Prove $11^{n} - 4^{n}$ is divisible by 7 for all natural numbers n.
I know this holds for the base case of n = 1. Am I supposed to use $11^{n + 1} - 4^{n + 1}$ for the induction step?
Any help would be greatly appreciated!
Thank you in advance!
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It might help you if you look at some similar problem from the past, for example, Prove $7$ divides $13^n- 6^n$ for any positive integer or Proof by induction: Show that $9^n-2^n$ for any $n$ natural number is divisible by $7$. More generally, many of questions tagged divisibility+induction might be problems of this type. – Martin Sleziak Dec 16 '16 at 00:43
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Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Martin Sleziak Dec 16 '16 at 00:44
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Thank you Martin! I will be sure to make my titles more informative! – Garrett Dec 16 '16 at 02:51
6 Answers
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We have that $11\equiv 4 \pmod{7}.$ Thus $11^n\equiv 4^n \pmod{7}, \forall n\in\mathbb{N}.$ That is, $11^n-4^n$ is divisible by $7.$
If you want to use induction note that
$$11^{n+1}-4^{n+1}=(7+4)\cdot11^n-4\cdot4^n=7\cdot 11^n+4\cdot(11^n-4^n).$$ Now $7\cdot 11^n$ and $11^n-4^n$ are multiples of $7$ (the second one using the induction hypothesis).
mfl
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1The first proof using modular arithmetic also uses induction, e.g. in the proof of the Congruence Power Rule. The second proof is exactly the same proof, but with the modular arithmetic eliminated, as I explain at length in this answer. – Bill Dubuque Dec 16 '16 at 00:55
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Without induction: $11^n-4^n=(11-4)(11^{n-1}+11^{n-2}\cdot 4 + \cdots + 4^{n-1})$ where $11-4=7$.
dxiv
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Instead of induction, write $11^n$ as $(7+4)^n$ and expand using the binomial theorem. After you subtract away the $4^n$ term, every remaining term has at least one factor of $7$.
hmakholm left over Monica
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If you have proved the base case, now move on to the induction step.
Assume $n=k$, and that $7|(11^k - 4^k)$. This means that $11^k - 4^k = 7M$
Now prove your induction step.
So that: $7|(11^k - 4^k) \to 7|(11^{k+1} - 4^{k+1})$
K Split X
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First, show that this is true for $n=1$:
$11^{1}-4^{1}=7$
Second, assume that this is true for $n$:
$11^{n}-4^{n}=7k$
Third, prove that this is true for $n+1$:
$11^{n+1}-4^{n+1}=$
$11(11^{n})-\color\green{4}(4^{n})=$
$11(11^{n})-(\color\green{11-7})(4^{n})=$
$11(11^{n})-11(4^{n})+7(4^{n})=$
$11(\color\red{11^{n}-4^{n}})+7(4^{n})=$
$11(\color\red{7k})+7(4^{n})=$
$7(11k)+7(4^{n})=$
$7(11k+4^{n})$
Please note that the assumption is used only in the part marked red.
barak manos
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@Joffan: Thank you for the edit. For my "induction" answers, I have my own regular template which I prefer to keep :) – barak manos Dec 16 '16 at 10:07
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Another way to to do it without induction is to note that the mod cycles for powers of $11$ and $4$ modulo congruent to $7$ correspond to each other. $$\begin{equation}\begin{aligned}11 &\equiv 4 \pmod 7\\ 11^2 &\equiv 2 \pmod 7\\ 11^3 &\equiv 1 \pmod 7 \\ 11^4 &\equiv 4 \pmod 7 \\ &\vdots\end{aligned}\end{equation}$$ $$\begin{equation}\begin{aligned}4 &\equiv 4 \pmod 7\\ 4^2 &\equiv 2 \pmod 7\\ 4^3 &\equiv 1 \pmod 7 \\ 4^4 &\equiv 4 \pmod 7 \\ &\vdots\end{aligned}\end{equation}$$
So for any $n$, $11^n \equiv 4^n \equiv r \pmod 7$, so $11^n - 4^n \equiv r - r \pmod 7 \implies 11^n - 4^n \equiv 0 \pmod 7$ as desired.
q.Then
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