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as per the title, i need to prove by induction that the expression "$7^{2n} - 2^n$" is always divisible by $47$.

base case is fine: $7^{2*1} - 2^1 = 47$.

Induction step gets lost quickly, any ideas?

step 1: $7^{2(n+1)} - 2^{(n+1)}$

step 2: $7^2 * 7^{2n} - 2^1 * 2^{n}$

step 3: ???

step 4: profit.

Jackie
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    This meme never gets old. –  Nov 09 '17 at 19:25
  • Induction? Why? Just note that$$7^{2n}-2^n=49^n-2^n=(49-2)\times(49^{n-1}+49^{n-2}\times2+\cdots+49\times2^{n-2}+2^{n-1}).$$ – José Carlos Santos Nov 09 '17 at 19:29
  • @JoséCarlosSantos could possibly be an exercise in the technique of proofs by induction – Matthew Leingang Nov 09 '17 at 19:29
  • @MatthewLeingang Probably. But I think that it is an abomination to provide, when introducing students to proofs by induction, exercices which admit a more natural proof without induction rather than with. – José Carlos Santos Nov 09 '17 at 19:31
  • Isn't $$ (49-2)\mid (49^n-2^n)$$ pretty trivial? – Jack D'Aurizio Nov 09 '17 at 19:34
  • @JoséCarlosSantos but don't blame the messenger :-) – Matthew Leingang Nov 09 '17 at 19:34
  • @MatthewLeingang Hey! I didn't downvote the question. ;-) – José Carlos Santos Nov 09 '17 at 19:35
  • @JackD'Aurizio Sure it is. – José Carlos Santos Nov 09 '17 at 19:36
  • @JoséCarlosSantos Oh, I wasn't accusing you of downvoting. I hadn't even noticed the downvote. – Matthew Leingang Nov 09 '17 at 19:38
  • @MatthewLeingang Actually, I only noticed it after writing what I wrote. – José Carlos Santos Nov 09 '17 at 19:39
  • I see an elementary number theory tag- are you familiar with modular congruence? Granted, this also sort of subverts the induction part of the proof. – Kevin Long Nov 09 '17 at 19:40
  • "Isn't $(49−2)∣(49^n−2^n)$ pretty trivial?" It may or may not be. But recognizing that is the pivotal key takes insight and experience and, probably, practice. – fleablood Nov 09 '17 at 19:44
  • @KevinLong im sorry to have to admit this, but i saw the tag on a couple of questions which were similar to mine, and added it with the intention that it would help circulate my question in the relevant fields. So in short, no i am not familiar with modular congruence. – Jackie Nov 09 '17 at 19:49
  • modular congruence is great! $a \equiv b \mod n$ means i) $a$ and $b$ have the same remainder when divided by $n$ or ii)$n|(a-b)$ or ii)$a = kn+b;n\in\mathbb Z$. so $n|a\iff a\equiv 0\mod n$. It's easy to show modular equations "carry" through addition, multiplication, and powers so if $a\equiv b \mod n$ the $a^k\equiv b^k \mod n$. So $7^{2n}-2^n=49^n-2^n\equiv 2^n-2^n\mod 47 \equiv 0 \mod 7$. Can't get much easier than that.... once you learn modular congruence. – fleablood Nov 09 '17 at 20:21
  • ... or in other words $7^{2n} - 2^n=49^n-2^n=(47+2)^n-2^n$. Now $(a+b)^n = a^n + na^{n-1}b+{n\choose 2}a^{n-1}b^2 + ...... + {n\choose 2}a^2{b-2} + nab^{n-1} + b^n$ and $a$ divides all* those terms but the last so $(47+2)^n = 47^n + n47^{n-1}2 + ...... + 472^{n-1}+ 2^n= 47K+2^n$ where $K = ... $whatever. So $7^{2n}-2^n=49^n-2^n=(47-2)^n - 2^n=47K + 2^n-2^n = 47K$. – fleablood Nov 09 '17 at 20:26
  • thanks @fleablood, you really helped my understanding beyond the scope of my original question. It pains me that I cannot accept two answers as solutions here. Thanks again! – Jackie Nov 09 '17 at 20:45

6 Answers6

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Step 1 is not “$7^{2(n+1)} - 2^{n+1}$.” You're thinking in terms of expressions and you want to think in terms of sentences. This is one half of the conclusion to an if-then statement you want to prove:

For any $n$, if $7^{2n} - 2^n$ is divisible by $47$, then $7^{2(n+1)} - 2^{n+1}$ is also divisible by $47$.

Step 0 is the base case, which you have shown.

Step 1 is “Suppose there exists a positive integer $n$ such that $7^{2n} = 2^n$.”

Step 2 is: you need to show that $7^{2(n+1)} - 2^{n+1}$ is divisible by $47$.

You've got a few answers on how to show that. Essentially it's the old add-and-subtract-the-same-thing trick. \begin{align*} 7^{2n+2} - 2^{n+1} &= 7^2\cdot 7^{2n} - 2\cdot 2^{n} \\&= 7^2 \cdot 7^{2n} - 7^2 \cdot 2^n + 7^2 \cdot 2^n - 2\cdot 2^n \\&= 49 (7^{2n} - 2^{n}) + 47 \cdot 2^n \end{align*} If $7^{2n} - 2^{n} = 47k$ for some integer $k$, then $$ 49 (7^{2n} - 2^{n}) + 47 \cdot 2^n = 49\cdot 47k + 47 \cdot 2^n = 47(49k + 2^n) $$ in other words, also a multiple of $47$.

Step 3 is “Therefore, by induction, $7^{2n} - 2^n$ is divisible by $47$ for all positive integers $n$.”

If you follow these steps, and apply them to all proofs by induction, you will indeed profit.

egreg
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Matthew Leingang
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$$7^{2n}-2^n=49^n-2^n=(49-2)(49^{n-1}+...+2^{n-1})$$ is divisible by $47$.

Michael Rozenberg
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it is $$(7^2)^n-2^n=49^n-2^n\equiv 2^n-2^n\equiv 0 \mod 47$$

Dr. Sonnhard Graubner
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Hint: For the induction step:

$7^{2n+2}-2^{n+1}=49\cdot 7^{2n}-2\cdot 2^{n}=47\cdot 7^{2n}+2(7^{2n}-2^n)$

paw88789
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Matthew Leingang’s answer is very good, but it can be simplified: there is no need to guess what to add and subtract.

The induction hypothesis that $7^{2n}-2^n$ is divisible by $47$ can be written as $$ 7^{2n}-2^n=47k $$ for some $k$. Therefore $7^{2n}=47k+2^n$ and so \begin{align} 7^{2(n+1)}-2^{n+1} &=7^2\cdot 7^{2n}-2^{n+1}\\[4px] &=49\cdot(47k+2^n)-2^{n+1}\\[4px] &=47\cdot49k+49\cdot2^n-2^{n+1}\\[4px] &=47\cdot 49k+(49-2)\cdot2^n\\[4px] &=47(49k+2^n) \end{align} is divisible by $47$.

egreg
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Step 1: $7^{2(n+1)} - 2^{n+1}$

Step 2: $7^2*7^{2n} - 2*2^n$.

Step 3: $(47 + 2)7^{2n} - 2*2^n=47*7^{2n} + 2(7^{2n} - 2^n)$

Step 4: Profit.

(BTW... you miswrote this as $7^{2n} - 2^{2n}$ which is not true.)

fleablood
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