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I was wondering if there are any $n$ that for $ \frac {2^{n+1}-1}{n+1}$ this number is natural, so:

$$k=\frac {2^{n+1}-1}{n+1} \qquad k,n \in \mathbb{N} $$

let $n+1=t$

$$ k\times t=2^{t}-1 $$

Now we have that $k\;$and $\;t$ are odd, so $n\;$ is even. I tried to rewrite it into form $k,t=2a+1$, but it didn't work. So i'm curious is there even a solution to this problem because i made it by myself

1qwertyyyy
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  • Actually $k=0$, $n=0$ is a solution... I guess you are looking for less trivial ones... – Alberto Saracco Dec 19 '19 at 21:52
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    @AlbertoSaracco In some cases, it's assumed that being a natural number, i.e., being in $\mathbb{N}$, means being positive, while in other cases it just means non-negative. I believe the former is what is meant here. – John Omielan Dec 19 '19 at 21:54
  • I go for Peano's definition. Axiom 1: 0 is a natural number. https://en.wikipedia.org/wiki/Peano_axioms#Formulation – Alberto Saracco Dec 19 '19 at 21:55
  • @AlbertoSaracco oh yes i do – 1qwertyyyy Dec 19 '19 at 21:55
  • But in my case there's divison thaht changes a little bit – 1qwertyyyy Dec 19 '19 at 21:59
  • @lulu There are likely many dupes - we weren't very good at handling duplication in the early days. But that one is already listed in the "Linked" question on the dupe I posted. – Bill Dubuque Dec 19 '19 at 22:02
  • @BillDubuque Sure...I included the second one because it had a much upvoted solution, which I thought might be useful to the OP. (Full disclosure: I didn't study the solution, so I'm not sure it differs materially from the other available solutions). – lulu Dec 19 '19 at 22:05
  • @lulu It's exactly the same method as is used in the answers in the older dupe I linked. All these old answers are highly voted - as would be expected given their very old age. – Bill Dubuque Dec 19 '19 at 22:09
  • @BillDubuque Yes, you are right. There's nothing distinct about that solution..I'll delete my reference to it. – lulu Dec 19 '19 at 22:11
  • another equivalent is if $n\cdot 2^{n+1}+2^{n+1}+n-1$ is squarefree. –  Dec 20 '19 at 03:02

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