Prove $2^n + 2^m ≠ 2^p$

Question

Prove : $2^n + 2^m ≠ 2^p$

For any $n,m$ ($n≠m$) and $p$ being positive integers.

I hadn't yet studied very much mathematics yet but I came across it when I had to build a python code. Any help is appreciated.

Well, try an example or two. It only takes one counterexample to disprove a claim. — lulu, Aug 10 '22 at 16:13
Write $2^n+2^m=2^n(1+2^{m-n})$ and note that the second factor is not a power of $2$. So it is never power of two for $m>n$. — Dietrich Burde, Aug 10 '22 at 16:14
@lulu Hello , I built a python code to found them but couldn't. Can you please give me such examples? — Get_ Maths, Aug 10 '22 at 16:14
@lulu Oops , I am extremely sorry . I built code to find examples which would only prove instead of disprove. — Get_ Maths, Aug 10 '22 at 16:16
Sorry, my actual question was regarding the proof only instead of disproof. — Get_ Maths, Aug 10 '22 at 16:17
the argument provided by @DietrichBurde Shows that it is never the case that you get a power of $2$ this way (trusting that $n\neq m$). — lulu, Aug 10 '22 at 16:18
The proof is trivial. We may assume that $m>n$ by symmetry. Then $1+2^{m-n}$ is odd, and we are done. — Dietrich Burde, Aug 10 '22 at 16:19
@DietrichBurde Yes and 2^m + 2^n is even . I understand it . I wished that I could get those pair(n,m). Then I could have done a great reform in science . However I failed . — Get_ Maths, Aug 10 '22 at 16:21
The argument is not that $2^n+2^m$ is even. It is that $2^n(1+2^{m-n})$ cannot be a power of two, because $1+2^{m-n}$ is odd. — Dietrich Burde, Aug 10 '22 at 16:22
@DietrichBurde While I am completely satisfied by your answer , can you please give me some other proofs too? — Get_ Maths, Aug 10 '22 at 16:26

score 1 · Answer 1 · answered Aug 10 '22 at 16:52

1

You might enjoy this view of the question, which provides an alternative proof.

Writing $2^m$ in binary, we have a $1$ followed by $m$ $0$'s. For example:

$2^1: 10 \\ 2^5: 100000$

How can we add two numbers of this form, and obtain a third one of the same form? If the $1$'s in each number are in different locations, then we get a number with two $1$'s in it:

$10 + 100000 = 100010$

The only way to get a sum with only a single $1$ in it is if the $1$'s in the two summands line up, giving a $0$ and a "carry" digit:

$10 + 10 = 100$

Does this help?

answered Aug 10 '22 at 16:52

G Tony Jacobs

31,218

Yes , it helped me. – Get_ Maths Aug 10 '22 at 17:11
Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Aug 10 '22 at 17:22
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@BillDubuque If I had known this was a dupe answer to a dupe FAQ, I would not have posted it. – G Tony Jacobs Aug 10 '22 at 17:23
One way to "know this" is to search, e.g. like this. Of course it is highly improbable for such a basic question to have never been posted here before. – Bill Dubuque Aug 10 '22 at 17:28
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@BillDubuque, what do you recommend for someone whose chief reason for being here is to help people who are asking for help, in real time when possible? Should I just channel that energy into linking them, in a friendly, welcoming, non-scolding way, to a high-quality previous instance of the question? The tension regarding site goals is getting to where I want less and less to do with this site. Can't we just do away with rep, beyond a certain level, or something? – G Tony Jacobs Aug 10 '22 at 17:52
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I can certainly empathize, since I felt similarly in the past. Although it might not be obvious at first glance, it can be very rewarding organizing the site, e.g. it sparks one to think more deeply about questions in order to understand how variations on common exercises can be handled through (slight) generalizations of prior answers. This leads to iterative improvements of prior answers (hopefully eventually to "proofs from the book") and - by increasing knowledge of answerers - to better answers to future questions. None of this occurs if we continually repost dupe answers. – Bill Dubuque Aug 10 '22 at 18:22

Prove $2^n + 2^m ≠ 2^p$

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