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Tell HN: 2^4 == 4^2, extended to rationals

news.ycombinator.com

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Re: Tell HN: 2^4 == 4^2, extended to rationals

#11

Post by AnimalMuppet »

gus_massa wrote: post #2

Integer version: https://news.ycombinator.com/item?id=35635491 (520 points | 2 days ago | 156 comments) It (implicitly) shows why the e number is important and why it appears as the limit of your example. Did you prove that this are all the case? Or this are only the cases you found and the general case is still open?

No, I did not prove that this was all the cases - not even that this was all the rational cases. I tried, years ago (I literally made this post by trying to recall what I had found over 40 years ago), but I did not find a proof that this was all the possibilities.

Re: Tell HN: 2^4 == 4^2, extended to rationals

#12

Post by yarg »

mahalex wrote: post #10
nmmnthrowaway wrote: post #8

Earlier quoted context omitted.

That is correct. Why are there no others?

Because Catalan’s conjecture.

> Catalan's conjecture was proven by Preda Mihăilescu in April 2002.

Cool, so the conjecture's a theorem.

https://xn--uni-gttingen-8ib.academia.edu/PredaMihailescu

Re: Tell HN: 2^4 == 4^2, extended to rationals

#19

Post by AnimalMuppet »

AnimalMuppet wrote: post #11
gus_massa wrote: post #2

Integer version: https://news.ycombinator.com/item?id=35635491 (520 points | 2 days ago | 156 comments) It (implicitly) shows why the e number is important and why it appears as the limit of your example. Did you prove that this are all the case? Or this are only the cases you found and the general case is still open?

No, I did not prove that this was all the cases - not even that this was all the rational cases. I tried, years ago (I literally made this post by trying to recall what I had found over 40 years ago), but I did not find a proof that this was all the possibilities.

After some time to think, yes, I'm pretty sure that it's all solutions. The shape of the function in the article means that there is only one b for any a. The formula I gave gives a solution, so it must be the solution. If you can find an n to generate an a, the formula must give you the correct b.

(As n goes from 0 to infinity, a goes from 1 to e, so the formula covers the entire range of possibilities.)

In my formula, however, n may be irrational, even for some rational a. I haven't proven that.

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