Showing that $\sum_{j=0}^{2n-1}{\cos^n(\frac{j\pi}{2n})(2\cos(\frac{2j\pi}n)+1)\cos(\frac{j\pi}2-\frac{2j\pi}n)}$ is never an integer for $n>10$

Question

I want to show that

$$f(n) = \sum_{j=0}^{2n-1}{\cos^n\left( \frac{j \pi}{2n}\right) \left( 2\cos \left( \frac{2 j \pi}{n} \right) + 1\right) \cos \left( \frac{j \pi}{2} - \frac{2 j \pi}{n} \right)}$$

is an integer only for $n \in \{1,2,3,4,5, 10\}$.

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What I have tried so far is:

• Calculate $f(n)$ for each $n$ up to $16$. The values above are the only values I have found that are integers.

• At $n = 16$, the value is $\frac{1941}{2048}$, just under $1$. By computing $f(n)$ for $16$ and up, I can see that it is always decreasing and therefore is $\lt 1$.

If I can show $\forall n \in \mathbb{N}, n\ge 16, f(n) \lt 1$, then it follows that it can't be an integer.

How do I show $f(n)$ is always decreasing (once $n$ gets big enough)?

Answer 1

Claim: for $n\geq 3$, $$ f(n)= 2^{-n}\cdot 2n\left(\binom{n}{4}+\binom{n}{2}+\binom{n}{0}\right) $$This immediately gives that for $n\geq 16$, $0<f(n)<1$; then the only integer solutions are the ones shown in the problem statement. Start with $$ f(n) = \sum _{j=0}^{2 n-1} \left(2 \cos \left(\frac{2 \pi j}{n}\right)+1\right) \cos \left(\frac{\pi j}{2}-\frac{2 \pi j}{n}\right)\cos ^n\left(\frac{\pi j}{2 n}\right) $$A key observation: each term is real. Use $\cos(\theta)= 1/2 (e^{i\theta}+e^{-i \theta})$: $$ f(n)=2^{-n} \sum _{j=0}^{2 n-1} \left(2 \cos \left(\frac{2 \pi j}{n}\right)+1\right) \cos \left(\frac{\pi j}{2}-\frac{2 \pi j}{n}\right)\left(\exp \left(\frac{\pi i j}{2 n}\right)+\exp \left(\frac{-\pi i j}{2 n}\right)\right)^n $$ $$ =2^{-n} \sum _{j=0}^{2 n-1} \left(2 \cos \left(\frac{2 \pi j}{n}\right)+1\right) \cos \left(\frac{\pi j}{2}-\frac{2 \pi j}{n}\right)(-i)^j \left(1+\exp \left(\frac{\pi i j}{n}\right)\right)^n $$Now use the Binomial Theorem and switch the order of summation: $$ f(n) = 2^{-n} \sum _{j=0}^{2 n-1} \left(2 \cos \left(\frac{2 \pi j}{n}\right)+1\right) \cos \left(\frac{\pi j}{2}-\frac{2 \pi j}{n}\right) (-i)^j \sum _{k=0}^n \binom{n}{k} \exp \left(\frac{\pi i j k}{n}\right) $$ $$ =2^{-n} \sum _{k=0}^n \binom{n}{k} \sum _{j=0}^{2 n-1} \left(2 \cos \left(\frac{2 \pi j}{n}\right)+1\right) \cos \left(\frac{\pi j}{2}-\frac{2 \pi j}{n}\right) (-i)^j \exp \left(\frac{\pi i j k}{n}\right) $$Playing around with the summand gives $$ \cos \left(\frac{\pi j}{2}-\frac{2 \pi j}{n}\right) (-i)^j \exp \left(\frac{\pi i j k}{n}\right) $$ $$ = \cos \left(\frac{\pi j (n-4)}{2 n}\right) \left(\cos \left(\frac{\pi j (n-2 k)}{2 n}\right)-i \sin \left(\frac{\pi j (n-2 k)}{2 n}\right)\right) $$But now, by the fact that everything in the problem is real, we can discard the imaginary term. Thus $$ f(n) = \sum _{k=0}^n 2^{-n} \binom{n}{k} \sum _{j=0}^{2 n-1}\left(2 \cos \left(\frac{2 \pi j}{n}\right)+1\right) \cos \left(\frac{\pi j (n-4)}{2 n}\right) \cos \left(\frac{\pi j (n-2 k)}{2 n}\right) $$Use the product-to-sum formula for cosine: $$ f(n) = \sum _{k=0}^n 2^{-n} \binom{n}{k} \sum _{j=0}^{2 n-1}\frac{1}{2} \left(\cos \left(\frac{\pi j (k-4)}{n}\right)+\cos \left(\frac{\pi j (k-2)}{n}\right)+\cos \left(\frac{\pi j k}{n}\right)+\cos \left(\frac{\pi j (k-n+4)}{n}\right)+\cos \left(\frac{\pi j (-k+n-2)}{n}\right)+\cos \left(\frac{\pi j (n-k)}{n}\right)\right) $$ $$ =\sum _{k=0}^n 2^{-n-1} \binom{n}{k} \underbrace{\sum _{j=0}^{2 n-1} \cos \left(\frac{\pi j (k-4)}{n}\right)+\cos \left(\frac{\pi j (k-2)}{n}\right)+\cos \left(\frac{\pi j k}{n}\right)}_{S_1} $$ $$ +\sum _{k=0}^n 2^{-n-1} \binom{n}{k} \underbrace{\sum _{j=0}^{2 n-1}\cos \left(\frac{\pi j (k-n+4)}{n}\right)+\cos \left(\frac{\pi j (-k+n-2)}{n}\right)+\cos \left(\frac{\pi j (n-k)}{n}\right)}_{S_2} $$Now I claim $S_1=S_2$. Indeed, if you sum $S_2$ in the opposite direction, you get exactly $S_1$. Thus we can do away with it, leaving: $$ f(n)=\sum _{k=0}^n 2^{-n} \binom{n}{k}\sum _{j=0}^{2 n-1} \cos \left(\frac{\pi j (k-4)}{n}\right)+\cos \left(\frac{\pi j (k-2)}{n}\right)+\cos \left(\frac{\pi j k}{n}\right) $$Here's the rub: the inner sums cancel by symmetry, except when $k=4,2,0$, when they are $2n$ (because $\cos(0)=1$). Then we have $$ f(n) = 2^{-n}\cdot 2n\left(\binom{n}{4}+\binom{n}{2}+\binom{n}{0}\right) $$I double-checked the closed-form against the original series for $3\leq n\leq 16$ and everything worked!

Answer 2

Here's an alternate solution.

Lemma: Let $0\le j,k < n$. Then $$ \sum_{j=0}^{n-1} \cos(2\pi k j/n) = \begin{cases}n,&k=0\\0,&\text{else}\end{cases} $$ This is clear geometrically: the non-unit roots of unity sum to zero, and when $k=0$ there are $n$ terms of $1$.

Take $f(n)$ and multiply through by $2^n$:

$$2^n f(n) = \sum_{j=0}^{2n-1}{\left(2\cos\left( \frac{j \pi}{2n}\right)\right)^n \left( 2\cos \left( \frac{2 j \pi}{n} \right) + 1\right) \cos \left( \frac{j \pi}{2} - \frac{2 j \pi}{n} \right)}$$ $$= \sum_{j=0}^{2n-1}{\left(2\cos\left( \frac{j \pi}{2n}\right)\right)^n \left( 2\cos \left( \frac{4 j \pi}{2n} \right) + 1\right) \cos \left( \frac{nj \pi}{2n} - \frac{4 j \pi}{2n} \right)}$$ By symmetry, it suffices to sum over $0\le j \le n-1$ and then double: $$= 2\sum_{j=0}^{n-1}{\left(2\cos\left( \frac{j \pi}{2n}\right)\right)^n \left( 2\cos \left( \frac{4 j \pi}{2n} \right) + 1\right) \cos \left( \frac{nj \pi-4j\pi}{2n} \right)}$$Denote $X=\frac{j\pi}{2n}$: $$= 2\sum_{j=0}^{n-1}\left(2\cos( X)\right)^n \left( 2\cos (4X) + 1\right) \cos \left( (n-4)X\right)$$Now we are going to use the product to sum rules to turn these products into a sum of frequencies. This is going to be great for two reasons:

• By the lemma, we need only worry about the constant term, as all the other frequencies will die off.
• In fact, the constant term is a polynomial in $n$. Once we figure out its degree and sample that many points plus one, we can reverse-engineer it.

Claim: the constant term is a polynomial of degree four in $n$. Indeed, this is because of the factors of $4$ that appear in the arguments of the cosines.

Writing out the constants produces the sequence

$$\{1,2,4,8,16,31,57,99,\ldots\}$$

This is a well-known sequence, whose terms represent the number of regions when $n$ points in general position on a circle are connected, as seen here.

In summary,

$$ 2^n f(n) = 2n \left(\binom{n}{4}+\binom{n}{2}+\binom{n}{0}\right),$$ and the conclusion follows.