CHAPTER 7

THE CYCLOTOMIC POLYNOMIALS

7.1. We define the n^{th} cyclotomic polynomial, Φ_{n}(x), as the product of the monic irreducible factors of x^{n}−1 which are not factors of x^{k} −1 for k ∈ {1, ..., n−1}. There are other ways one can define Φ_{n}(x). This particular definition seems to be the simplest in that it can be explained rather readily to a junior high school or high school student familiar with the rudiments of basic algebra. Observe that every irreducible factor of x^{n} − 1 in ℤ[x] necessarily has leading coefficient ±1. If w(x) is such a factor, then so is −w(x). We have restricted our attention to only the monic irreducible factors of x^{n} − 1 in defining Φ_{n}(x); thus, only one of w(x) and −w(x) is considered in the definition. The first 10 values of Φ_{n}(x) are:

Φ_{1}(x) = x − 1, Φ_{2}(x) = x + 1, Φ_{3}(x) = x^{2} + x + 1, Φ_{4}(x) = x^{2} + 1,

Φ_{5}(x) = x^{4} +x^{3} +x^{2} +x+1, Φ_{6}(x) = x^{2} −x+1, Φ_{7}(x) = x^{6} +x^{5} +x^{4} +x^{3} +x^{2} +x+1,

Φ_{8}(x) = x^{4} + 1, Φ_{9}(x) = x^{6} + x^{3} + 1, Φ_{10}(x) = x^{4} − x^{3} + x^{2} − x + 1. Our first theorem in this chapter shows that there is only one monic irreducible factor

of x^{n} − 1 which is not a factor of x^{k} − 1 for some k ∈ {1, ..., n − 1}. In other words, Φ_{n}(x) is irreducible. This result is due to Kronecker [1]. Our proof will be based on a proof of Landau [1]. Note that when n is a prime, the irreducibility of Φ_{n}(x) follows by Eisenstein’s Criterion (see Problem (7.1)); Gauss [1] first established this particular case of the theorem.

Theorem 23. Φ_{n}(x) is irreducible for all positive integers n. Proof. The roots of x^{n} − 1 are e^{2πim/n} where m ∈ {0, 1, …, n − 1}. If gcd(m, n) = d > 1, then there are integers m′ and n′ such that m = dm′ and n = dn′ so that e^{2πim/n} = e^{2πim′/n′} is a root of x^{k} − 1 where k = n′ = n/d < n. In this case, we get that e^{2πim/n} is a root of gcd(x^{n} − 1, x^{k} − 1) and, hence, a root of an irreducible factor of x^{n} − 1 which is not among the irreducible factors defining Φ_{n}(x). Thus, the roots of Φ_{n}(x) are among the numbers of the form e^{2πim/n} where m ∈ {0, 1, …, n − 1} and gcd(m, n) = 1. In particular, observe that e^{2πi/n} is a root of Φ_{n}(x) since it is not a root of x^{k} − 1 for k ∈ {1, …, n − 1}.

Let ζ = e^{2πij/n} with j a non-negative integer. Let f(x) ∈ ℤ[x] such that f(x) is monic and f(ζ) = 0. By the above comments, it suffices to show that if m ∈ {0, 1, …, n − 1} and gcd(m, n) = 1, then f(ζ^{m}) = 0 (and, in fact, we only need to establish this for j = 1). Since ζ is a root of x^{n} − 1, ζ is a root of a monic irreducible polynomial in ℤ[x]. Let d be the degree of this monic irreducible polynomial. By Problem (7.7), for each positive integer k, there is a unique element R_{k}(x) of ℤ[x] which is ≡ 0 or of degree < d such that f(ζ^{k}) = R_{k}(ζ); furthermore, if p is a prime, then every coefficient of R_{p}(x) is divisible by p.

Since ζ^{n} = 1, we get that for every positive integer k, R_{k}(x) = R_{k+n}(x). Thus, the set of coefficients of the polynomials R_{1}(x), R_{2}(x), … is finite. Let A denote the maximum of the absolute values of these coefficients. By the previous paragraph, if p is a prime > A, then we must have that R_{p}(x) ≡ 0 so that f(ζ^{p}) = 0 for every prime p > A. It suffices at this point to use Dirichlet’s Theorem concerning primes in arithmetic progressions, but we will avoid the use of Dirichlet’s Theorem as follows. The above all held with ζ = e^{2πij/n} where j is any non-negative integer. By applying the above observations several times while appropriately replacing j by suitable multiples of j, one gets that f(ζ^{w}) = 0 for every positive integer w which has all of its prime factors > A. Let m ∈ {0, 1, …, n − 1} such that gcd(m, n) = 1. Since ζ^{n} = 1, we get that ζ^{m} = ζ^{w}, where

w = m + n ∏_{\begin{matrix}p prime \\ p≤A,p∤m \end{matrix}} p.

Since gcd(m, n) = 1, one gets that w is not divisible by any prime ≤ A. Hence, f(ζ^{m}) = f(ζ^{w}) = 0. This completes the proof. ■

The following is an easy consequence of the above proof. Corollary. Let n be a positive integer. Then

Φ_{n}(x) = ∏_{\begin{matrix}1≤j≤n \\ gcd(j,n)=1 \end{matrix}} (x − e^{2πij/n}) .

The Corollary implies that the degree of Φ_{n}(x) is the number of positive integers relatively prime to n and ≤ n. In other words,

deg Φ_{n}(x) = ϕ(n),

where ϕ denotes Euler’s ϕ−function. A different formula for Φ_{n}(x) is often useful, and our next goal is to establish such a formula. We make use of Lemma 2 to Theorem 18 in Chapter 4. Theorem 24. Let n be a positive integer. Then

Φ_{n}(x) = ∏_{d|n} (x^{d} − 1)^{μ(n/d)} .

Proof. Whenever d divides n, the linear factors of x^{d} − 1 in ℂ[x] are all of the form x − e^{2πij/n} for some j ∈ {0, 1, …, n − 1}. For a fixed j ∈ {0, 1, …, n − 1}, x − e^{2πij/n} is a factor of x^{d} − 1 if and only if n/ gcd(j, n) divides d or, in other words, if and only if n/d divides gcd(j, n). Thus, the factor x − e^{2πij/n} appears on the right-hand side above with the exponent

∑_{\begin{matrix}d|n \\ (n/d)| gcd(j,n) \end{matrix}} μ (\frac{n}{d}) .

Observe that as d runs over the divisors of n so does k = n/d. Hence,

∑_{\begin{matrix}d|n \\ (n/d)| gcd(j,n) \end{matrix}} μ (\frac{n}{d}) = ∑_{k| gcd(j,n)} μ(k).

By Lemma 2 to Theorem 18, x − e^{2πij/n} is a factor on the right-hand side above if and only if gcd(j, n) = 1 and then x − e^{2πij/n} appears with the exponent 1. By the Corollary to Theorem 23, the result follows. ■

7.2. To illustrate an application of the cyclotomic polynomials, we give the next result first stated by Euler (cf. Dickson [2, Vol. I, p. 415]). It is an “easy” case of Dirichlet’s Theorem that if a and b are relatively prime positive integers, then there exist infinitely many primes in the arithmetic progression a + kb.

Theorem 25. Let n be a positive integer. Then there are infinitely many primes p such that p ≡ 1 (mod n).

Proof. By Problem (7.8), it suffices to show that if p is a prime dividing Φ_{n}(a) for some integer a, then p|n or p ≡ 1 (mod n). Fix an integer a and a prime p dividing Φ_{n}(a) with p ∤ n. Note that p cannot divide a since otherwise the fact that p|Φ_{n}(a) would imply that p divides the constant term in Φ_{n}(x); this is impossible since the constant term of Φ_{n}(x) divides the constant term of x^{n} − 1 and, hence, 1.

Next, we show that a has order n modulo p. Once this has been established, we will be through since the order of a modulo p must divide p − 1 which would imply that n|(p − 1) so that p ≡ 1 (mod n).

Observe that since p|Φ_{n}(a) and Φ_{n}(a) divides a^{n} − 1, we have that a^{n} ≡ 1 (mod p). Assume that there is a positive integer k < n such that a^{k} ≡ 1 (mod p). Let d = gcd(k, n). Then d ≤ k < n. There are integers u and v such that ku + nv = d. Recalling that p does not divide a (so that a and its powers have inverses modulo p), we get that

a^{d} = a^{ku+nv} = (a^{k})^{u} × (a^{n})^{v} ≡ 1 (mod p).

Since d = gcd(k, n) divides n, we get that x^{d} −1 divides x^{n} −1. By the definition of Φ_{n}(x),

(x^{d} − 1) Φ_{n}(x)| (x^{n} − 1) .

Since a^{d} − 1 ≡ 0 (mod p) and Φ_{n}(a) ≡ 0 (mod p), we get that (x − a)^{2} divides x^{n} − 1 modulo p. This contradicts that a is non-zero modulo p and

\frac{d}{d x} (x^{n} − 1) = nx^{n−1}

has 0 as its only root modulo p (where here we have used that p ∤ n). Hence, a has order n modulo p, and the proof is complete. ■

Note that it is not difficult to modify the above proof to establish that p is a prime with p ≡ 1 (mod n) if and only if p ∤ n and p is a prime divisor of Φ_{n}(a) for some integer a.

7.3. In this section, we begin with the following result due to Kronecker [2]. Theorem 26. If f(x) ∈ ℤ[x] is monic, is irreducible, and has all its roots on {z : |z| = 1}, then f(x) is a cyclotomic polynomial. Proof. Let α be such that f(α) = 0. If we can establish that α is a root of some cyclotomic polynomial, then since both cyclotomic polynomials and f(x) are irreducible, f(x) will be cyclotomic. Thus, it suffices to show that there exists a positive integer m such that α^{m} = 1.

Let α_{1}, α_{2}, …, α_{n} be the complete list of roots of f(x) with α_{1} = α. Using elementary symmetric functions (cf. Uspensky [1]), it is easy to deduce that (x−α_{1}^{k})(x−α_{2}^{k}) ⋯ (x−α_{n}^{k}) is in ℤ[x] for every positive integer k. We can avoid the use of elementary symmetric functions, however, by restricting consideration to polynomials of the form

f_{k}(x) = (x − α2^{k}_{1})(x − α2^{k}_{2}) ⋯ (x − α2^{k}_{n}).

Then one easily deduces that

f_{1}(x^{2}) = (−1)^{n}f(x)f(−x) ∈ ℤ[x].

Since f_{1}(x^{2}) is a polynomial in x^{2} with integer coefficients, it follows that f_{1}(x) has integer coefficients. An easy induction argument now implies that f_{k}(x) ∈ ℤ[x] for every positive integer k.

Since f_{k}(x) is monic and each root of f_{k}(x) has absolute value ≤ 1, we conclude that the coefficient of x^{j} in f_{k}(x) is ≤ \binom{n}{j} (by observing, for example, that the coefficient of x^{j} in f_{k}(x) must be less than or equal to the coefficient of x^{j} in (x+1)^{n}). Since n is fixed, this implies that the set {f_{k}(x) : k ≥ 1} is finite. Let F (x) denote the least common multiple of the elements of {f_{k}(x) : k ≥ 1}. Since α^{2}, α^{4}, α^{8}, … are all roots of F (x), there exist integers r and s with 1 ≤ r < s and α2^{r} = α2^{s}. Since |α| = 1 ≠ 0, we get that α^{m} = 1 with m = 2^{s} − 2^{r}, completing the proof. ■

Before continuing, we make a comment about what Theorem 26 is not saying. Consider the polynomial f(x) = x^{4} − 2x^{3} + x^{2} − 2x + 1. It is easy to verify that f(x) is not divisible by a cyclotomic polynomial. Is it possible that f(x) has roots with absolute value 1? Yes, and as we shall see, it does. Theorem 26 asserts that all of its roots cannot have absolute value 1. To see that f(x) has roots on the unit circle in the complex plane, observe that f(α) = 0 if and only if

(α + \frac{1}{α})^{2} − 2 (α + \frac{1}{α}) − 1 = 0,

from which one can easily deduce that the roots of f(x) are

\frac{1 + \sqrt{2} ± \sqrt{2\sqrt{2} − 1}}{2} and \frac{1 − \sqrt{2} ± i\sqrt{2\sqrt{2} + 1}}{2}.

The last two roots are imaginary, and a quick check indicates that they have absolute value 1.

There are a variety of results related to Theorem 26 (cf. Cassels [1], Smyth [1], and Lloyd-Smith [1]). In particular, we mention the following result of Dobrowolski [1]. Theorem 27. Let ϵ > 0, and let n be a sufficiently large integer. If f(x) ∈ ℤ[x] is monic, non-cyclotomic, and irreducible of degree n, then there is a root α of f(x) such that

|α| > 1 + \frac{2 − ϵ}{n} (\frac{log log n}{log n})^{3} .

It has been conjectured by Schinzel and Zassenhaus [2] that the factor (log log n/ log n)^{3} can be replaced by an absolute constant. We will establish Theorem 28. Let n be a positive integer, and let f(x) ∈ ℤ[x] be monic, non-cyclotomic, and irreducible of degree n. Furthermore, suppose that f(0) ≠ 0 and that f(x) has no reciprocal roots (i.e., that f(α) = 0 implies f(1/α) ≠ 0). Then there is a root α of f(x) such that

|α| > 1 + \frac{1}{10n}.

In other words, the conjecture of Schinzel and Zassenhaus is true when f(x) has no reciprocal roots. Theorem 28 was stated in such a way as to make its connection to Theorem 27 and the conjecture of Schinzel and Zassenhaus, but observe that the conditions that f(x) is non-cyclotomic and that f(x) is irreducible may be omitted from the theorem without changing the content of the result. Theorem 28 was established by Cassels [1] and is proven below. However, first we deal with some preliminaries. Lemma 1. Let δ ∈ (0, 1), and let x_{j}, for j ∈ {1, 2, …, m}, be real numbers satisfying

(7.1)  0 < x_{j} ≤ 1 + δ for j ∈ {1, 2, …, m}

and

(7.2)  ∏_{j=1}^{m} x_{j} = 1.

Then

(7.3)  ∏_{j=1}^{m} |x_{j} − 1| < (δe)^{m}.

Proof. Let x_{1}, …, x_{m} be real numbers satisfying (7.1) and (7.2), and assume that (7.3) is not true. For the time being, suppose that there are i and j ∈ {1, 2, …, m} such that x_{i} < 1, x_{j} < 1, and x_{i} ≠ x_{j}. For such an i and j fixed, consider

x′_{k} = \{ \begin{matrix}\sqrt{x_{i}x_{j}} & if k = i or j \\ x_{k} & otherwise. \end{matrix}

Then

0 < x′_{k} ≤ 1 + δ for k ∈ {1, 2, …, m},

and

∏_{k=1}^{m} x′_{k} = ∏_{k=1}^{m} x_{k} = 1.

Observe that

|x_{i} − 1| |x_{j} − 1| = (1 − x_{i})(1 − x_{j}) = 1 − (x_{i} + x_{j}) + x_{i}x_{j}

and

|x′_{i} − 1| |x′_{j} − 1| = (1 − \sqrt{x_{i}x_{j}})(1 − \sqrt{x_{i}x_{j}}) = 1 − 2\sqrt{x_{i}x_{j}} + x_{i}x_{j}.

Since x_{i} ≠ x_{j}, we get that x_{i} + x_{j} > 2\sqrt{x_{i}x_{j}}. Therefore,

∏_{k=1}^{m} |x_{k} − 1| < ∏_{k=1}^{m} |x′_{k} − 1|.

Since x_{1}, …, x_{m} do not satisfy (7.3), we get that x′_{1}, …, x′_{m} do not satisfy (7.3). By repeating the above several times if necessary, we get that we can replace the real numbers x_{1}, …, x_{m} by a new collection of m real numbers satisfying (7.1) and (7.2) and not (7.3) and having the property that any two elements of the new collection of real numbers which are < 1 are equal. For notational reasons, we assume as we can that x_{1}, …, x_{m} already have the latter property.

Observe that if x_{j} = 1 for every j ∈ {1, …, m}, then (7.3) would be true, giving a contradiction. Thus, since (7.1) and (7.2) hold, there must be an i and j ∈ {1, …, m} such that

x_{i} < 1 < x_{j} ≤ 1 + δ.

Using an argument similar to the above, one can show that we can further assume that any such x_{j} = 1 + δ. To do this, note that

((1 + δ) − x_{i}) ((1 + δ) − x_{j}) > 0 if x_{j} ≠ 1 + δ,

and set

x′_{k} = \{ \begin{matrix}1 + δ & if k = j \\ x_{i}x_{j}(1 + δ)^{−1} & if k = i \\ x_{k} & otherwise. \end{matrix}

By the above observations, there will be a certain number, say m − s, of the x_{j} equal to 1 + δ and the remaining s will be equal to 1 − η where

(1 − η)^{s}(1 + δ)^{m−s} = 1.

Then

−s log(1 − η) ≤ m log(1 + δ)

so that

sη < s (η + \frac{η^{2}}{2} + \frac{η^{3}}{3} + ⋯) ≤ m (δ − \frac{δ^{2}}{2} + \frac{δ^{3}}{3} − ⋯) < mδ.

Thus, η < mδ/s, and we get that

∏_{k=1}^{m} |x_{k} − 1| = η^{s}δ^{m−s} ≤ (\frac{m}{s})^{s} δ^{m}.

Since m/s > 1, we get that

\frac{m}{s} < (\frac{m}{s})^{e} ≤ e^{m/s}

(see Problem (7.10)). Thus, (m/s)^{s} < e^{m}, and we get that

∏_{k=1}^{m} |x_{k} − 1| < (eδ)^{m},

concluding the proof. ■

We will put off the proof of the next lemma for the moment and discuss first how to obtain a certain Corollary to the lemma and how to establish Theorem 28 from the Corollary. Lemma 2. Let m be an integer > 1, and let ρ be a real number > 1 satisfying

(7.4)  cos (\frac{π}{m}) < \frac{ρ^{2}}{ρ^{4} + 1 − ρ^{2}} .

Suppose that z_{1}, …, z_{m} ∈ ℂ and that

(7.5)  |z_{j}| ≤ ρ for all j ∈ {1, …, m}.

Then

(7.6)  ∏_{\begin{matrix}1≤j,k≤m \\ j≠k \end{matrix}} |z_{j}\overline{z_{k}} − 1| ≤ (\frac{ρ^{2m} − 1}{ρ^{2} − 1})^{m} .

Corollary. Let m be an integer > 1, and let

(7.7)  1 < ρ ≤ 1 + \frac{1}{10m}.

Suppose that z_{1}, …, z_{m} ∈ ℂ satisfying (7.5). Then

(7.8)  ∏_{\begin{matrix}1≤j,k≤m \\ j≠k \end{matrix}} |z_{j}\overline{z_{k}} − 1| ≤ m^{m}ρ^{2m(m−1)}.

Proof (assuming Lemma 2). One easily checks that for x > 1, x^{2} + x^{−2} − 1 is increasing so that for 1 < x < 1 + (1/(10m)), the value of x^{2} + x^{−2} − 1 is at most

(1 + \frac{1}{10m})^{2} + (1 + \frac{1}{10m})^{−2} − 1

= (1 + \frac{2}{10m} + \frac{1}{100m^{2}} ) + (1 − \frac{2}{10m} + \frac{3}{100m^{2}} − \frac{4}{1000m^{3}} + ⋯) − 1

< 1 + \frac{4}{100m^{2}} .

Thus, the value of 1/(x^{2} + x^{−2} − 1) is greater than

\frac{1}{1 + \frac{1}{25m^{2}}} > 1 − \frac{1}{25m^{2}} > 1 − \frac{π^{2}}{2m^{2}} + \frac{π^{4}}{24m^{4}} > cos (\frac{π}{m}) .

From (7.7), we get that

cos (\frac{π}{m}) < \frac{1}{ρ^{2} + ρ^{−2} − 1} = \frac{ρ^{2}}{ρ^{4} + 1 − ρ^{2}} .

Thus, the conditions of Lemma 2 are satisfied. Hence,

∏_{\begin{matrix}1≤j,k≤m \\ j≠k \end{matrix}} |z_{j}\overline{z_{k}} − 1| ≤ (\frac{ρ^{2m} − 1}{ρ^{2} − 1})^{m}
= (ρ^{2m−2} + ρ^{2m−4} + ⋯ + ρ^{2} + 1)^{m}
≤ (ρ^{2m−2}m)^{m} = m^{m}ρ^{2m(m−1)},

establishing the corollary. ■

Proof of Theorem 28 (assuming Lemma 2). Let w(x) = ∑_{j=0}^{m} a_{j}x^{j} with a_{0} ≠ 0 and a_{m} = 1, and suppose that the roots of w(x) satisfy that

(7.9)  |α_{j}| ≤ 1 + \frac{1}{10m} ∀j ∈ \{1, …, m\}.

First, we show that |a_{0}| = 1. Assume that |a_{0}| ≥ 2. Then observe that

(1 + \frac{1}{2m})^{m} = 1 + \frac{1}{2} + \frac{m(m − 1)}{2} \frac{1}{4m^{2}} + ⋯ < 1 + \frac{1}{2} + \frac{1}{4} + ⋯ = 2.

Also,

|∏_{j=1}^{m} α_{j}| = |a_{0}| ≥ 2.

Therefore, we get that there is a j ∈ {1, …, m} such that

|α_{j}| ≥ 2^{1/m} > 1 + \frac{1}{2m},

contradicting (7.9). Hence, |a_{0}| = 1. Thus,

(7.10)  ∏_{j=1}^{m} |α_{j}| = 1.

Consider

P = ∏_{1≤i,j≤m}(α_{i}α_{j} − 1).

If P = 0, then w(x) has reciprocal roots, and we are through. We therefore assume P ≠ 0. By the definition of P, P is a symmetric function of the roots α_{1}, …, α_{m} of w(x), and hence P ∈ ℤ − {0}. Thus,

(7.11)  ∏_{1≤i,j≤m} |α_{i}α_{j} − 1| ≥ 1.

Observe that

∏_{1≤i,j≤m} |α_{i}α_{j} − 1| = ∏_{1≤i,j≤m} |α_{i}\overline{α_{j}} − 1| = ( ∏_{1≤i≤m} |α_{i}\overline{α_{i}} − 1| )( ∏_{\begin{matrix}1≤i,j≤m \\ i≠j \end{matrix}} |α_{i}\overline{α_{j}} − 1| ).

Also,

α_{i}\overline{α_{i}} = |α_{i}|^{2} ≤ (1 + \frac{1}{10m})^{2} = 1 + \frac{1}{5m} + \frac{1}{100m^{2}} ≤ 1 + \frac{1}{4m}.

Therefore, by Lemma 1, with δ = 1/(4m), we get by (7.10) that

∏_{1≤i≤m} |α_{i}\overline{α_{i}} − 1| ≤ (\frac{e}{4m})^{m} .

Also, by the foregoing Corollary, we get that

∏_{\begin{matrix}1≤i,j≤m \\ i≠j \end{matrix}} |α_{i}\overline{α_{j}} − 1| ≤ m^{m} (1 + \frac{1}{10m})^{2m(m−1)}

= m^{m} ((1 + \frac{1}{10m})^{10m})^{(m−1)/5} < m^{m}e^{m/5}.

Combining the above, we get that

∏_{1≤i,j≤m} |α_{i}α_{j} − 1| < (\frac{e}{4m})^{m} m^{m}e^{m/5} = (\frac{e^{6/5}}{4} )^{m} < 1.

This contradicts (7.11); hence, the assumption that P ≠ 0 is invalid and the proof is complete. ■

To complete this section, we now proceed to prove Lemma 2. To do so, we first make some further preliminaries.

Lemma 3. Let r, α, and β be real numbers, and set λ = (α + β)/2. Suppose that

r ≥ 1, α ≥ 0, β ≥ 0, λ < 3π/2,

and

cos(λ) < \frac{r}{r^{2} + 1 − r} .

Then

(7.12)  |re^{iα} − 1| |re^{iβ} − 1| ≤ |re^{iλ} − 1|^{2} ,

with equality if and only if α = β = λ. Proof. We suppose as we may that α ≥ β. Define μ = α − λ, and observe that

α = μ + λ and β = λ − μ.

Let

L = cos λ and M = cos μ.

Note that μ ≥ 0; and if μ = 0, then α = β = λ and (7.12) holds. We need only show now that if μ > 0, then (7.12) holds with strict inequality. Therefore, we suppose that μ > 0. Observe that μ = α − λ ≤ λ < 3π/2. Also, if λ ≥ 0, then 0 ≤ λ ≤ π/2. We get either

(7.13)  L < 0 and − 1 ≤ M < 1

or

(7.14)  0 ≤ L ≤ M < 1 and L < \frac{r}{r^{2} + 1 − r} .

Now, squaring the left-hand side of (7.12), we get that

|re^{iα} − 1|^{2} |re^{iβ} − 1|^{2} = (re^{iα} − 1) (re^{−iα} − 1) (re^{iβ} − 1) (re^{−iβ} − 1)
= ((r^{2} + 1) − 2r cos(α)) ((r^{2} + 1) − 2r cos(β)) .

Using this together with

cos(α) + cos(β) = cos(λ + μ) + cos(λ − μ)

= (cos(λ) cos(μ) − sin(λ) sin(μ)) + (cos(λ) cos(μ) + sin(λ) sin(μ)) = 2 cos(λ) cos(μ) = 2LM

and

cos(α) cos(β) = (cos(λ + μ)) (cos(λ − μ))
= cos^{2}(λ) cos^{2}(μ) − sin^{2}(λ) sin^{2}(μ)
= cos^{2}(λ) cos^{2}(μ) − (1 − cos^{2}(λ)) (1 − cos^{2}(μ))
= cos^{2}(λ) + cos^{2}(μ) − 1 = L^{2} + M^{2} − 1,

we get that the square of the left-hand side of (7.12) is

(7.15)  (r^{2} + 1)^{2} − 4r (r^{2} + 1) LM + 4r^{2} (L^{2} + M^{2} − 1) .

Observe that if we set α = λ and β = λ in the above calculations of the square of the left-hand side of (7.12), we will obtain the square of the right-hand side of (7.12). In other words, we get the latter by setting M = 1 in (7.15). Thus, to finish the proof of the lemma, it suffices to show that

(r^{2} + 1)^{2} − 4r (r^{2} + 1) LM + 4r^{2} (L^{2} + M^{2} − 1) < (r^{2} + 1)^{2} − 4r (r^{2} + 1) L + 4r^{2} (L^{2}) .

This is the same as establishing that

4r^{2} (M^{2} − 1) < 4r (r^{2} + 1) L(M − 1)

or, upon reducing and noting that M − 1 < 0,

r(M + 1) > (r^{2} + 1) L.

This is clear in the case that (7.13) holds. In the case that (7.14) holds, we rewrite the above as

r > (r^{2} + 1) L − rM.

This inequality follows directly from

r > (r^{2} + 1 − r) L

and

(r^{2} + 1 − r) L ≥ (r^{2} + 1) L − rM,

completing the proof. ■

Lemma 4. Let m > 1 be an integer, and let θ_{1}, θ_{2}, …, θ_{m} be real numbers satisfying

0 ≤ θ_{j} ≤ 2π for 1 ≤ j ≤ m.

Let

w = \frac{1}{2m} (θ_{1} + ⋯ + θ_{m}) .

Consider r > 1 such that

(7.16)  | cos(w)| < \frac{r}{r^{2} + 1 − r} .

Then

(7.17)  ∏_{1≤j≤m} |re^{iθ_{j}} − 1| ≤ |re^{2iw} − 1|^{m}

with equality if and only if θ_{1} = ⋯ = θ_{m} = 2w. Proof. Observe that if r > 1, then r/(r^{2} + 1 − r) < 1. Hence, the lemma vacuously holds if w = 0 or π. Also, the lemma follows if w = π/2 since then we get that

|re^{iθ_{j}} − 1| ≤ r + 1 = |re^{2iw} − 1|

with equality if and only if θ_{j} = π = 2w. If π/2 < w < π, then replace θ_{j} with 2π − θ_{j} and w by π − w to reduce the lemma to a case in which

0 < w < π/2.

Thus, we suppose as we may that the latter holds for w.

For fixed w, by continuity and compactness considerations, the left-hand side of (7.17) obtains its upper bound for some choice of θ_{1}, …, θ_{m} as in the lemma. We fix θ_{1}, …, θ_{m} so that this upper bound is obtained and note that now it suffices to prove (7.17) with the θ_{j} so chosen. If θ_{1} = ⋯ = θ_{m}, then we are through; thus, we suppose as we may that θ_{1} and θ_{2} satisfy

0 ≤ θ_{1} < 2w < π and 2w < θ_{2} ≤ 2π.

Set α = θ_{1}, β = θ_{2}, and λ = (α + β)/2. Then

λ = \frac{1}{2} (θ_{1} + θ_{2}) < \frac{1}{2} (π + 2π) = \frac{3}{2}π

and either λ ≥ π/2 so that

cos(λ) ≤ 0 < \frac{r}{r^{2} + 1 − r}

or 0 < w < θ_{2}/2 < λ < π/2 so that from (7.16)

cos(λ) ≤ cos (θ_{2}/2) < cos(w) < \frac{r}{r^{2} + 1 − r} .

Thus, from Lemma 3, we get that

|re^{iα} − 1| |re^{iβ} − 1| < |re^{iλ} − 1|^{2} .

By considering θ′_{1} = θ′_{2} = λ = (θ_{1} + θ_{2})/2 and θ′_{j} = θ_{j} for j ∈ {3, …, m}, we get that the above inequality contradicts that θ_{1}, …, θ_{m} were chosen so that the left-hand side of (7.17) was maximal. Hence, for θ_{1}, …, θ_{m} so chosen, we must have that θ_{1} = ⋯ = θ_{m} = 2w so that (7.17) holds with equality. This completes the proof. ■

Lemma 5 (The Maximum Modulus Principle). Let f(z) ∈ ℂ[z], and let ρ ≥ 0. Then

max {|f(z)| : |z| ≤ ρ} = max {|f(z)| : |z| = ρ} .

Proof. Observe that the maximums exist above since f(z) is continuous. Let z_{0} be such that |z_{0}| ≤ ρ and

|f(z_{0})| = max {|f(z)| : |z| ≤ ρ} .

Furthermore, suppose that |z_{0}| is maximal with the above conditions (noting that this is in fact possible). If |z_{0}| = ρ, then we’re done. Assume therefore that |z_{0}| < ρ. Let r = ρ − |z_{0}|. Then the average value of |f(z)|^{2} on the circle {z : |z − z_{0}| = r} is < |f(z_{0})|^{2}.

Let g(z) = f(z + z_{0}) = ∑_{j=0}^{n} b_{j}z^{j}. Then the above implies that

∑_{j=0}^{n} |b_{j}|^{2}r^{2j} = \frac{1}{2π} ∫_{0}^{2π} (∑_{j=0}^{n} b_{j}r^{j}e^{jiθ}) (∑_{j=0}^{n} \overline{b_{j}}r^{j}e^{−jiθ}) dθ

= \frac{1}{2π} ∫_{0}^{2π} g (re^{iθ}) \overline{g (re^{iθ})} dθ = \frac{1}{2π} ∫_{0}^{2π} |g (re^{iθ})|^{2} dθ

= \frac{1}{2π} ∫_{0}^{2π} |f (z_{0} + re^{iθ})|^{2} dθ < |f(z_{0})|^{2} = |g(0)|^{2} = |b_{0}|^{2} ,

giving a contradiction. Thus, the result follows. ■

Proof of Lemma 2. Fix j ∈ {1, …, m} and complex numbers z_{1}, …, z_{j−1}, z_{j+1}, …, z_{m} as in the lemma. Then

∏_{\begin{matrix}1≤t,k≤m \\ t≠k \end{matrix}} |z_{t}\overline{z_{k}} − 1| = ( ∏_{\begin{matrix}1≤t,k≤m \\ t≠k,t≠j,k≠j \end{matrix}} |z_{t}\overline{z_{k}} − 1|) ∏_{\begin{matrix}1≤k≤m \\ k≠j \end{matrix}} (|z_{j}\overline{z_{k}} − 1||\overline{z_{j}}z_{k} − 1|)

= ( ∏_{\begin{matrix}1≤t,k≤m \\ t≠k,t≠j,k≠j \end{matrix}} |z_{t}\overline{z_{k}} − 1|) ∏_{\begin{matrix}1≤k≤m \\ k≠j \end{matrix}} |z_{j}\overline{z_{k}} − 1|^{2}.

Hence, we may view

(7.18)  ∏_{\begin{matrix}1≤t,k≤m \\ t≠k \end{matrix}} |z_{t}\overline{z_{k}} − 1|

as the absolute value of a polynomial in z_{j}. By Lemma 5, given that |z_{j}| ≤ ρ, we get that (7.18) obtains its maximum for some z_{j} with |z_{j}| = ρ. Letting j vary now, we see that to finish the proof, we need only consider the case when

|z_{1}| = |z_{2}| = ⋯ = |z_{m}| = ρ.

By reordering the z_{j}’s if necessary, we suppose as we may that

z_{j} = ρeiϕ^{j} for j ∈ {1, …, m},

where

0 ≤ ϕ_{1} ≤ ϕ_{2} ≤ ⋯ ≤ ϕ_{m} < 2π.

Now, (7.18) becomes

∏_{\begin{matrix}1≤t,k≤m \\ t≠k \end{matrix}} |ρ^{2}ei(ϕ^{t} − ϕ^{k}) − 1| = ∏_{1≤s≤m−1} P_{s},

where

P_{s} = ∏_{\begin{matrix}1≤t,k≤m \\ t≡k+s (mod m) \end{matrix}} |ρ^{2}ei(ϕ^{t} − ϕ^{k}) − 1|.

Observe that if s and t are known, then k is uniquely determined by the conditions 1 ≤ k ≤ m and t ≡ k + s (mod m). For fixed s ∈ {1, …, m − 1} and fixed t ∈ {1, …, m}, consider the uniquely determined k as in the product above. Define

θ_{t} = θ_{t}(s) = \{ \begin{matrix}ϕ_{t} − ϕ_{k} & if t > k \\ ϕ_{t} − ϕ_{k} + 2π & if k > t \end{matrix}

and note that k > t in this definition if and only if t ∈ {1, …, s}. Set

r = ρ^{2}

and

w = sπ/m.

Then, for s ∈ {1, …, m − 1},

θ_{1} + ⋯ + θ_{m} = 2πs = 2mw

and

| cos(w)| = |cos (\frac{sπ}{m} )| ≤ cos (\frac{π}{m}) < \frac{ρ^{2}}{ρ^{4} + 1 − ρ^{2}} = \frac{r}{r^{2} + 1 − r} .

Thus, we may apply Lemma 4 to get that

P_{s} ≤ |ρ^{2}e^{2πis/m} − 1|^{m}

so that (7.18) is bounded above by

∏_{1≤s≤m−1} |ρ^{2}e^{2πis/m} − 1|^{m} = |\frac{ρ^{2m} − 1}{ρ^{2} − 1}|^{m} ,

completing the proof. ■

7.4. In this section, we investigate the size of the coefficients of Φ_{n}(x). Recall the values of Φ_{n}(x) for n ∈ {1, 2, …, 10} given at the beginning of the chapter. If we continue calculating up to Φ_{104}(x), the coefficients obtained will remain in the set {−1, 0, 1} suggesting at the very least that the coefficients of Φ_{n}(x) do not get very large. This is in fact not the case, and it is the purpose of this section to mention two results in this direction. We shall only prove the first. It is a consequence of the second but seemingly much easier to establish. The proof given here is due to I. Schur (cf. E. Lehmer [1], her first mathematical publication). Theorem 29. Given B, there exists a positive integer n such that Φ_{n}(x) has at least one coefficient with absolute value > B. Proof. Let n = p_{1}p_{2} … p_{k}, where k is an odd positive integer and p_{1}, p_{2}, …, p_{k} are primes satisfying

p_{1} < p_{2} < ⋯ < p_{k} < p_{1} + p_{2}.

Note that there are infinitely many such n for any given k (see problem (AII.1)). To prove the theorem, it is sufficient to prove that the coefficient of xp^{k} in Φ_{n}(x) is 1 − k. Using Theorem 24 and calculating Φ_{n}(x) modulo xp^{k} + 1, we get that modulo xp^{k} + 1

Φ_{n}(x) ≡ (∏_{j=1}^{k} (xp^{j} − 1)) /(x − 1)

≡ (xp^{k} − 1 + xp^{k} − 2 + ⋯ + x + 1) (xp^{1} − 1) (xp^{2} − 1) ⋯ (xp^{k−1} − 1)
≡ (xp^{k} − 1 + xp^{k} − 2 + ⋯ + x + 1) (−xp^{k−1} − xp^{k−2} − ⋯ − xp^{1} + 1)

The fact that the coefficient of xp^{k} in Φ_{n}(x) is 1 − k follows, and the proof is complete. ■

The above proof shows that certain positive integers n with sufficiently many distinct prime factors are such that Φ_{n}(x) has a coefficient whose absolute value exceeds B for any preassigned B. Migotti (cf. E. Lehmer [1]) showed that if n is the product of 2 primes, then the coefficients of Φ_{n}(x) are all from the set {0, ±1}. E. Lehmer [1] showed that as n runs through the positive integers which are the product of 3 distinct primes, the coefficients of Φ_{n}(x) get arbitrarily large.

We end this section by stating what is undoubtedly one of the nicest result on the subject. Let M_{n} denote the maximum of the absolute values of the coefficients of Φ_{n}(x). Erdős conjectured that for every constant c, one has that M_{n} ≥ c for almost all n. In other words, the number of n ≤ x for which M_{n} < c is o(x). The conjecture was first resolved by Maier [1] in a much stronger form. He showed the following:

Theorem 30. Let ϵ(n) be any function defined for all positive integers and satisfying lim_{n→∞} ϵ(n) = 0. Then

M_{n} ≥ n^{ϵ(n)}

for almost all n.

Thus, the conjecture follows by taking, for example, ϵ(n) = 1/ log log n. We note that the result of Maier [1] is in fact even stronger than that given by Theorem 30.

7.5. In this section, we discuss the factorization of Φ_{n}(x) modulo a prime. We will establish

Theorem 31. Let n be a positive integer, and let p be a prime. Write n = p^{k}m where k is a non-negative integer and gcd(p, m) = 1. Let f be the least positive integer such that p^{f} ≡ 1 (mod m). Then Φ_{n}(x) factors as a product of ϕ(m)/f incongruent irreducible polynomials modulo p of degree f each raised to the ϕ(p^{k}) power.

Before presenting the proof of Theorem 31, we give two examples. Further examples can be found in the exercises as well as in the Corollaries following the proof.

Example 1. Consider f(x) = Φ_{10}(x) = x^{4} − x^{3} + x^{2} − x + 1. If p = 2, then m = 5 and f = 4 in Theorem 31 so that f(x) is irreducible modulo 2. If p = 5, then m = 2 and f = 1 so that f(x) factors as a linear polynomial raised to the 4th power modulo 5. In fact, we have

f(x)(x + 1) ≡ x^{5} + 1 ≡ (x + 1)^{5} (mod 5)

so that by Theorem 3 (unique factorization in ℤ_{5}[x]) we get that f(x) ≡ (x + 1)^{4} (mod 5). If p is a prime ≡ 1 (mod 10), then m = 10 and f = 1 so that f(x) factors as a product of 4 distinct linear polynomials modulo p. If p is a prime ≡ 3 or 7 (mod 10), then m = 10 and f = 4 so that f(x) is irreducible modulo p. Finally, if p is a prime ≡ 9 (mod 10), then m = 10 and f = 2 so that f(x) factors as a product of 2 distinct irreducible quadratic polynomials modulo p.

Example 2. Consider f(x) = Φ_{8}(x) = x^{4} + 1. If p = 2, then m = 1 and f = 1 in Theorem 31 so that f(x) factors as a linear polynomial raised to the 4th power modulo 2. Here we have

f(x) ≡ x^{4} + 1 ≡ (x + 1)^{4} (mod 2).

If p is a prime ≡ 1 (mod 8), then m = 8 and f = 1 so that f(x) factors as a product of 4 distinct linear polynomials modulo p. If p is a prime ≡ 3, 5, or 7 (mod 8), then m = 8 and f = 2 so that f(x) factors as a product of 2 distinct irreducible quadratic polynomials modulo p. Observe that this implies the result of Problem (4.1) that x^{4} + 1 is reducible modulo every prime. (Also, see Corollary 1 below.)

These examples demonstrate what is apparent from the statement of the theorem, namely that the factorization of Φ_{n}(x) modulo a prime p is completely determined by the residue class to which p belongs modulo n. We also note that Theorem 31 can be used to give an alternative proof of Theorem 25. More specifically, Theorem 31 implies that if p is a prime which does not divide n and is such that p|Φ_{n}(a) for some integer a, then p ≡ 1 (mod n).

Proof of Theorem 31. The main tool we use to obtain Theorem 31 is Theorem 17 of Chapter 4. We begin, however, by making use of Problem (7.3). If, in the statement of Theorem 31, k ≥ 1, then Problem (7.3) (a) implies that

Φ_{n}(x) = Φ_{pm}(xp^{k−1}),

and Problem (7.3) (b) implies that

Φ_{pm}(xp^{k−1})Φ_{m}(xp^{k−1}) = Φ_{m}(xp^{k}).

On the other hand,

Φ_{m}(xp^{k−1}) ≡ Φ_{m}(x)p^{k−1} (mod p) and Φ_{m}(xp^{k}) ≡ Φ_{m}(x)p^{k} (mod p).

Hence, we deduce that

Φ_{n}(x) ≡ Φ_{m}(x)^{ϕ(p^{k})} (mod p).

We assumed above that k ≥ 1, but we note that this last congruence is trivially true in the case that k = 0. To establish the theorem, then, it suffices to show Φ_{m}(x) factors modulo p as a product of ϕ(m)/f incongruent irreducible polynomials of degree f.

By the definition of f, we see that m divides p^{f} − 1. Hence, x^{m} − 1 divides xp^{f} − 1 − 1, and we obtain that Φ_{m}(x) divides (xp^{f} − 1 − 1)x = xp^{f} − x. By Theorem 17, each irreducible factor g(x) of Φ_{m}(x) modulo p is such that its degree, say r, divides f. We show that for each such g(x), r = f. Assume for some such g(x), we have r < f. Then by Theorem 16 or Theorem 17, g(x) divides xp^{r} − x modulo p. In fact, since Φ_{m}(x) divides x^{m} − 1, the constant term of Φ_{m}(x) is ±1 and so the constant term of g(x) is non-zero modulo p. Thus, g(x) is not a multiple of x modulo p and g(x) divides xp^{r} − 1 − 1 modulo p. In particular, x has an inverse (modd p, g(x)). The definition of f implies that m does not divide p^{r} − 1 so that d = gcd (m, p^{r} − 1) < m. Let u and v be integers satisfying mu + (p^{r} − 1) v = d. Then

x^{d} − 1 ≡ xmu + (p^{r} − 1) v − 1 ≡ (x^{m})^{u} (xp^{r} − 1)^{v} − 1 ≡ 0 (modd p, g(x)).

Therefore, g(x) divides x^{d} − 1 modulo p.

Observe that d < m implies Φ_{m}(x) is by definition relatively prime to x^{d} −1. Therefore, Φ_{m}(x) and x^{d} − 1 are relatively prime divisors of x^{md} − 1 in ℤ[x]. Let h_{1}(x) ∈ ℤ[x] with x^{md} − 1 = Φ_{m}(x) (x^{d} − 1) h_{1}(x). Since g(x) is a common divisor of Φ_{m}(x) and x^{d} − 1 modulo p, we get that for some h_{2}(x) ∈ ℤ[x]

x^{md} − 1 ≡ Φ_{m}(x) (x^{d} − 1) h_{1}(x) ≡ g(x)^{2}h_{2}(x) (mod p).

Since d divides p^{r} − 1, p does not divide d. The definition of m implies that p does not divide m. Hence, taking derivatives above, we deduce that g(x) is an irreducible factor of x^{md−1} modulo p and yet g(x) is not a multiple of x modulo p. This is a contradiction which implies that r = f. Thus, every irreducible factor of Φ_{m}(x) modulo p is of degree f.

Since deg Φ_{m}(x) = ϕ(m), it remains only to show that if g(x) is an irreducible factor of Φ_{m}(x) modulo p, then g(x)^{2} does not divide Φ_{m}(x) modulo p. Assume g(x)^{2} divides Φ_{m}(x) modulo p. Then there is an h(x) ∈ ℤ[x] such that x^{m} − 1 ≡ g(x)^{2}h(x) (mod p). Following the argument above, we get in this case that g(x) must be an irreducible factor of x^{m−1} modulo p, resulting in a contradiction and, hence, completing the proof. ■

Let n be a positive integer. Observe that if a prime p does not have order ϕ(n) modulo n, then the above theorem implies Φ_{n}(x) is reducible modulo p. In particular, if there are no primitive roots modulo n (i.e., no integers a for which the order of a is ϕ(n) modulo n), then Φ_{n}(x) is reducible modulo every prime. On the other hand, using the above theorem in conjunction with Dirichlet’s Theorem on primes in arithmetic progression, one can easily deduce that if there exists a primitive root modulo n, then Φ_{n}(x) is irreducible modulo some prime. The n for which a primitive root modulo n exists are 1, 2, 4, and numbers of the form p^{k} or 2p^{k} where p is an odd prime and k a positive integer; thus, we can summarize the comments here with Corollary 1. Let n be a positive integer. Then Φ_{n}(x) is reducible modulo every prime p if and only if n is not among the numbers of the form 1, 2, 4, p^{k}, or 2p^{k} where p denotes an odd prime and k denotes a positive integer.

We consider now the possibility that Φ_{n}(x) is Eisenstein with respect to some prime p. Then Φ_{n}(x) would factor modulo p as a constant times a linear polynomial raised to the power ϕ(n). Using the notation of Theorem 31, we would necessarily have that f = 1 and ϕ(m) = 1. Therefore, m = 1 or 2. This implies that either n = p^{k} or n = 2p^{k} for some prime p. In these cases, Theorem 31 can be used to establish that Φ_{n}(x) is Eisenstein with respect to p (or see Problem (7.5) and Problem (7.6)). Hence, we get

Corollary 2. Let n be a positive integer. Then Φ_{n}(x) is Eisenstein with respect to a prime p if and only if n = p^{k} or n = 2p^{k} for some positive integer k.

7.6. There are numerous results concerning cyclotomic polynomials and it would be impossible in one chapter to give them a thorough treatment. In this section, we briefly mention a few other results without proofs. The results are classical and can be found in Narkiewicz [1].

The field ℚ(ζ_{n}) is called a cyclotomic field. All the roots of x^{n} − 1 are in ℚ(ζ_{n}). Thus, we can refer to the galois group G = Gal(ℚ(ζ_{n})/ℚ) associated with the polynomial Φ_{n}(x). The galois group G is isomorphic to ℤ_{n}^{∗} (the multiplicative group of integers modulo n). The elements of G can be described as follows. Let j ∈ {1, …, n − 1} with gcd(j, n) = 1. Define σ_{j} as the automorphism of ℚ(ζ_{n}) satisfying σ_{j}(ζ_{n}) = ζ_{n}^{j} and σ_{j}(u) = u for all u ∈ ℚ. Then the σ_{j}’s are precisely the elements of G. Observe that in particular |G| = ϕ(n).

The ring of algebraic integers in ℚ(ζ_{n}) clearly contains ℤ[ζ_{n}]. In fact, the ring of integers can be shown to be ℤ[ζ_{n}]. The units in ℤ[ζ_{n}] are described in a convenient form by a result known as Kummer’s Lemma (which Kummer used to establish his classical result that Fermat’s Last Theorem holds for any “regular” prime exponent). It is

Theorem 32. Every unit in ℤ[ζ_{n}] can be written in the form rζ_{n}^{k} where r is real and k is an integer.

Problems

(7.1) (a) What is the value of Φ_{p}(x) when p is a prime?

(b) Expand Φ_{p}(x + 1) as a polynomial in x, and deduce that Φ_{p}(x) is irreducible.

(7.2) Prove that if Φ_{n}(x) is Eisenstein with respect to p, then p divides n. (Hint: First show that if f(x) and g(x) are in ℤ[x] and each does not have a multiple root, then R(f, f′)R(g, g′) divides R(fg, (fg)′). Be sure to justify that R(fg, (fg)′)/(R(f, f′)R(g, g′)) is an integer.)

(7.3) Let n be a positive integer, and let p be a prime.

(a) Prove that if p divides n, then Φ_{pn}(x) = Φ_{n}(x^{p}).

(b) Prove that if p does not divide n, then Φ_{pn}(x) = Φ_{n}(x^{p})/Φ_{n}(x).

(7.4) Prove that if n is an odd integer ≥ 3, then Φ_{2n}(x) = Φ_{n}(−x).

(7.5) Without using any material after Section 7.1 except the problems above, prove that if k is a positive integer and p is a prime, then Φ_{p^{k}}(x) is Eisenstein with respect to p.

(7.6) Without using any material after Section 7.1 except the problems above, prove that if k is a positive integer and p is a prime, then Φ_{2p^{k}}(x) is Eisenstein with respect to p.

(7.7) Let f(x) be a monic irreducible polynomial of degree n, and let α denote a root of f(x).

(a) Prove that for every positive integer k, there is a unique polynomial g(x) ∈ ℤ[x] which is ≡ 0 or of degree < n such that f(α^{k}) = g(α).

(b) In (a), if k = p where p is a prime, then prove that every coefficient of g(x) is divisible by p. (Hint: Consider f(x^{p}) − f(x)^{p}.)

(7.8) Let f(x) be a non-constant polynomial with integer coefficients. Prove that there are infinitely primes p for which p divides f(m) for some integer m.

(7.9) Modify the proof of Kronecker’s Theorem in Section 3 to give an easy proof that there is a positive function ϵ(n) such that if f(x) ∈ ℤ[x] is a monic irreducible polynomial of degree n with all of its roots having absolute value ≤ 1 + ϵ(n), then f(x) is cyclotomic.

(7.10) Prove that if x > 0, then x^{e} ≤ e^{x} (a result used in the proof of Lemma 1 to Theorem 28).

(7.11) (a) Is it possible to load 2 dice in such a way that each face of each die has a rational probability of coming facing up on a roll and such that if both dice are rolled, then the sum of the 2 numbers rolled is equally likely to be each of 2, 3, …, 12?

(b) Do part (a) with each face of each die having a “real” probability of coming facing up on a roll.

(7.12) Let n and k be positive integers. Prove that Φ_{n}(x^{k}) is a product of distinct cyclotomic polynomials. (Note: The product may consist of just one factor.)

(7.13) Let n and k be positive integers. Prove that Φ_{n}(x^{k}) is irreducible if and only if every prime divisor of k is a prime divisor of n.

(7.14) Prove that for each prime p, Φ_{200}(x) can be written as a product of at least 4 factors modulo p.