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I'm stuck on the following proof by induction: $$8\mid3^n +7^n -2$$

And this is how far I've gotten:

$$\begin{aligned}3&\cdot3^n+7\cdot7^n-2\\3&(3^n+7^n-2)+7^n(7-3)-2\end{aligned}$$

Any help on where to go after this would be great!

PinkyWay
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Dimitri
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5 Answers5

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Note that by hypothesis $$8|3^n+7^n-2$$ then $8|3^{n+1}+7^{n+1}-2$ if and only if also $$8|(3^{n+1}+7^{n+1}-2)-(3^{n}+7^{n}-2)=3^n(3-1)+7^n(7-1)=\underbrace{2\cdot 3^n+2\cdot 7^n-4}_{\mathrm{divisible\ by\ 8}}+\underbrace{4+4\cdot 7^n}_{\mathrm{divisible\ by\ 8}}.$$

Valent
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You did a mistake:

$$3\cdot3^n+7\cdot7^n-2=3(3^n+7^n-2)+7^n(7-3)+4$$

Bow by $P(n)$ you know that $8|3(3^n+7^n-2)$. Thus, to complete the problem, you have to prove that

$$8|7^n\cdot 4+4$$

you can do this either by induction or by simply observing that $7^n+1$ is even, thus $4(7^n+1)$ is a multiple of $8$.

Added $P(n+1)$ contains $3\cdot3^n+7\cdot7^n-2$. Our goal is to create $3(3^n+7^n-2)$.

So, we start by adding and subtracting the missing terms:

$$3\cdot3^n+7\cdot7^n-2 =3 \cdot3^n +\left( 3 \cdot 7^n +3 \cdot (-2)- 3 \cdot 7^n -3 \cdot (-2) \right)+7\cdot7^n-2\\ =3 \left(\cdot3^n + 7^n -2\right)- 3 \cdot 7^n +6 +7\cdot7^n-2\\ =3 \left(\cdot3^n + 7^n -2\right)+7\cdot7^n- 3 \cdot 7^n +6 -2$$

An alternate way to figure this is the following:

You have

$$A=3\cdot3^n+7\cdot7^n-2$$ and need $$B=3 \cdot \left(3^n + 7^n -2\right)$$

Then you can write

$$A=B+(A-B) \,.$$

Thus, what you need to add to $3 \cdot \left(3^n + 7^n -2\right)$ is $$3\cdot3^n+7\cdot7^n-2-3 \cdot \left(3^n + 7^n -2\right)=3\cdot3^n+7\cdot7^n-2-3\cdot3^n-3\cdot7^n-6\\ =7^n(7-3)+4$$

N. S.
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  • How did you get $+4$ at the end? – Dimitri Dec 06 '13 at 22:37
  • @Dimitri $3(3^n+7^n-2)=3^{n+1}+3\cdot 7^n -6$. – N. S. Dec 06 '13 at 23:00
  • Okay, now why is it that we factor out a 3 on the first side and the other side we factor out $7^n $. I don't understand the mechanics behind that. – Dimitri Dec 06 '13 at 23:14
  • @Dimitri It is your computation, how did you go from $$3\cdot3^n+7\cdot7^n-2$$ to $$3(3^n+7^n-2)+7^n(7-3)-2 ,?$$ – N. S. Dec 06 '13 at 23:15
  • The computations have been mimicked by other similar problems, and I keep doing new ones to see why they're done in this fashion, hence is why I'm getting stuck most of the time. – Dimitri Dec 06 '13 at 23:17
  • @Dimitri Will add the computation to my answer then. – N. S. Dec 06 '13 at 23:18
  • This is good thank you. Would this go aswell?

    $3^{n+1}+7^{n+a}-2=8a $ for some integer $a$ $$\equiv 3 \cdot3^n+7\cdot7^n-2=8a\ \equiv3(3^n+7^n-2)+7 \cdot7^n-2 \ \equiv 3(8a-7^n+2)+7\cdot7^n-2 \text { substituted from our inductive hypothesis } \ \equiv 24a -3\cdot 7^n+6+7\cdot7^n-2=24a+4\cdot7^n+4\ \text{ factor out a 4 =} , 4(8a+7^n+1) $$

     – Dimitri Dec 07 '13 at 00:55
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By direct induction following the initial step:

$$3^{n+1}+7^{n+1}-2=7(3^n+7^n-2)-\left(4\cdot3^n-12\right)$$

But the first summand in the right expression above is divisible by eight by ind. hypothesis, and the second one is

$$12(3^{n-1}-1)=12(3-1)(3^{n-2}+3^{n-3}+\ldots+3+1)$$

and this is also a multiple of eight (and observe the expression is correct for $\;n\ge 2$)

DonAntonio
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Let $\displaystyle f(n)=3^n+7^n-2$

$\displaystyle\implies f(m+1)-f(m)=3^{m+1}+7^{m+1}-2\left(3^m+7^m-2\right)$ $\displaystyle=2\cdot3^m+6\cdot7^m=8\cdot3^m+6(7^m-3^m)$

Now we know $\displaystyle7^m-3^m$ is divisible by $7-3=4$ (Proof)

$\displaystyle\implies f(m+1)-f(m)$ is divisible by $8$

$\displaystyle\implies 8$ will divide $f(m+1)$ if $8$ divides $f(m)$

Now show that the base case i.e. for $n=1$ holds true

Community
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lab bhattacharjee
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Hints:

$$3^{2n}=9^n\equiv 1 \; (\mathrm{mod}\; 8)$$ $$7^{2n}=49^n\equiv 1 \; (\mathrm{mod}\; 8)$$

and

$$3^{2n+1}=3\cdot 9^n\equiv 3 \; (\mathrm{mod}\; 8)$$ $$7^{2n+1}=7\cdot 49^n\equiv 7 \; (\mathrm{mod}\; 8)$$

Jean-Claude Arbaut
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