+3^k)9-3^k.3=6k+3^k(9-3)=6k+3^k\cdot6=6\lambda$

OR

Screw induction. It's divisible by $3$ and $\text{odd-odd=even}$

Answer 3

Note that:$$9^{(n+1)}-3^{(n+1)}=3(9^n-3^n)+6\cdot9^n$$ both divisible by 6.

Answer 4

True for $n=1$.

If true for $n$,

$\begin{array}\\ (9^{n+1}-3^{n+1})-(9^n-3^n) &=(9^{n+1}-9^n)-(3^{n+1}-3^n)\\ &=9^n(9-1)-3^n(3-1)\\ &=8\cdot 9^n-2\cdot 3^n\\ &=(6+2)\cdot 9^n-2\cdot 3^n\\ &=6\cdot 9^n+2\cdot 9^n-2\cdot 3^n\\ &=6\cdot 9^n+2(9^n- 3^n)\\ \end{array} $

and both terms are divisible by $6$.

Q.E.D.

Answer 5

Simplest non-inductive proof:

$a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b + ... + a n^{n-2} + b^{n-1})$.

Therefore $a-b$ divides $a^n-b^n$.

Let $a=9$ and $b=3$.

Answer 6

Hint $\ $ Suppose that $\ \color{#c00}{9^n = 3^n\! + 6k}.\ $ Then

$\qquad\ \ \begin{eqnarray} 9^{n+1}&=\,&\quad 9\cdot \color{#c00}{9^n}\\ &=& (3\!+\!6)(\color{#c00}{3^n\!+6k})\\ &=&\quad 3^{n+1}\! + 6(\cdots)\quad\text{is the induction step.}\end{eqnarray} $

Remark $\ $ Essentially it is congruence multiplication, i.e.

$\qquad {\rm mod}\ 6\!:\,\ 9\equiv 3,\ 9^n\equiv 3^n \,\Rightarrow\, 9^{n+1}\equiv 3^{n+1}$

a special case of using the $\ $ Congruence Product Rule $\ \ A\equiv a,\ B\equiv b\,\Rightarrow\, AB\equiv ab\ $ in order to inductively prove the sought Congruence Power Rule. $\ A\equiv a\,\Rightarrow\, A^n\equiv a^n,\, $

Answer 7

Here's a hint: since $3\mid 9^n-3^n$, all you really need to show is that $9^n-3^n$ is even. And you don't need induction to do it, just factoring and common sense: $9^n-3^n = (3^n)^2 - 3^n = 3^n(3^n-1)$. Why is this even?

Answer 8

Another proof follows from the following (which can be proven using induction): $$x^n - y^n = (x-y)\sum_{k=0}^{n-1} x^k y^{n-1-k}$$ So $(x-y)\ | \ (x^n-y^n)$ for each $n \geq 1$. (Note this immediately implies that $9-3 = 6$ divides $9^n - 3^n$ for each $n\geq 1$, by setting $x = 9$, $y = 3$, and is in fact more general than the problem you were assigned.)

Proof:

$n=1$: Clear

$n\geq 2$:

\begin{align*} x^{n+1} - y^{n+1} &= x x^n - y y^n\\ &= x x^n - x y^n + x y^n - y y^n\\ &= x(x^n - y^n) + y^n (x - y)\\ &= (x-y)\left(x\left(\sum_{k=0}^{n-1} x^k y^{n-1-k}\right) + y^n\right)\\ &= (x-y)\left(y^n + \sum_{k=0}^{n-1} x^{k+1} y^{n-1-k}\right)\\ &= (x-y)\sum_{k=0}^n x^k y^{n-k} \end{align*}

(The last equality is the only one which requires any extra work, but this is clear once you expand it out.)

Answer 9

Inductive step:

suppose 9^k-3^k = 6a we can write 9^k =3^k+6a

Now 9^(k+1)-3^(k+1)

= 9^k *9 -3*3^k

= 9(3^k+6a)-3(3^k)

=(9-3)*3^k +9*6a

= 6(3^k+9a)

proved