When is the number of areas obtained by cutting a circle with $n$ chords a power of $2$?

Question

Also posted on the Math Stackexchange: When is the number of areas obtained by cutting a circle with $n$ chords a power of $2$?

Introduction

Recently, a friend told me about the following interesting fact:

Place $n$ points on a circle and draw a line between every pair of points. Suppose that no three lines intersect at one point. Then the number of regions which are separated by the lines is equal to the sum of the first five numbers in the $n-1$st row of Pascal's triangle!

See this image image (from Wikipedia). Here, $n$ is the number of points, $c$ is the number of lines and $r_G$ is the number of regions:

Here is a great video by 3Blue1Brown on this subject: Circle Division Solution. The series is A000127 in the OEIS.

Preliminary results

The following is known (see again Wikipedia for instance):

For $n$ points, the number of resulting regions is $$1+\binom n2+\binom n4 = \sum_{i=0}^4 \binom{n-1}i=\text{sum of first } 5 \text{ numbers in $n$th row of Pascal's triang.}=\frac{1}{24}n(n^3-6n^2+23n-18)+1.$$

In particular, for $n\in\{1,2,3,4,5,10\}$, the number of areas is a power of $2$.

My question

Is it true that, for any other $n$, the number of areas is not a power of two?

Some attempts

First off, we can simply check that for $n\in\{6,7,8,9\}$, the number of areas is not a power of two. So the question is equivalent to: Is it true that, for any $n\geq 11$, the number of areas is not a power of $2$?

The following Proposition is easy to prove:

Proposition. For $n> 5$, we have that $f(n)< 2^{n-1}$, where $f(n)$ denotes the number of regions.

Proof. For $n>5$ we have $$f(n)=\sum_{i=0}^{n-1} \binom{n-1}i-\sum_{i=5}^{n-1}\binom{n-1}i = 2^{n-1}-\sum_{i=5}^{n-1}\binom{n-1}i<2^{n-1}.\square$$

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However, this only proves that $f(n)\neq 2^{n-1}$ for any $n>6$. There could still be some $m\in\mathbb N$ with $m<n$ such that $f(n)=2^m$.

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User Pazzaz at MSE said that he checked all cases up to $n=10^{10}$ and none of them were powers of $2$.

Answer 1

Denoting $k:=2^{\lfloor m/2\rfloor}$, we get two cases to consider: $k^2 = f(n)$ and $2k^2 = f(n)$, or making the coefficients integer: $$(12k)^2 = 144f(n)\qquad\text{and}\qquad (12k)^2 = 72f(n).$$

These equations have finite number of integer solutions by Siegel's theorem.

Numerically these equations can be solved in Magma with IntegralQuarticPoints() function.

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For the first equation, Magma gives the following integral points $(n,12k)$ (up to a sign of $k$):

[ [ 36, 2928 ], [ 5, 48 ], [ 3, -24 ], [ 1, 12 ], [ -12, 456 ], [ 10, -192 ], [ -2, -36 ], [ -237, 139344 ], [ 0, -12 ] ]

which correspond to the following solutions in $n$ and $m$: $$(n,m)\in \{ (5, 4), (3, 2), (1, 0), (10, 8) \}.$$

Similarly, for the second equation we get that the only solutions are $(n,m) \in \{ (2,1), (4,3) \}$.

Hence, there are no other solutions besides those mentioned by OP.

Answer 2

In this paper by Schinzel and Tijdeman it is proven that if a polynomial $P(x)$ with rational coefficients has at least 2 distinct zeros then the equation $$y^m=P(x),\;\;\;\; x,y\in\mathbb{Z}, \;\;\;\;\;|y|\geq2$$ implies that $m<c(P)$ where $c(P)$ is an effectively computable constant.

They deduce a corollary (3) which says that if polynomial $P(x)$ with rational coefficients has at least 3 simple zeros then the equation $y^m=P(x)$ has only finitely many solutions with $m\geq2$ and $|y|\geq2$ and these solutions can be found effectively.

Since $f(x)=\frac{1}{24} \left(n^4-6 n^3+23 n^2-18 n+24\right)$ has all roots simple and $f(n)=2^r$ implies $r\geq2$ for $n>2$ we can apply the theorem to show that $f(n)=2^m$ has only finitely many solutions and they can be found effectively.

So this at least reduces the problem to a finite search.

Answer 3

I can prove that there are only finitely many solutions as a consequence of Theorem 2 in The $p$-adic generalization of the Thue-Siegel-Roth theorem (Ridout, Mathematika 2010). Specializing this theorem slightly, it states

Let $F(x,y)$ be a homogenous irreducible polynomial with integral coefficients of degree $n \geq 3$. Let $p$ be a prime and let $G_p(x,y)$ be the greatest power of $p$ dividing $F(x,y)$. Let $\kappa > 2$. Then there are only finitely many solutions to $$\frac{|F(x,y)|}{G_p(x,y)} < \max(|x|,|y|)^{n-\kappa}$$ with $GCD(x,y)=1$.

Putting $y=1$, we deduce

Let $f(x)$ be an irreducible polynomial with integral coefficients of degree $n \geq 3$. Let $p$ be a prime and let $g(x)$ be the greatest power of $p$ dividing $f(x)$. Let $\kappa > 2$. Then there are only finitely many solutions to $$\frac{|f(x)|}{g_p(x)} < |x|^{n-\kappa}.$$

It is quite probable that the version with $y=1$ is much easier, but I didn't find a source in the literature for the easier version.

Now, let $$\binom{x}{0} + \binom{x}{1} + \binom{x}{2} + \binom{x}{3} + \binom{x}{4} = \frac{f(x)}{24}.$$ The polynomial $f(x)$ is $x^4-2 x^3+11 x^2+14 x+24$, which is easily checked to be irreducible. So there are only finitely many solutions to $$\frac{|f(x)|}{g_2(x)} < \max(|x|)^{1.9}.$$ But, if $f(x)$ is a power of $2$, then $\tfrac{|f(x)|}{g_2(x)} = 3$. So this can only happen finitely often.