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Prove that $ \phi(n) =11 \cdot 3^n + 3 \cdot 7^n - 6 $ is divisible by 8 for all $n \in N$.

Base: $ n = 0 $

$ 8 | 11 + 3 - 6 $ is obvious.

Now let $\phi(n)$ be true we now prove that is also true for $ \phi(n+1)$.

So we get $ 11 \cdot 3^{n+1} + 3 \cdot 7^{n+1} - 6$ and I am stuck here, just can't find the way to rewrite this expression so that I can use inductive hypothesis or to get that one part of this sum is divisible by 8 and just prove by one more induction that the other part is divisible by 8.

For instance, in the last problem I had to prove that some expression a + b + c is divisible by 9. In inductive step b was divisible by 9 only thing I had to do is show that a + c is divisible by 9 and I did that with another induction, and I don't see if I can do the same thin here.

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Suppose $11*3^n + 3*7^n - 6 = 8k$

The $11*3^{n+1} + 3*7^{n+1} - 6 = 11*3^n*3 + 3*7^n*7 - 6$

$=3(11*3^n + 3*7^n-2) + 4*3*7^n $

$= 3(11*3^n + 3*7^n - 6) + 4*3*7^n + 12$

$= 3(8k) + 4(3*7^n + 3)$; $3*7^n$ is odd and $3$ is odd so $(3*7^n + 3)$ is even.

$= 3(8k) + 8(\frac{3*7^n + 3}2) = 8(3k + \frac{3*7^n + 3}2)$.

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Actually I like and am inspired by Bill Dubuques answer.

We want to prove $\phi(n) = 11*3^n + 3*7^n - 6 \equiv 0 \mod 8$

And we know $\phi(n) = 11*3^n + 3*7^n - 6 \equiv 3*3^n + 3*(-1)^n -6 = 3^{n+1} + 3*(-1)^n - 6 \mod 8$.

So it's a matter of showing $f(n) = 3^{n+1} + 3(-1)^n \equiv 6 \mod 8$.

And if we notice $f(n+2) = 3^{n+3} + 3(-1)^{n+2} = 3^{n+1}*9 + 3(-1)^{n} \equiv 3^n + 3(-1)^{n}= f(n) \mod 8$.

So it's now just a matter of showing for $f(0) \equiv f(1) \equiv 6 \mod 8$.

Which is easily verified $3^1 + 3*(-1)^0 =3+3= 6$ and $3^2 + 3*(-1)^1 = 9 -3 = 6$

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  • $\begingroup$ Yap, that's it. Thank you, just what I was looking for. $\endgroup$ – stigma6 Jan 27 '17 at 7:15
  • $\begingroup$ You want a secret confession? When I began answering I didn't know how it would turn out. But I know I had to factor $f(n+1) = manipulate(f(n)) = manipulate(8k)$ and I had faith that $manipulate(8k) = 8j$ and I just chewed on it to see what would happen. That's how a lot of figuring induction proofs. You know $f(n) = Property(t)$ and $f(n+1) = manipulate(f(n))=manipulate(Property(t))$ and know you want $manipulate(Property(t))=Property(s)$. Then you just gum and chew away until you get it. $\endgroup$ – fleablood Jan 27 '17 at 7:36
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${\rm mod}\ 8\!:\ f(n\!+\!2)\equiv f(n)\,$ by $\,a\equiv 3,7\,\Rightarrow\,a^{\large 2}\equiv 1\,\Rightarrow\,a^{\large n+2}\equiv a^{\large n}.\,$ Thus $\,8\mid f(n)\iff 8\mid f(n\!+\!2),\,$ hence by (strong/parity) induction, it is true for all $n$ $\iff$ it is true for the base cases $\,n=0,1.$

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    $\begingroup$ @Downvoter The downvote is puzzling. If something is not clear then please feel welcome to ask questions and I will be happy to elaborate. This is one of the easiest ways to prove it, $\endgroup$ – Bill Dubuque Jan 26 '17 at 22:40
  • $\begingroup$ If congruences are unknown then we can eliminate them, viz. show that $8$ divides $\,f(n\!+\!2)-f(n)\,$ because $8$ divides $\,3^{\large n+2}-3^{\large n} = (3^{\large 2}-1)3^{\large n} = 8\cdot 3^n,\,$ and similarly $8$ divides $\,7^{\large n+2}-7^{\large n}.\ \ $ $\endgroup$ – Bill Dubuque Jan 26 '17 at 23:04
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Setup the same as your current work:

$\dots$

$\dots = 11\cdot 3^{n+1}+3\cdot 7^{n+1}-6 = 11\cdot 3\cdot 3^{n}+ 3\cdot 7\cdot 7^n - 6$

$=33\cdot 3^n + 21\cdot 7^n - 6 = (11+22)\cdot 3^n + (3 + 18)\cdot 7^n - 6$

$=\underbrace{11\cdot 3^n + 3\cdot 7^n - 6}_{\text{should be familiar}} + \underbrace{22\cdot 3^n + 18\cdot 7^n}_{\text{unknown}}$

Now, what can we say about $22\cdot 3^n+18\cdot 7^n$? Anything? You say in a previous example, you had to run a second induction proof to finish, might that be useful here?

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    $\begingroup$ fyi: your answer was downvoted 1 second before mine, so it doesn't appear that the downvoter took the time to carefully consider the answers. $\endgroup$ – Bill Dubuque Jan 26 '17 at 22:50

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