Coefficients of equations whose roots are primitive nth roots are -1, 0, 1 for n < 105.

Question

I came across an enigmatic comment without a specific reference in the book "Theory of Equations" by Joseph Thomas, p. 90 (see my paragraph following the quote for the context of the phrase 'resulting equation').

In addition, it is known that for $n < 105$ every coefficient in the resulting equation has one of the three values -1, 0, 1.

The context is the equation satisfied by all the primitive nth roots of unity. For example (x^6 -1) can be divided by (x^3 -1) and by x^2 - 1 to eliminate non-primitive cube and square roots. However this eliminates factor (x-1) twice because 1 is both a cube root and a square root. Thus a final (single) multiplication by (x - 1) to add it back in gives the equation x^2 - x + 1, satisfied by the sole primitive 6th root of unity. A similar procedure can be followed for other values of n.

I could not locate a reference showing this intriguing fact. Can anyone either explain it or provide a reference (or both)?

Answer 1

This is a consequence of several simple facts. If we denote the $n-th$ cyclomatic polynomial by $F_n(X)$, then we have

$1) F_p(X) = X^{p-1} + X^{p-2} + \dots + X+1$ if $p$ is a prime number.

$2) F_{2m}(X) = F_m(-X) $ for every odd integer $m$.

$3) F_m(X^p) = F_m(X)F_{mp}(X)$ for all primes $p$ not dividing $m$.

$4) F_n(X) = F_m(X^{n/m})$ if $m$ is the product of all primes dividing $n$.

$5)$ for primes $p \ne q$, all coefficients of $F_{pq}(X)$ have absolute $1$.

from these facts, one can conclude that for $F_n$ to have coefficients with absolute greater than $1$, at least three distinct primes should divide $n$ (except for $2$). Then, the smallest candidate is $3\times 5 \times 7 = 105$. Some calculation shows that the coefficient of $X^{42}$ in $F_{105}(X)$ is actually $-2$.

update:

$F_n(X)$ is the minimal polynomial of $e^{\frac{2\pi i}{n}}$ over $\mathbb{Q}$, and is of order $\phi(n)$ (Euler totient function). Also, $$F_n(X) = \Pi_{(k, n) = 1, \space 1\le k\le n} (X-e^{\frac{2\pi i}{n}k}).$$

To prove the facts mentioned above, the main idea is to show that every root of LHS is a root of RHS (and vice versa). Then the result follows, since $F_n(X)$ is always monic and has $n$ distinct roots.

for $5)$, try using $3)$ with $m=q$, and look at it as an identity between power series.

Answer 2

This is well known, see the wikipedia article on Roots of unity, with references. The first exception is the cyclotomic polynomial $\Phi_{105}$. Before all coefficients are $0,1,-1$, e.g.

\begin{align} Φ_1(z) & = z − 1 \\ Φ_2(z) & = (z^2 − 1)⋅(z − 1)−1 = z + 1 \\ Φ_3(z) &= (z^3 − 1)⋅(z − 1)−1 = z^2 + z + 1\\ Φ_4(z) &= (z^4 − 1)⋅(z^2 − 1)−1 = z^2 + 1\\ Φ_5(z) &= (z^5 − 1)⋅(z − 1)−1 = z^4 + z^3 + z^2 + z + 1\\ Φ_6(z) &= (z^6 − 1)⋅(z^3 − 1)−1⋅(z^2 − 1)−1⋅(z − 1) = z^2 − z + 1\\ Φ_7(z) &= (z^7 − 1)⋅(z − 1)−1 = z^6 + z^5 + z^4 + z^3 + z^2 +z + 1\\ Φ_8(z) &= (z^8 − 1)⋅(z^4 − 1)−1 = z^4 + 1\\ \end{align}

Reference: Emma Lehmer, On the magnitude of the coefficients of the cyclotomic polynomial, Bulletin of the American Mathematical Society 42 (1936), no. 6, pp. 389–392.