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As said in the title you have to show by induction that 1+$2^{2^n}+2^{2^{n+1}}$ is divisible by 7. So you start with n=0, that gives 1+2+4=7. So the start is shown. Let 1+$2^{2^n}+2^{2^{n+1}}$ be divisble by 7 for a n. I've tried several attempts but I ended up in a mess. For example the latest attempt:

We have $1+2^{2^{n+1}}+2^{2^{n+2}}$

Adding $2^{2^n}-2^{2^n}$ gives you

$1+2^{2^{n+1}}+2^{2^n}+2^{2^{n+2}}-2^{2^n}$

As the first three summands resemble our Assumption, only $2^{2^{n+2}}-2^{2^n}$ needs to be proved as a multiple of seven.

But I am stuck at trying to show this.

As stated that might be a incorrect approach by myself, so I'm not really sure whether or not that is going in the correct direction. In the end I'd really appreciate some help on this question!

Bill Dubuque
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babemcnuggets
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  • You're doing well so far. Can you factor $2^{2^{n+2}}-2^{2^n}?$ – saulspatz Feb 14 '18 at 02:01
  • $2^{2^{n+2}}-2^{2^{n}}$ would be $(2^{2^{n+1}}+2^{2^{n-1}})*(2^{2^{n+1}}-2^{2^{n-1}})$ or are you getting on something else? – babemcnuggets Feb 14 '18 at 03:50
  • All I was getting at was that if you want want to show that some number is divisible by something, a good approach is to always factor the number. I was thinking of pulling out a factor of $2^{2^n}$ This gets you a factor of the form $2^{3k}-1,$ which looks a lot easier to deal with. Let me say, I haven't worked this out. You said you were stuck, and I just trying to suggest something to try. It may be that one has to consider the form of $k$ in the last formula. – saulspatz Feb 14 '18 at 04:00
  • Ahh, you're right. You could show that $2^{3k}-1$ can be divided by 7 by induction aswell which is not that long! I think that is pretty much the way I always wanted to go. I also see where I made a mistake in my writtings. I pulled $2^{2^{n}}$ out as you did and kept on working with $(2^{2^{n+1}}*2^{2{n}}-1)$, which wasn't efficent at all. Thank you, that way I really learned something again. – babemcnuggets Feb 14 '18 at 04:17
  • Or now that I think of it, $2^{3k}-1 = (2^3)^k-1$ and since $(x-1) \mid (x^k-1),$ this is divisible by $2^3-1 = 7.$ – saulspatz Feb 14 '18 at 16:22

2 Answers2

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If $2^{2^n}=x$

$f(n+1)=1+x^2+x^4,f(n)=1+x+x^2$

$1+x^2+x^4=(1+x^2)^2-x^2=?$

lab bhattacharjee
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Starting with $n = 0$, we have $1+2^{2^0}+2^{2^{0+1}} = 7$, which is clearly divisible by $7$. Now suppose inductively that $n \ge 1$ and argument holds for all $n$. Then for $n+1$, we have $$1+2^{2^{n+1}}+2^{2^{n+2}} = 1+2^{2\cdot2^n}+2^{2\cdot2^{n+1}} = 1+(2^{2^n})^2+(2^{2^{n+1}})^2$$ Now and and subtract $(2^{2^n})^2$, we have $$1+2(2^{2^n})^2+(2^{2^{n+1}})^2-(2^{2^n})^2 = [1+(2^{2^n})^2]^2-(2^{2^n})^2$$ $$ = [1+(2^{2^n})^2+2^{2^n}]\cdot[1+(2^{2^n})^2-2^{2^n}]$$ $$ = [1+2^{2^{n+1}}+2^{2^n}]\cdot[1+2^{2^{n+1}}-2^{2^n}]$$

Here, notice that by inductive hyphothesis, $7|(1+2^{2^{n+1}}+2^{2^n})$. Therefore $7$ divides the whole expression and argument holds for $n+1$. Therefore by induction, argument holds for all $n$.

ArsenBerk
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