Values of $\Phi_n(-1)$

Question

Let

$$\Phi_n(x) = \prod_{0<k\leq n, \gcd(k,n)=1}(x-e^{\frac{2\pi i k}{n}})$$

be the $n$-th cyclotomic polynomial. By observation it seems that cyclotomic polynomials when evaluated at $x=-1$ give the following

$$\Phi_n(-1) = \begin{cases} 2 \quad \text{if} \quad n = 2^k, k>1\in \mathbb{N}\\ p \quad \text{if} \quad n = 2p\\ 1 \quad \text{otherwise} \end{cases}$$

where $p$ denotes a prime $(> 2)$. The first few cases are $$\Phi_1(-1) = -2$$ $$\Phi_2(-1) = 0$$ $$\Phi_3(-1) = 1$$ $$\Phi_4(-1) = 2$$ $$\Phi_5(-1) = 1$$ $$\Phi_6(-1) = 3$$ $$\Phi_7(-1) = 1$$ $$\Phi_8(-1) = 2$$ however we will only consider $n>2$ for the conjecture.

This I thought was quite weird at first as it seems like a very simple pattern for something as complicated as the coefficients of cyclotomic polynomials. I managed to prove the first two cases, and will share the short proofs below, but am unsure of how to tackle the last case. I don't have any way to check for high values of $n$ other than calculation, and so I apologise in advance if there is a counterexample to this conjecture that can easily be checked with a computer. Below are the proofs for $n= 2^k$ and $n=2p$,

$n = 2^k$

By Möbius inversion we get $$\Phi_n(x) = \prod_{d\mid n}(x^d-1)^{\mu\left(\frac{n}{d}\right)}$$

Since $\mu(2^n) = 0$ for $n>1$ with $\mu(2)=-1$, therefore $$\Phi_{2^n}(x) = (x^{2^n}-1)(x^{2^{n-1}}-1)^{-1} = x^{2^{n-1}}+1$$ $$\Phi_{2^n}(-1) = 2$$

$n = 2p$

We can use the following identity $$\Phi_{2p}(x) = \sum_{k=0}^{p-1}(-1)^kx^k$$ $$ = \sum_{k=0}^{p-1}1 = p$$

But for the 'otherwise' case, I don't know how to approach it. It seems to general to approach in any specific way using identities or by looking at the values of $\mu(d)$ as that would require more information about the divisors of $n$. Any help or a proof/counterexample is appreciated!

Answer 1

We use the notation $ f(x) \mid_{x = a } $ to denote we are evaluating $f(x)$ at $ x= a$. This allows me to list out $f(x)$, then evaluate it.

All that we use is $ \prod _{d\mid n } \Phi_d(x) = x^n - 1$.

1. $ \Phi_1( -1) = x-1 \mid_{x=-1} = -2$
2. $ \Phi_2 (-1) = \frac{x^2 - 1 } { x-1} \mid_{x=-1} = x+1 \mid_{x=-1} = 0 $
3. $\Phi_p ( -1) = 1 $ for primes $p$.
   • $ \Phi_1 (-1) \times \Phi_p (-1) = x^p - 1 \mid_{x=-1} = - 2$
4. $ \Phi_n ( -1) = 1$ for odd $ n > 1$
   • Proof by strong induction on the number of divisors $k$ of $n$.
   • Base case: $ k = 2$ is done above.
   • Suppose $n$ has $k+1$ divisors. Then

$$-2 = x^n - 1 \mid _{x=-1} = \Phi_1(-1) \times \Phi_n (-1) \times \prod_{d \mid n, d\neq 1, d\neq n} \Phi_d(-1) = (-2) \times \Phi_n(-1) \times 1 .$$

When naively extending this argument to even $n$, we run into an issue where $\Phi_2(-1) = 0 $ and $ x^{n} - 1 = 0 $, so we can't determine what $\Phi_n(-1)$ will be.
To deal with this, we account explicitly for those zeros via $$ \frac{ x^{n} - 1 } {x +1 } \mid_{x=-1} = \prod_{d \mid n, d \neq 2 } \Phi_{d} ( - 1 ).$$ So, to get to $ \Phi_n(-1)$, we have to evaluate the LHS.
Let $ n = 2^k m$, where $m$ is odd. Then,

$$ \begin{array} {l l ll } \frac{ x^{n} - 1 } { x + 1 }\mid_{x=-1} & = \frac{ x^n - 1 } { x^{2^k} - 1 }\mid_{x=-1} & \times & \frac{ x^{2^k} - 1 } { x + 1 }\mid_{x=-1} \\ & = (x^{2^k(m-1) } + x^{2^k(m-2) } + \ldots + x^{2^k } + 1 ) & \times & (x-1)(1+x^2)(1+x^4) \ldots ( 1 + x^{2^{k-1} } ) \\ & = m & \times & -2^{k} \\ & = - n \\ \end{array}$$

Corollaries:

6. Quick observation that since $\frac{ x^{n} - 1 } {x +1 } \mid_{x=-1}$ is non-zero, hence so are the individual terms $\prod_{d \mid n, d \neq 2 } \Phi_{d} ( - 1 )$. (This should be obvious beforehand.) Thus, we don't have any more issues to deal with, and can iteratively calculate $ \Phi_n(-1)$.

7. $\Phi_{2^k} ( -1) = 2 $ for $ k > 1 $

   • Proof by induction. $-2^k = \prod_{i \neq 1 } \Phi_{2^k} (-1) = (-2) \times 2^{k-2} \times \Phi_{2^k} (-1) $.

8. $ \Phi_{2p^k} (-1) = p$ for odd prime $ p$

   • Similar proof by induction.
   • Note that this differs from OP's conjecture. We can check that $\Phi_{18} (-1) = x^6 - x^3 + 1 \mid_{x=-1} = 3$.

9. $\Phi_{2n} (-1) = 1 $ for any non-prime power $ n > 1$.

   • Essentially do a proof by strong induction per point 4.
   • Notice that for $ n = \prod p_i ^ { a _ i } $, we have $\Phi_1 (-1) \times \prod_{p^a \mid m, p^a \neq 1 } \Phi_{2p^a} ( -1 ) = -n$, and all of the other non $\Phi_{2n}(-1)$ terms contribute $1$ to the product by induction, hence $ \Phi_{2n} (-1) = 1 $.

Answer 2

First note that $\Phi_n(-1)=0$ if and only if $n=2$. This is immediate from your definition $$\Phi_n(X)=\hspace{-10pt}\prod_{\substack{1\leq d\leq n\\\gcd(d,n)=1}}\hspace{-10pt}(X-e^{\frac{2\pi d}{n}i}),$$ because $e^{\frac{2\pi d}{n}i}=-1$ implies $\tfrac dn=\tfrac12$ given that $1\leq d\leq n$, and then $\gcd(d,n)=1$ implies $d=1$ and $n=2$.

For the proof we only require a few of the most basic identities for cyclotomic polynomials:

1. If $n$ is an odd positive integer then $\Phi_{2n}(X)=\Phi_n(-X)$.
2. If $k$ is a positive integer and $p$ an odd prime number then $\Phi_{p^k}(X)=\sum_{i=0}^{p-1}X^{jp^{k-1}}$.
3. If $n=p^km$ is a positive integer with $m$ coprime to $p$ then $\Phi_n(X)=\Phi_{pm}(X^{p^{k-1}})$.
4. If $p$ is a prime number and $m$ is coprime to $p$ then $\Phi_{pm}(X)=\frac{\Phi_m(X^p)}{\Phi_m(X)}$.

If you are not familiar with these results, it is an instructive exercise to prove them. They can also be found on the Wikipedia page on cyclotomic polyonomials, the key observation being that $$X^n-1=\prod_{d\mid n}\Phi_d(X).$$

Now let $n$ be a positive integer. Let $p$ be the largest prime factor of $n$, and write $n=p^km$ with $k$ maximal, so that $m$ is coprime to $p$. Then by identities $(3)$ and $(4)$ we have $$\Phi_n(X)=\Phi_{pm}(X^{p^{k-1}})=\frac{\Phi_m(X^{p^k})}{\Phi_m(X^{p^{k-1}})}.$$ If $m>2$ then $p$ is odd and we can plug in $\Phi_m((-1)^{p^i})=\Phi_m(-1)\neq0$ to find that $\Phi_{p^km}(-1)=1$.

If $n=2p^k$ for some odd prime $p$ then by identities $(1)$ and $(2)$ we have $$\Phi_n(X)=\Phi_{p^k}(-X)=\sum_{i=0}^{p-1}(-X)^{ip^{k-1}},$$ and so plugging in $X=-1$ yields $\Phi_n(-1)=p$.

If $n=p^k$ for a prime number $p$, then plugging in $X=-1$ in identity $(2)$ shows that

• If $p=2$ and $k=1$ then $\Phi_n(-1)=0$, as mentioned before.
• If $p=2$ and $k>2$ then $\Phi_n(-1)=2$.
• If $p>2$ then $\Phi_n(-1)=1$.