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Prove that $2^n+1$ is not a cube for any $n\in\mathbb{N}$.

I managed to prove this statement but I would like to know if there any other approaches different from mine.

If existed $k\in\mathbb{N}$ such that $2^n+1=k^3$ then $k=2l+1$ for some $l\in\mathbb{N}$. Then $(2l+1)^3=2^n+1 \iff 4l^3+6l^2+3l=2^{n-1}$. As I am looking for an integer solution, from the Rational Root Theorem $l$ would need to be of the form $2^j$ for $j=1,...,n-1$. But then

$$4(2^j)^3+6(2^j)^2+3\times2^j=2^{n-1} \iff 2^{2j+2}+3(2^{j+1}+1)=2^{n-1-j}$$

the LHS is odd which implies that $j=n-1$. Absurd.

Thank you in advance.

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    $\begingroup$ One other easy approach: if $2^n+1$ is a cube, then there's some $s$ with $2^n=s^3-1$. Factor the RHS and consider what the parities of the two factors can be (keeping in mind that both must be powers of $2$ themselves). $\endgroup$ Aug 20, 2020 at 0:13
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    $\begingroup$ You don’t need the rational root theorem. You have that $l$ is a factor of $2^{n-1},$ and hence must be of the form $2^j.$ Also, there is no immediate reason that $j=0$ is not allowed. Also, what is $k?$ $\endgroup$ Aug 20, 2020 at 0:54
  • $\begingroup$ @ThomasAndrewz $j=0$ is just a simple calculation. I have edited, thank you. $\endgroup$
    – user723846
    Aug 20, 2020 at 1:00
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    $\begingroup$ are you familiar with Mihăilescu's theorem? $\endgroup$ Aug 20, 2020 at 1:05

6 Answers 6

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Here is a different approach.

Modulo $7$, there aren't so many cubes, so that can be a good setting to investigate such problems:

$2^n+1\equiv 2, 3, $ or $5\pmod7$, but $m^3\equiv0, 1, $ or $6\pmod 7$.

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  • $\begingroup$ I'd never seen the relative paucity of cubes mod 7; that's a good thing to remember. Thank you! $\endgroup$ Aug 20, 2020 at 16:27
  • $\begingroup$ You're welcome, @StevenStadnicki; there's also a paucity of cubes mod $9$, though that didn't help in this problem; see my answer to this question $\endgroup$ Aug 20, 2020 at 16:39
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    $\begingroup$ When $k$ divides $\varphi(n)$, the map $x \to x^k$ mod $n$ will only hit a fraction of residues. In this case, $3$ divides $\varphi(7)=7-1$. For example, the same thing happens with $5$th powers mod $11$. $\endgroup$
    – 2'5 9'2
    Aug 20, 2020 at 19:46
  • $\begingroup$ Alex, it seems like any greater odd power. For $3$, we have $3 \mid \varphi(7) = 7-1$. For $5$, we have $5 \mid \varphi(11) = 11-1$. For $7$, we have $7 \mid \varphi(15) = 15-1$. Ad infinitum. $\endgroup$ Jan 19, 2021 at 12:46
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Here is a parity-based solution that avoids the rational root test.

If $2^n+1=m^3$, then $2^n=m^3-1=(m-1)(m^2+m+1)$, so $m-1=2^k$ for some $k\le n$, and

$$2^n+1=\left(2^k+1\right)^3=2^{3k}+3\cdot2^{2k}+3\cdot2^k+1\,.$$

Then $2^n=2^k\left(2^{2k}+3\cdot2^k+3\right)$, so $2^{n-k}=2^{2k}+3\cdot2^k+3$ is odd and greater than $1$, which is impossible.

Added: As one can see from the comments below, there are many ways to continue this argument after the first line. I took what I think of as the follow-your-nose approach, i.e., the most obvious, straightforward one, not necessarily the neatest. (And speaking of neatest, I quite like the one by rtybase.) Then again, folks’ noses don’t always point in the same direction. :-)

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    $\begingroup$ Or $m^2+m+1=2^k \Rightarrow m^2+m=2^k-1$, then LHS is even and RHS is odd, except when $k=0$. $\endgroup$
    – rtybase
    Aug 20, 2020 at 0:33
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    $\begingroup$ @rtybase: Yep, that works too. $\endgroup$ Aug 20, 2020 at 0:37
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    $\begingroup$ Or $$gcd(m-1,m^2+m+1)=gcd(m-1, (m-1)m+2m-2+3=gcd(m-1,3)=1$$ or $=3$. Hence it is 1. Since $m-1<m^2+m+1$, $m-1=1, m^2+m+1=2^k$, a contradiction. $\endgroup$
    – markvs
    Aug 20, 2020 at 1:26
  • $\begingroup$ It ca not be a perfect square too. I do not think it can be any power of an integer. $\endgroup$
    – sirous
    Aug 20, 2020 at 4:55
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    $\begingroup$ @sirous: $2^3+1=9=3^2$ $\endgroup$ Aug 20, 2020 at 5:24
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Invoking an argument more powerful than needed for this:

there cannot be any solutions to $2^n+1=m^3$ (i.e., $m^3-2^n=1$) by Mihăilescu's theorem,

which states that $2^3$ and $3^2$ are the only two powers of natural numbers

whose values are consecutive.

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Suppose $2^n + 1 = k^3$. Then $2^n = k^3 - 1 = (k^2 + k + 1)(k - 1)$. So both factors are even ($k = 2$ doesn't work; the first factor is at least $3^2 + 3 + 1 = 13$, it can't be 1). But the first factor is always odd, contradiction.

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Let $$2^n=m^3-1\\\implies 2^n=(m-1)(m^2+m+1)\\\implies(m-1)=2^a\text{ and }(m^2+m+1)=2^b\\\implies3m=(m^2+m+1)-(m-1)^2=2^b-2^{2a}$$ Now, since $m$ is odd, we must have $a=0$ or $b=0$. But $(m-1)<(m^2+m+1)$ implies $a=0$. This implies $m=2$ a contradiction since $m$ must be odd.

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Let's set the cubes to $8m^3$ and $8m^3+12m^2+6m+1$. As $8m^3$ is even and it doesn't works for $n=0$, that's impossible. For the second one, ignoring the $1$ you can factor it to $2m(4m^2+6m+3)$. Since there ain't any natural in which $4m^2+6m+3=1$ it is impossible to be a $2^n$ for natural $n$.

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