Verify by induction that $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N} \quad \forall n \ge 1 \in \mathbb{N}$

Basis: $P(1) \Rightarrow \frac{33}{11} \in \mathbb{N}$.

Induction: if the statement holds for some $n$ $\Rightarrow$ holds for $n+1$

I would appreciate some small hints because I'm stuck at $P(n+1) = \frac{6^{2(n+1)} - 3^{n+1}}{11}$

  • $\begingroup$ If you convert $P(n)$ to the equivalent statement, $36^n-3^n\equiv0 \pmod{11}$, the problem becomes a lot easier. I'm not recommending to do it that way, because I suspect it would defeat the purpose of the exercise you were given, but it's a motivation to study "modulo $m$" arithmetic at some other time. $\endgroup$ – David K Jan 10 '17 at 14:40
  • $\begingroup$ Modular arithmetic is fun but sadly I had to solve it by induction. $\endgroup$ – Lovecraft Jan 10 '17 at 14:46
  • $\begingroup$ I gathered that from the way the question was worded. At least you can easily see the thing you are supposed to prove is true (unlike some problems where someone copied something wrong and ended up asking you to prove a false statement). $\endgroup$ – David K Jan 10 '17 at 14:49

Because $\displaystyle \frac{6^{2n}-3^n}{11}$ is a natural number, we can write for some $k \in \mathbb N, 6^{2n} =3^n +11k $. Then our proof becomes $$\frac {6^{2n+2}-3^{n+1}}{11} = \frac {36 (3^n +11k )-3 (3^{n})}{11} =\frac {33 (3^n) +11 (36k)}{11} \in \mathbb N $$ And thus the proof is completed. Hope it helps.

  • $\begingroup$ where did the $6^{2n}$ go?? $\endgroup$ – user394255 Jan 10 '17 at 14:16
  • $\begingroup$ @A.Molendjik We know that $\frac {6^{2n}-3^{n}}{11}$ is an integer. So keep this expression as k and we get $6^{2n} =3^{n} +11k$. Now substitute for this in the proof. $\endgroup$ – Rohan Jan 10 '17 at 14:18
  • $\begingroup$ @David K Thank you for pointing out that typo. $\endgroup$ – Rohan Jan 10 '17 at 14:34

You don't really need induction, or just use some simple production, as $$6^{2n}-3^n=3^n(12^n-1)$$

  • 1
    $\begingroup$ And if one must use induction, it's easier to do it when the formula is written this way. In fact, you really only have to prove that $(12^n-1)/11$ is an integer. $\endgroup$ – David K Jan 10 '17 at 14:35
  • $\begingroup$ @DavidK Thanks! $\endgroup$ – S. Y Jan 10 '17 at 15:18

We have $P(n)=\frac{6^{2n}-3^n}{11} \in \mathbb{N}$.

You have proven that the base case is true.

Now, assume true for $n=k$.

$$6^{2k}-3^k=11p \tag{1}$$

Where $p \in \mathbb{N}$.

True for $n=k+1$.

$6^{2(k+1)}-3^{k+1}=6^{2k+2}-3^{k+1}=36 \cdot 6^{2k}-3 \cdot 3^k \tag{2}$

Now, here comes the trick.

We can substitute from $6^{2k}$ from equation $(1)$ into equation $(2)$.

$$36 \cdot (11p+3^k)-3 \cdot 3^k=11\cdot 36p+36\cdot 3^k-3 \cdot 3^k=11 \cdot 36p-33 \cdot 3^k \tag{3}$$

Since $11\cdot 36p$ is obviously divisible by $11$, and $33 \cdot 3^k$ can be divided by $11$ to give $3 \cdot 3^k$, the whole of equation $(3)$ is divisible by $11$.

Hence, we are done.


So the statement $P(n)$ could also be formulated that $11$ divides $6^{2n} -3^n$. So $P(n+1)$ would consider \begin{align} 6^{2(n+1)} - 3^{n+1} & = \\ 6^{2n} \times 36 - 3 \times 3^n & = \\ 3(12 \times 6^{2n} - 3^n )&= \\ 3(11\times 6^{2n} + 6^{2n} - 3^{n}) & = \ldots \end{align} Can you use $P(n)$ in order to prove $P(n+1)$?



From $$x^{n+1}-y^{n+1}=(x^n+y^n)(x-y)+xy(x^{n-1}+y^{n-1})$$

We have

$$6^{2n+2}-3^{n+1}=(6^{2n}+3^{n})(6^2-3)+108 (6^{2n-2}+3^{n-1})$$

We use induction: If $n-1$ is true, $n+1$ is true.


${\rm mod}\ 11\!:\ 6^{\large 2}\!\equiv 3\,\Rightarrow\, (6^{\large 2})^{\large N}\!\equiv 3^{\large N} $ by the Congruence Power Rule (it has a trivial inductive proof).

Remark $\ $ Some of the other answers essentially do the same, except they eliminate the congruence language, and directly inline the proof of the Power Rule (or its inductive basis - the Congruence Product Rule), i.e. they are presented in divisbility "assembly language" instead of higher-level arithmetical language. This viewpoint is discussed at length in this answer.

  • $\begingroup$ Thank you for your interesting remark. $\endgroup$ – Lovecraft Jan 10 '17 at 19:29

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