I'm stuck on the induction step where substitution is difficult.
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2^3^n is ambiguous as the box in the title tries to tell you. Should $2^3$ be calculated first or $3^n$? – mathreadler May 10 '17 at 21:44
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If it is as you edited, then J.OK is right. An integer power of 2 is never divisible by anything else than powers of 2 and 1. – mathreadler May 10 '17 at 21:47
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Perhaps you mean $2^{3^n}+1$ is divisible by $3^{n+1}$. – Robert Israel May 10 '17 at 21:47
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@mathreadler I guessed for the edit. I feel (2^3)^n does not make sense,as he could just write 8^n - but either way, it is not divisible by 3. – Jay Zha May 10 '17 at 21:48
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@RobertIsrael Crap yeah i accidentally for got to add the one – Gerard L. May 10 '17 at 21:49
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Now it makes a lot more sense – Jay Zha May 10 '17 at 21:49
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@YujieZha: there are many cases where one can have nested powers and then making clear which order things should be done is important to avoid not only misunderstanding but also confusing oneself when trying to make sense of the problem. – mathreadler May 10 '17 at 21:50
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@mathreadler yea, agree. That's fair enough. – Jay Zha May 10 '17 at 22:02
3 Answers
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Fill in details (algebraic little ones)
$$2^{3^{n+1}}+1=2^{3^n\cdot3}+1=\left(2^{3^n}\right)^3+1\;\;(**)$$
Now, we assume (inductive hypothesis) that
$$2^{3^n}+1=k\,3^{n+1}\implies2^{3^n}=k\,3^{n+1}-1$$
and thus we get:
$$(**)=(k3^{n+1}-1)^3+1=k^33^{3n+3}-k^23^{2n+3}+k3^{n+2}-1+1=$$
$$=k3^{n+2}\left(k^23^{2n+1}-k3^{n+1}+1\right)$$
and we're done.
DonAntonio
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Just wondering, can't it be much shorter with using $\equiv -1$ and $(-1)^3=-1$ sort of argument\s? – user76568 May 10 '17 at 22:07
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For $n=1$, both are $9$, so it is true.
Suppose it works for $n$, i.e. $2^{3^n}+1$ is divisible by $3^{n+1}$, then $2^{3^n}+1 = k\times 3^{n+1}$
For $n+1$:
$$2^{3^{n+1}}+1 = (2^{3^{n}})^3+1 = (k \times 3^{n+1} - 1)^3 + 1 =$$ $$k^3\times 3^{3n + 3} - 3\times k^2\times 3^{2n+2} + 3\times k\times 3^{n+1} - 1 + 1$$ $$ = 3^{n+2}(k^3\times 3^{2n+1}-k^2\times 3^{n} +k)$$
Jay Zha
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Answer to first question : Prove that : $2^{3^n}$ is divisble by $3^{n+1}$
This result is false. $3$ never divides a power of $2$
J. OK
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Take a look at the edit from the OP within the first five minutes of s/he posting it. – amWhy May 10 '17 at 21:50
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@amWhy the answer was faster than the edit in this case. Could also blame the answerer for sloppily written question. The fastest-gun-in-the-west culture that has evolved tends to make these things happen over and over. – mathreadler May 10 '17 at 21:53
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The answer was correct with the original post of the OP. That's why I upvoted it. And the editing wasn't that fast... – DonAntonio May 10 '17 at 21:55
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1@mathreadler my point is that IF folks wouldn't run to answer a trivial what is trivially true, and wait, as do other potential answers, for clarification from the asker, this situation could have been avoided. – amWhy May 10 '17 at 21:55
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3I think that downvoting was what a correct answer to a given question, no matter how fast, trivial or whatever the question was, is pretty close to be as overwhelmingly ridiculous as one could ever expect to see. Maybe not upvoting, but downvoting? Oh, well... – DonAntonio May 10 '17 at 21:57
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@DonAntonio: No worries; I did NOT downvote, so stop making an as* out of u and me (assume). – amWhy May 10 '17 at 22:00
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Yes I agree with you, I did not do any downvoting either. I just stated from my perspective what mechanism that I think causes this pattern of behaviour that happens over and over and often upsets at least 2 parties. – mathreadler May 10 '17 at 22:00
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@amWhy You must be suffering from a very strong paranoia: I didn't even hinted at you and, in fact, I didn't honestly think you were one of the downvoters. I guess you should stop doing that thing to yourself. – DonAntonio May 10 '17 at 22:01
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@DonAntonio and upvoting to counter folks downvotes is as immature a thing as I've ever seen! – amWhy May 10 '17 at 22:02
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@amWhy You continue guessing and making that thing of yourself you mentioned before: I was the very first upvoter of this post from the very beginning, long before the discussion about quick answer and etc. even began. – DonAntonio May 10 '17 at 22:03
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Okay, @DonAntonio I tried to post this immediately after my last comment, and yes, I did assume you were replying to my comment immediately above your comment. I'm sorry for that. No ill feelings, okay? – amWhy May 10 '17 at 22:03
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@amWhy Not at all. This is already almost like talkback-sport for both of us, I guess...:) – DonAntonio May 10 '17 at 22:04
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