# The cyclotomic polynomials

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 The cyclotomic polynomials Notes by G.J.O. Jameson 1. The definition and general results We use the notation e(t) = e 2πit. Note that e(n) = 1 for integers n, e(s + t) = e(s)e(t) for all s, t. e( 1 ) = 1 and 2 Consider the polynomial x n 1. The complex factorisation is obvious: the zeros of the polynomial are e(k/n) for 1 k n, so x n 1 = n k=1 [ ( )] k x e. (1) n One of these factors is x 1, and when n is even, another is x + 1. The remaining factors can be combined in pairs of the form [x e(k/n)][x e( k/n)] = x 2 2x cos 2πk n + 1 to give the factorisation into linear and quadratic real factors. We will describe a constructive factorisation into polynomials with integer coefficients, and then present some applications to number theory. We write (a, b) for the gcd of a and b, and E n = {k : 1 k n, (k, n) = 1}. By definition, the number of members of E n is Euler s φ(n). The cyclotomic polynomials Φ n are defined for all n 1 by Φ n (x) = k E n (This is the usual notation; be careful to distinguish Φ n and φ(n)!) [ ( )] k x e. (2) n It is clear that Φ n is a monic polynomial (with, apparently, complex coefficients) of degree φ(n). We note some elementary cases: n = 1: E 1 = {1}, hence Φ 1 (x) = x 1. n = 2: E 2 = {1} and e( 1 2 ) = 1, hence Φ 2(x) = x + 1. n = 4: E 4 = {1, 3} and e( 1 4 ) = i, hence Φ 4(x) = (x i)(x + i) = x For prime p, Φ p (x) = xp 1 x 1 = 1 + x + + xp 1. 1

2 Proof. Consider the factorisation of x p 1 given by (1) (with n = p). To obtain Φ p (x), remove the factor x 1 corresponding to k = p. Let n 3. Denote by E n the members k of E n with k < n 2. Then E n consists of the pairs (k, n k) with k E n (note that 1 2 n E n). Hence φ(n) is even and E n has 1 2 φ(n) members. Since e[(n k)/n] = e( k/n), we have the following factorisation of Φ n (x) into quadratic real factors: Φ n (x) = k E n [x e(k/n)][x e( k/n)] = (x 2 2x cos 2πk + 1). (3) n k E n We deduce some properties of Φ n (x) as a function of a real variable For all n 3, (i) Φ n (x) has real coefficients; (ii) Φ n (0) = 1; (iii) Φ n (x) > 0 for all real x; (iv) Φ n (x) is strictly increasing for x > 1; (v) for all x > 0, (x 1) φ(n) < Φ n (x) < (x + 1) φ(n). Proof. (i) and (ii) are obvious. For 1 < a < 1, write f a (x) = x 2 2ax + 1. Then f a (x) = (x a) 2 + (1 a 2 ) > 0 for all x, and is strictly increasing for x 1: hence (iii) and (iv). Also, (x 1) 2 < f a (x) < (x + 1) 2 for all x > 0: hence (v). 1.3 PROPOSITION. Let n 2. Let φ(n) = m and Φ n (x) = m r=0 c rx r. Then: (i) Φ n (x) = x m Φ n (1/x) for x 0, (ii) c m r = c r for 0 r m. Proof. This is trivial for n = 2, so assume n > 2. Write cos(2πk/n) = a k. By (3), ( ) 1 x m Φ n = ( x 2 1 2a k x x + 1 ) = (x 2 2a x 2 k x + 1) = Φ n (x). k E n Hence m r=0 c rx r = m r=0 c rx m r = m r=0 c m rx r for all x 0, so c m r = c r for each r. k E n We now explain how the cyclotomic polynomials provide a factorisation of x n LEMMA. Let n = n 1 d. Then Φ d (x) = {[x e(k/n)] : 1 k n, (k, n) = n 1 }. Proof. By definition, Φ d (x) = r E d [x e(r/d)]. Now r/d = (n 1 r)/n and (r, d) = 1 iff (n 1 r, n) = n 1 ; also, r d iff n 1 r n. Writing k = n 1 r, we see that Φ d (x) equates to the stated product. 2

3 1.5 THEOREM. For all n 1, x n 1 = d n Φ d (x). (4) Proof. Every k such that 1 k n has (k, n) = n 1 for some divisor n 1 of n. When d runs through the divisors of n, so does n 1 = n/d. Hence n Φ d (x) = [x e(k/n)] = x n 1. d n k=1 1.6 COROLLARY. For n 2, {Φd (x) : d n, d > 1} = 1 + x + + x n 1. at x = 1. Proof. In (4), divide both sides by Φ 1 (x) = x 1. The statement follows by continuity It follows at once from (4) that if m divides n, then {Φd (x) : d n, d m} = xn 1 x m 1. (5) A typical deduction (which will be generalised by later results) is: 1.7. For prime p, Φ p k(x) = yp 1 y 1 = 1 + y + + yp 1, where y = x pk 1. In particular, Φ 2 k(x) = x 2k Proof. In (5), take m = p k 1 and n = p k. The only divisor of n that is not a divisor of m is p k. The statement follows. Example. Let p, q be distinct primes. By 1.5, x pq 1 = Φ 1 (x)φ p (x)φ q (x)φ pq (x). With 1.1, this gives In particular, Φ pq (x) = (xpq 1)(x 1) (x p 1)(x q 1). (6) Φ 2p (x) = xp + 1 x + 1 = 1 x + x2 + x p THEOREM. For all n, Φ n (x) has integer coefficients. Proof. Assume that Φ r (x) has integer coefficients for all r < n. By (4), x n 1 = Φ n (x)g n (x), where g n (x) = d n, d<n 3 Φ d (x).

4 Write Φ n (x) = m r=0 c rx r (where m = φ(n)) and g n (x) = n m s=0 a sx s. By the induction hypothesis, each a s is an integer. Also, a 0 = g n (0) = 1. If any of the coefficients c r are not integers, let c k be the first one that is not. Then the coefficient of x k in the product Φ n (x)g n (x) is c 0 a k + c 1 a k c k 1 a 1 c k, which is not an integer. But this product is x n 1, so this is a contradiction. Alternative proof. By the division algorithm in Z[x], since g n (x) is monic, there are polynomials q(x), r(x), with integer coefficients, such that x n 1 = q(x)g n (x) + r(x) and either r = 0 or deg r < deg g n. So g n (x)[φ n (x) q(x)] = r(x), hence Φ n (x) = q(x). Hence Φ n (a) is an integer when a is an integer. Also, for positive integers a and n 2, Φ n (a) Φ n (0) 1 mod a. Note. It can be shown that Φ n (x) is irreducible in Z[x], so that (4) is the complete factorisation of x n 1 in this ring (e.g. [vdw, p ]). This is important in Galois theory, but not for the topics considered here. 1.9 LEMMA. If n is odd, then the elements {n + 2r : r E n } (considered mod 2n) comprise E 2n. Proof. Suppose that (r, n) = 1 and that d divides n + 2r and 2n. Then d is odd, since n + 2r is odd, so d divides n. Hence d 2r, so d r. So d = 1. Hence (n + 2r, 2n) = 1. If n + 2r n + 2s mod 2n, then r s mod n, so the elements n + 2r (r E n ) are distinct mod 2n. Since φ(2n) = φ(n), the statement follows If n 3 is odd, then Φ 2n (x) = Φ n ( x). Proof. Recall that e( 1 + t) = e(t). By the Lemma, 2 Φ 2n (x) = [ ( )] n + 2r x e = 2n r E n r E n Since φ(n) is even, this equals r E n [ x e(r/n)] = Φ n ( x). [ ( r x + e. n)] Alternatively, one can give an induction proof of 1.10, using 1.5. Möbius inversion of (4). The expression in (6) can be extended to general n. Recall that the Möbius function µ is defined by µ(1) = 1, µ(n) = ( 1) k if n is square-free with k prime factors and µ(n) = 0 for other n. The Möbius inversion theorem states: if β n = d n α d for all n 1, then α n = d n β dµ(n/d) for n 1. By applying this to log a n and log b n, we deduce: if a n > 0 and b n = d n a d for all n, then a n = d n bµ(n/d) d inverted as follows: 4 for all n. So (4) can be

5 1.11 THEOREM. For all n 1 and x ±1, Φ n (x) = d n (x d 1) µ(n/d). (7) Proof. This follows in the way just stated for x > 1, since then x d 1 > 0 for each d. The statement can be rewritten in the form Φ n (x)q(x) = P (x), where P (x) and Q(x) are polynomials. Since it holds for x > 1, it holds for all real x. A variant, valid also when x = 1, is as follows: let f n (x) = 1 + x + + x n 1 (n 2) and f 1 (x) = 1. By Möbius inversion of 1.6, with Φ 1 (x) replaced by 1, we have for n 2: Φ n (x) = f d (x) µ(n/d). d n We can now show that once Φ n is known for square-free values of n, it can be derived for other n by a simple substitution THEOREM. Let n = p k 1 1 p k p kr r. Write p 1 p 2... p r = n 0 and n 1 = n/n 0. Then Φ n (x) = Φ n0 (x n 1 ). Proof. The divisors of n 0 are the numbers d = r j=1 pε j j, with each ε j in {0, 1}. Denote this set of divisors by d. Then Φ n0 (x) = d D(x d 1) µ(n 0/d). The divisors e of n that have µ(n/e) 0 are the numbers r j=1 pa j j with a j k j 1 for each j. These are exactly the numbers n 1 d (d D), and n/n 1 d = n 0 /d. Hence Φ n (x) = d D(x n 1d 1) µ(n 0/d) = Φ n0 (x n 1 ) COROLLARY. If p is prime and does not divide m, then Φ mp k(x) = Φ mp k 1(x p ) = Φ mp (x pk 1 ). Proof. Let mp k 1 = n, mp k = n and write n = n 0 n 1 and n = n 0n 1 as in Then n 0 = n 0 and n 1 = pn 1. By 1.12, Φ n (x) = Φ n (x p ). The second equality follows by iteration, or similarly. In particular, Φ 4m (x) = Φ 2m (x 2 ) (and so is an even function) for any m. Example. Let p, q be distinct primes. Then Φ p q(x) = Φ k pq (y), where y = x pk 1. In particular, Φ 2p k(x) = (y p + 1)/(y + 1). Also, Φ 2 q(x) = (y q + 1)/(y + 1), where y = x k 2k 1. 5

6 With the results now at our disposal, we can write down Φ n (x) for a selection of values of n: n Φ n (x) n Φ n (x) 1 x 1 2 x x 2 + x x x 4 + x 3 + x 2 + x x 2 x x x 6 + x x 4 x 3 + x 2 x x 4 x x x 6 x x 8 x 6 + x 4 x x 24 x Example. The factorisation of x 6 1 given by (4) is (x 1)(x + 1)(x 2 + x + 1)(x 2 x + 1). To determine Φ n (x) when n has two or more odd prime factors, one has to calculate coefficients, as in the following example. Example. Determine Φ 15 (x). Denote it by 8 r=0 c rx r. By (6), (c 8 x 8 + c 7 x c 1 x + c 0 )(x 3 1)(x 5 1) = (x 15 1)(x 1). Equating coefficients, we find c 0 = 1, c 1 = 1, c 2 = 0, c 3 = c 0 = 1, c 4 = c 1 = 1. By 1.3, c 8 r = c r. Hence Φ 15 (x) = x 8 x 7 + x 5 x 4 + x 3 x + 1. Another deduction from 1.11 is the following identity, which turns out to be very useful: 1.14 PROPOSITION. If p is prime, p does not divide m and k 1, then for all x, Φ mp k(x) = Φ m(x pk ) Φ m (x pk 1 ). Proof. Write mp k = n. The divisors j of n with µ(n/j) 0 are the numbers dp k and dp k 1 for divisors d of m. So Φ n (x) = P 1 (x)p 2 (x), where P 1 (x) = d m(x dpk 1) µ(m/d) = Φ m (x pk ), P 2 (x) = d m(x dpk 1 1) µ(pm/d) = 1 Φ m (x pk 1 ), since µ(pm/d) = µ(m/d). The statement is a polynomial identity, valid for all real x. 6

7 1.15. For n 2, { p if n = p Φ n (1) = k (p prime, k 1), 1 otherwise. Proof. The statement for n = p k is clear from 1.7. Otherwise, n can be expressed as mp k with p m and m > 1, k 1. By 1.14, Φ n (1) = Φ m(1) Φ m (1) = 1. We sketch two alternative proofs, both illuminating. Proof 2. By 1.12, it is enough to show that Φ n (1) = 1 for n of the form p 1 p 2... p k, where k 2. Assume this for lower values of k. By 1.6, d n,d>1 Φ d(1) = n. Now Φ pj (1) = p j, and by hypothesis, these are the only proper divisors d of n with Φ d (1) 1. So the stated product is p 1 p 2... p k Φ n (1) = nφ n (1), hence Φ n (1) = 1. Proof 3. By the variant of 1.11, Φ n (1) = d n dµ(n/d), so log Φ n (1) = d n µ(n/d) log d. A well-known convolution identity equates this to the von Mangoldt function Λ(n), defined by: Λ(p k ) = log p for prime p and Λ(n) = 0 for other n. Since Φ n (x) is strictly increasing for x 1, it follows that Φ n (x) > 1 for all x > 1. Recall that Φ n (x) = x m Φ n (1/x). By differentiation, one finds that Φ n(1) = 1 2 mφ n(1) For integers a 1 and n 2, Φ n (a) is odd unless n = 2 k and a is odd. Proof. If a is even, then Φ n (a) is odd, since it is congruent to 1 mod a. If a is odd, then Φ n (a) Φ n (1) mod 2. By 1.15, Φ n (1) is odd except when n = 2 k. We now restate identity (7) in a more specific form, concentrating on square-free n: let n = p 1 p 2... p k, where p 1 < p 2 <... < p k, with k 2 and p 1 3 (for p 1 = 2, then apply 1.10). Let D, E respectively be the set of products of even and of odd numbers of the p j. Then D and E each have 2 k 1 members, because k r=0 ( 1)r( k r) = (1 1) k = 0. Let P (x) = d D (x d 1), Q(x) = e E(x e 1). Then Φ n (x) equals P (x)/q(x) if k is even and Q(x)/P (x) if k is odd. We can deduce some information about the first (and last) few coefficients: Let n = p 1 p 2... p k as above, and let Φ n (x) = m r=0 c rx r. Then c 0 = 1 and: 7

8 (i) if k is even, then c 1 = 1, c r = 0 for 2 r < p 1, c p1 = 1, c p1 +1 = 1; (ii) if k is odd, then c r = 1 for 1 r < p 1 and c p1 = c p1 +1 = 0. Proof. Let D = D \ {1}. The smallest element of D is p 1 p 2, so P (x) = (x 1) (x d 1) = 1 x x p1p2 + p(x), d D where p(x) is composed of higher powers of x. The two smallest elements of E are p 1 and p 2, so Q(x) = 1 x p 1 x p 2 + q(x), where q(x) is composed of higher powers. If k is even, then Φ n (x)q(x) = P (x). By equating coefficients, we obtain the values stated for r < p 1, also c p1 = c 0 = 1 and c p1 +1 = c 1 = 1. If k is odd, then Φ n (x)p (x) = Q(x). For 1 r < p 1, we see that c r c r 1 = 0, so c r = c 0 = 1. Also, c p1 c p1 1 = 1, so c p1 = 0, and c p1 +1 c p1 = 0. Recall from 1.3 that c m r = c r. So, for example, for even k, Φ n (x) is of the form x m x m 1 + x m p 1 x m p1 1 + x p1+1 + x p 1 x + 1. The results on c 1 can be summarised in the form c 1 = µ(n), since 1.12 shows that c 1 = 0 when n is not square-free. Also, it follows from the original definition that c m 1 = k E n e(k/n), so our statement is equivalent to k E n e(k/n) = µ(n), a special case of the Ramanujan sum. Note. From the examples seen so far, one might be led to suppose that the coefficients in Φ n (x) are always 1, 0 or 1. However, none of these examples have more than two distinct prime factors. For n = = 105, by solving for coefficients as above, one finds that c 7 = 2. Inequalities for Φ n (x). Recall from 1.2 that for n 3, Φ n (x) lies between (x 1) φ(n) and (x + 1) φ(n). With 1.14, this gives the following inequality, which looks rather special, but is just what is needed for the proof of Zsigmondy s theorem in Section Let n = mp k, where p is prime, k 1 and p does not divide m. Let x 2 and y = x pk 1. Then Further, Φ n (x) > [y p 2 (y 1)] φ(m). ( ) y p φ(m) 1 < Φ n (x) < y ( ) y p φ(m) + 1. y 1

9 Proof. By 1.14, Φ n (x) = Φ m(y p ) Φ m (y). Now (y 1) φ(m) < Φ m (y) < (y + 1) φ(m), and similarly for y p. The first statement follows. The second statement is derived by noting that y p 1 y p 2 (y 2 1). We now investigate the comparison between Φ n (x) and x m, where m = φ(n). (Any readers who are more interested in the number-theoretic applications can omit this and move on to section 2.) By 1.2, Φ n (x) (x + 1) m. We will show that (for x 2) this bound can be improved to x m or [x/(x 1)]x m, depending on whether the number of prime factors of n is even or odd. Since Φ n (x)/x m = Φ n (1/x), the statement Φ n (x) < x m for x 2 is equivalent to Φ n (x) < 1 for 0 < x 1, and it is rather neater to prove the statement in this second 2 form. We will apply the following simple Lemma to the P (x) and Q(x) introduced above LEMMA. If 1 n 1 < n 2 <... < n k, where k 2, then for 0 < x < 1, (1 x n 1 )(1 x n 2 )... (1 x n k ) > 1 xn 1 1 x. Proof. It is elementary that if 0 a j < 1 for each j, then (1 a 1 )(1 a 2 )... (1 a k ) > 1 (a 1 + a a k ). Hence the stated product is greater than 1 x n 1 x n 2 x n k > 1 x n 1 (1 + x + x 2 + ) = 1 xn 1 1 x PROPOSITION. Suppose that n has k distinct prime factors. Write φ(n) = m. If k is even, then 1 x < Φ n (x) < 1 for 0 < x 1 2 and (x 1)x m 1 < Φ n (x) < x m for x 2. If k is odd, then 1 < Φ n (x) < 1/(1 x) for 0 < x 1 2 and x m < Φ n (x) < xm+1 x 1 for x 2. Proof. We prove the statements for 0 < x 1 : the statements for x 2 then follow, 2 by the identity Φ n (x) = x m Φ n (1/x). Also, if we can prove the statements for square-free n, the statements for other n then follow, by 1.12 (in fact, 1 x can be replaced by 1 x n 1, in the notation of 1.12). So let n = p 1 p 2... p k, where p 1 < p 2 <... < p k. If k = 1, so n = p (say), then Φ p (x) = (1 x p )/(1 x), and it is clear that the stated inequalities hold (and that the upper inequality is asymptoticaly optimal). So we assume that k 2. Let D, E, P (x), Q(x) be as in 1.17, but we now write P (x) in the form d D (1 xd ). Recall that Φ n (x) is P (x)/q(x) if k is even and Q(x)/P (x) if k is odd. The stated inequalities follow if we can show that 1 x < P (x)/q(x) < 1 for 0 < x

10 Since 1 D, we have P (x) < 1 x. Meanwhile, the smallest member of E is p 1 2, so, by the Lemma, Hence for 0 < x 1 2, Q(x) > 1 x2 1 x. so P (x) < Q(x). Q(x) P (x) > x x2 1 x = x 2x2 1 x 0, Lemma, Also, Q(x) < 1 x p 1. Apart from 1, the smallest member of D is p 1 p 2, so, by the ( P (x) > (1 x) 1 xp 1p 2 ) = 1 x x p 1p 2. 1 x So we will have P (x) > (1 x)q(x) if x p 1p 2 < x p 1 (1 x), or x p 1(p 2 1) < 1 x. Now p 1 (p 2 1) 4, so it is sufficient if x 4 < 1 x, which is satisfied when 0 < x 1 2 when x 0.724). (and in fact Note. For k = 2, it is easily seen from (6) that 1 x < Φ n (x) < 1 for all x in (0, 1). However, in general the inequalities comparing Φ n (x) with 1 do not extend to (0, 1). Indeed, for any k 2 (even or odd), we have Φ n (1) = 1 and Φ n(1) > 0, so that Φ n (x) < 1 on some interval (1 δ, 1). Meanwhile, for n = 210 = , calculation (from (7), assisted by 1.10) shows that Φ n (0.9) > 1. In particular, 2 m 1 < Φ n (2) < 2 m if k is even and 2 m < Φ n (2) < 2 m+1 if k is odd. One can deduce a fact that is quoted sometimes: Φ n (2) > n for all n except 1 and 6. This is not really a natural comparison, and we will leave it aside. However, we include the following weaker statement, which can serve as an alternative to 1.18 for the purpose of Zsigmondy s theorem: Let n 2 and let p be the largest prime factor of n. Then Φ n (2) p, with strict inequality except when n = 6. Proof. It is enough to prove the statement for square-free n. For then if n = n 0 n 1 as in 1.12 (with n 1 > 1), it follows that Φ n (2) = Φ n0 (2 n 1 ) > Φ n0 (2) p, since Φ n0 (x) is strictly increasing. If n = p (prime), then Φ p (2) = 2 p 1 > p. So assume that n = p 1 p 2... p k, with k 2 and p the largest p j. Then φ(n) p 1, so Φ n (2) > 2 p 2, which is greater than p if p 5. This leaves only the case n = 2 3 = 6, and we note that Φ 6 (2) = 3. 10

11 2. Prime factorisation of Φ n (a) and number-theoretic applications We now consider the prime factorisaton of Φ n (a). We show that it is closely connected to the order of a modulo these primes. If (a, p) = 1, the order of a mod p, denoted by ord p (a), is the smallest n 1 such that a n 1 mod p. Recall: if ord p (a) = n and r is any positive integer such that a r 1 mod p, then r is a multiple of n. By Fermat s little theorem, if p is prime and ord p (a) = n, then n divides p 1, so p 1 mod n.. It is not hard to determine orders using modular arithmetic. As a background to the following results, we record the values of ord p (2) and ord p (3) for primes p 41: p ord p (2) ord p (3) Example. There is no prime p such that ord p (2) = 6. For then p would have to divide = 63 = However, ord 3 (2) = 2 and ord 7 (2) = 3. We shall see, as an application of cyclotomic polynomials, that this example is quite exceptional (Zsigmondy s theorem, below). We now start describing the connection. Note first that prime factors of Φ n (a) do not divide a, since Φ n (a) 1 mod a If p is prime and ord p (a) = n, then p divides Φ n (a). Proof. Then p divides a n 1 and does not divide a d 1 (so does not divide Φ d (a)) for any d < n. Since a n 1 = d n Φ d(a), it follows that p divides Φ n (a). As a first application, we have the following estimate: 2.2. Let a 2 and let N(n, a) be the number of primes p such that ord p (a) = n. Then N(n, a) φ(n) log a + 1. log(n + 1) Proof. Let these primes be p 1, p 2,..., p k. Then p j n + 1 for each j, and p 1 p 2... p k Φ n (a). By 1.20, Φ n (a) 2a φ(n). So (n + 1) k 2a φ(n), hence k log(n + 1) φ(n) log a + 1. Without cyclotomic polynomials, one would obtain a similar estimate with n replacing φ(n), since clearly p 1 p 2... p k a n 1. Primes p with ord p (a) = n are called Zsigmondy factors of Φ n (a) (they are also called primitive prime factors of a n 1). Note that they satisfy p 1 mod n. We now show that non-zsigmondy prime factors of Φ n (a) can occur. We will make repeated use of the following elementary lemma on congruences; 11

12 2.3 LEMMA. Suppose that p is prime and a 1 mod p. Then: (i) (a p 1)/(a 1) = 1 + a + + a p 1 is a multiple of p; (ii) a p 1 mod p 2 ; (iii) If p 3, then 1 + a + + a p 1 p mod p 2 (so is not a multiple of p 2 ). Proof. (i), (ii). 1 + a + + a p 1 p mod p, so is a multiple of p. Also, a p 1 = (a 1)(1 + a + + a p 1 ), in which both factors are multiples of p. is even), (iii) Let a = kp + 1. By the binomial theorem, a r 1 + rkp mod p 2. Hence (since p 1 p 1 p 1 a r p + kp r = p + 1k(p 2 1)p2 p mod p 2. r=0 r=0 Recall that Φ p (a) = 1 + a + + a p 1 for prime p. So if a 1 mod p (so that ord p (a) = 1), then p itself is a factor (but not a Zsigmondy factor) of Φ p (a). This remark can be generalised, as follows: 2.4 PROPOSITION. Suppose that p is prime and ord p (a) = m. Let n = mp k, where k 1. Then p divides Φ n (a). However, if n 3, then p 2 does not divide Φ n (a). Proof. We use (7). Let D be the set of divisors d of m with µ(m/d) 0. The divisors j of n with µ(n/j) 0 are dp k and dp k 1 for d D. Now (m, p t ) = 1, so m divides dp t only when d = m. Hence when d < m, p does not divide a dpt 1. By (7), Φ n (a) is of the form where b = a mpk 1 b p 1 b 1 d D, d<m f d (a) g d (a), and the terms f d (a), g d (a) are not multiples of p. Now b 1 mod p, so by 2.3(i), (b p 1)/(b 1) is a multiple of p. So p divides Φ n (a). By 2.3(iii), if p 3, then p 2 does not divide (b p 1)/(b 1), so p 2 does not divide Φ n (a). This statement also holds when p = 2 and n > 2: then m = 1, so n = 2 k for some k 2, and Φ n (a) = a 2k Now r 2 1 mod 4 for odd r, so Φ n (a) 2 mod 4. Note 1. For p, m, n as in 2.4, m divides p 1, so p is the largest prime factor of n. Note 2. The second statement in 2.4 fails for n = 2 (so p = 2, m = 1): Φ 2 (7) = 8 = 2 3. Note 3. Under the hypotheses of 2.4, p divides Φ mp r(a) for 0 r k, so p k+1 divides a n 1 (however, this can be proved quite easily without cyclotomic polynomials). Note 4. By a well-known result in number theory, if m divides p 1, then the condition 12

13 ord p (a) = m is satisfied by exactly φ(m) values of a mod p. For example, ord 5 (a) = 4 if a is congruent to 2 or 3 mod 5. We now establish a very striking fact: non-zsigmondy prime factors of Φ n (a) only occur in the way described in 2.4, so that for a given n, there is at most one such factor, the largest prime factor of n. Consider first the special case n = p. Let q be a non-zsigmondy prime factor of Φ p (a). Then a p 1 mod q, so ord q (a) divides p. By assumption, it is not p, so it is 1, hence a 1 mod q. Hence Φ p (a) = 1 + a + + a p 1 p mod q. Since Φ p (a) is a multiple of q, it follows that q = p: the only possible non-zsigmondy factor of Φ p (a) is p itself, and it is a factor exactly when a 1 mod p. This reasoning can be adapted without too much trouble to the general case. The neat version given here is due to Roitman [Roi]. 2.5 THEOREM. Let n 2, a 2. Let p be the largest prime factor of n, and let n = mp k. Then: (i) p is a prime factor of Φ n (a) iff ord p (a) = m (in which case m divides p 1); (ii) all other prime factors q of Φ n (a) have ord q (a) = n (hence q 1 mod n). Proof. We prove the statements together. Let q be a prime factor of Φ n (a) with ord q (a) = s < n. Then a n 1 mod q, so s divides n. Also, s divides q 1. Let r be a prime factor of n/s, and write n/r = h. Then s divides h, so a h 1 mod q. Write a h = c. By (5), Φ n (a) divides a n 1 a h 1 = cr 1 c 1 = 1 + c + + cr 1, which is congruent to r mod q, since c 1 mod q. But it is a multiple of q, so r = q. So q is the only prime factor of n/s, hence n = sq k for some k 1. Since s divides q 1, this shows that q is the largest prime factor of n, so q = p and s = m. So p is the only possible non-zsigmondy factor of Φ n (a), and if it is one, then ord p (a) = m. The converse was shown in 2.4. Example. Since Φ 4 (a) = a 2 + 1, the case n = 4 in 2.5 reproduces the well-known fact that any odd prime factor of a is congruent to 1 mod COROLLARY. If q is a prime factor of Φ n (a), then the following statements are equivalent: (i) ord q (a) = n, (ii) q does not divide n, (iii) q 1 mod n. 2.7 COROLLARY. For k 1, every prime factor q of Φ n (kn) is congruent to 1 mod n. 13

14 Proof. Since Φ n (kn) 1 mod kn, q does not divide n. By 2.6, q 1 mod n. This corollary has a pleasant application to prime numbers: 2.8. For every n 2, there are infinitely many primes congruent to 1 mod n. Proof. First, let p 1 be a prime factor of Φ n (n). Then p 1 1 mod n. Having found primes p 1, p 2,..., p r congruent to 1 mod n, let N = np 1 p 2... p r, and choose a prime factor q of Φ n (N). Then q 1 mod n, and q N, so q p j for each j. Probably the most important application of cyclotomic polynomials is the next result, usually known as Zsigmondy s theorem, though in fact it was first proved by Bang in THEOREM. For every a 2 and n 3 except the case a = 2, n = 6, there exists at least one prime q such that ord q (a) = n. Equivalently, there is a Zsigmondy factor of Φ n (a). Proof 1, using Assume, for a particular a and n, that there is no such q. Let p be the largest prime factor of n. By 2.5 and 2.4, p is the only prime factor of Φ n (a) and p 2 does not divide Φ n (a), so in fact Φ n (a) = p. In 1.21, we saw that if a = 2, this only occurs when n = 6. Since Φ n is strictly increasing, it never occurs with a 3. Proof 2, using Assume there is no such q. Again, it follows that Φ n (a) = p; further, n = p k m, where m divides p 1. If p = 2, then n = 2 k, where k 2, so Φ n (a) = a 2k > 2, a contradiction. Now suppose that p 3. By 1.18, Φ n (a) > b p 2, where b = a pk 1. So we require b p 2 < p. For p 5, this does not occur for any b 2. For p = 3, it occurs only for b = 2, which implies that a = 2 and k = 1. Also m is 1 or 2, so n is 3 or 6. However, Φ 3 (2) = 7, which is not a factor of 3 (or we can just note that ord 7 (2) = 3). So a = 2, n = 6 is the only case in which there is no Zsigmondy factor. Note on the case n = 2: This case is easily handled. To have ord q (a) = 2 for some prime q > 2, we require q to divide a 2 1 = (a + 1)(a 1) but not a 1, hence q must divide a + 1. Clearly, such a q exists provided that a is not of the form 2 k 1. Zsigmondy (in 1892) established a more general version of the theorem in which a n 1 is replaced by a n b n. Let us call a number pure (not a standard term) if it is of the form mp k, where p is prime and m divides p 1, and impure otherwise. Clearly, if n is impure, then for all a, 14

15 every prime factor of Φ n (a) is a Zsigmondy factor. Many numbers are impure, for example any of the form p j q k, where p < q and p j does not divide q 1 (e.g. 12, 15, 28, 36). Some examples of pure numbers are those of the form p k, 2p k, 4.5 k. For n = p k, the condition ord p (a) = m equates to a 1 mod p, and for n = 2p k, it equates to a 1 mod p. Rather simpler proofs of 2.4 and 2.5 can be given for these two cases. We illustrate these facts by the first few values of Φ 6 (a) = a 2 a + 1. In accordance with the remarks above, 3 is a prime factor exactly when a 2 mod 3, and all other prime factors are congruent to 1 mod 6. a Φ 6 (a) a Φ 6 (a) a Φ 6 (a) = = = = = 7 19 For further illustration, we list the values of Φ n (2) and φ n (3) (in factorised form) for selected small values of n. We record the largest prime factor of n, denoted as before by p. n p Φ n (2) Φ n (3) n p Φ n (2) Φ n (3) Note from the example Φ 5 (3) = 11 2 that Zsigmondy factors, unlike non-zsigmondy factors, can appear squared. Example. To find Φ 48 (2) and factorise it: By 1.12, Φ 48 (x) = Φ 6 (x 8 ) = x 16 x 8 +1, hence Φ 48 (2) = Any prime factors must be congruent to 1 mod 48. The first candidate is 97, and we find that = Write Φ n(a) for Φ n (a) with any non-zsigmondy factors removed. In other words, if n = mp k 3 and p divides Φ n (a), then Φ n(a) = Φ n (a)/p, otherwise Φ n(a) = Φ n (a). This is not a new polynomial, since (for a given n), the division by p occurs only for certain values of a. All prime factors q of Φ n(a) have ord q (a) = n, so that q 1 mod n. Hence Φ n(a) 1 mod n, and if m n, Φ m(a) and Φ n(a) have no prime factors in common, so are coprime. 15

16 Application to pseudoprimes Fix a > 1. We write F (a) for the set of positive integers m such that a m 1 1 mod m. By Fermat s theorem, F (a) includes all primes that are not divisors of a. Note that if m F (a), then (a, m) = 1. A composite member of F (a) is called an a-pseudoprime. We denote the set of such numbers by P S(a). If, for some n, we know that m divides a n 1 and m 1 mod n, then m F (a), since a n 1 mod m and m 1 is a multiple of n. These conditions are satisfied by m = Φ n(a) (and its divisors), so we have: For all a 2 and n 2, we have Φ n(a) F (a), so is in P S(a) if it is composite. The same applies to any composite divisor of Φ n(a). In particular, if p is prime and does not divide a 1, then by 2.5, Φ p(a) = Φ p (a) = (a p 1)/(a 1), and this number belongs to F (a). Similarly for Φ 2p (a) = (a p + 1)/(a + 1) if p does not divide a + 1. These facts were observed by Cipolla in For readers familiar with the notion, we mention that Φ n(a) is actually a strong a- pseudoprime if composite, since a has the same order n modulo each of its prime factors. The following further result was also noted by Cipolla in the case n = p If n 3 is odd, then Φ n(a)φ 2n(a) P S(a) except in the case a = 2, n = 3. Proof. By (4), the stated product divides a 2n 1. Now Φ n(a) 1 mod n and is odd (by 1.16), so Φ n(a) 1 mod 2n. By 2.9, both factors are greater than 1. Example. The cases n = 5, 7, 9 give the 2-pseudoprimes 11 31, , Of course, it follows that there are infinitely many a-pseudoprimes for each a 2. We now describe a refinement in which the number of prime factors is specified. It was first proved for the case a = 2 by Erdös in With a fixed, write m(p) = ord p (a). Consider a square-free number n = p 1 p 2... p k. Write n/p j = r j. Clearly, n P S(a) if and only if a n 1 1 mod p j, equivalently, m(p j ) divides n 1, for each j. Further, this condition is equivalent to m(p j ) dividing r j 1 for each j, since m(p j ) divides p j 1 and n 1 = p j r j 1 = (p j 1)r j + (r j 1) PROPOSITION. For every a 2 and k 2, there are infinitely many square-free a-pseudoprimes with k prime factors. 16

17 Proof. First we prove the case k = 2. Let p ( 3) be a prime not dividing a 1 or a + 1. Choose prime factors q 1 of Φ p (a) and q 2 of Φ 2p (a). By 2.5, q 1 1 mod p, hence also mod 2p, and q 2 1 mod 2p. Further, q 1 q 2 divides a 2p 1, so q 1 q 2 P S(a). Since m(q 1 ) = p, a different such pseudoprime is obtained for each p. Now assume the statement for a particular k. Take n = q 1 q 2... q k P S(a). Then m(q j ) divides n 1 for each j. By Zsigmondy s theorem, there is a prime q with m(q) = n 1. Then n 1 divides q 1, so m(q j ) divides qn 1 = q(n 1) + (q 1). So by the criterion stated above, qn P S(a). Also, q > n, from which it is easily seen that a different choice of n will generate a different qn. This proves the statement for k + 1. Application to Wieferich primes Fix a 2. If p is prime and not a divisor of a, then by Fermat s little theorem, a p 1 1 mod p. How many primes satisfy the stronger condition that a p 1 1 mod p 2? This is equivalent to p 2 being a-pseudoprime. For a = 2, such primes are known as Wieferich primes, and they do not occur frequently: computation has shown that there are only two less than 10 9, namely 1093 and For a = 3, it is easily seen that the condition is satisfied by p = 11, but the second example is 1,006,003 [Rib, p. 347]. It has not been proved that there are infinitely many Wieferich primes, or even (despite their apparent preponderance) that there are infinitely many non-wieferich primes. Without offering any progress on these questions, we show how they can be rephrased in terms of cyclotomic polynomials. We need the following lemma, which is well-known in the context of pseudoprimes: 2.13 LEMMA. If p is prime, a p 1 1 mod p 2 and ord p (a) = n, then a n 1 mod p 2. Proof. Then a a p mod p 2, so a n a pn mod p 2. But a n 1 mod p, so by Lemma 2.3, a pn 1 mod p Suppose that p is prime and not a divisor of a. Let ord p (a) = n. Then the following statements are equivalent: (i) a p 1 1 mod p 2, (ii) p 2 divides Φ n (a). Proof. If (ii) holds, then a n 1 mod p 2. But n divides p 1, so (i) holds. Conversely, assume (i). Then, by the Lemma, p 2 divides a n 1. Since ord p (a) = n, p does not divide Φ d (a) for any d < n. Since a n 1 = d n Φ d(a), it follows that p 2 divides Φ n (a). 17

18 Hence all the Zsigmondy factors of Φ n (2) (for any n) that do not appear squared are non-wieferich primes. From our table of values, we see that these include, for example, 11, 13, 17, 19, 23, 31, 41, 43, 73, 89, 127, 151, 257. To show that there are infinitely many non-wieferich primes, it would be sufficient to show that there are infinitely many values of n for which such factors occur. References [Rib] P. Ribenboim, The New Book of Prime Number Records, Springer (1995). [Roi] Moishe Roitman, On Zsigmondy primes, Proc. Amer. Math. Soc. 125 (1997) [vdw] B.L. van der Waerden, Modern Algebra, vol. 1, Frederick Ungar (1949). 18

### Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013

Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of

### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

### Die ganzen zahlen hat Gott gemacht

Die ganzen zahlen hat Gott gemacht Polynomials with integer values B.Sury A quote attributed to the famous mathematician L.Kronecker is Die Ganzen Zahlen hat Gott gemacht, alles andere ist Menschenwerk.

### U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra

U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory

### Introduction to Finite Fields (cont.)

Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number

### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY

January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.

### Notes on Factoring. MA 206 Kurt Bryan

The General Approach Notes on Factoring MA 26 Kurt Bryan Suppose I hand you n, a 2 digit integer and tell you that n is composite, with smallest prime factor around 5 digits. Finding a nontrivial factor

### MOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu

Integer Polynomials June 9, 007 Yufei Zhao yufeiz@mit.edu We will use Z[x] to denote the ring of polynomials with integer coefficients. We begin by summarizing some of the common approaches used in dealing

### Primality - Factorization

Primality - Factorization Christophe Ritzenthaler November 9, 2009 1 Prime and factorization Definition 1.1. An integer p > 1 is called a prime number (nombre premier) if it has only 1 and p as divisors.

### 7. Some irreducible polynomials

7. Some irreducible polynomials 7.1 Irreducibles over a finite field 7.2 Worked examples Linear factors x α of a polynomial P (x) with coefficients in a field k correspond precisely to roots α k [1] of

### The Division Algorithm for Polynomials Handout Monday March 5, 2012

The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following,

### Winter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com

Polynomials Alexander Remorov alexanderrem@gmail.com Warm-up Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)).

### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

### minimal polyonomial Example

Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We

### FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization

### Carmichael numbers and pseudoprimes

Carmichael numbers and pseudoprimes Notes by G.J.O. Jameson Introduction Recall that Fermat s little theorem says that if p is prime and a is not a multiple of p, then a p 1 1 mod p. This theorem gives

### Unique Factorization

Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon

### it is easy to see that α = a

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore

### 9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11.

9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n-1 x n-1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role

### 3 Factorisation into irreducibles

3 Factorisation into irreducibles Consider the factorisation of a non-zero, non-invertible integer n as a product of primes: n = p 1 p t. If you insist that primes should be positive then, since n could

### 11 Ideals. 11.1 Revisiting Z

11 Ideals The presentation here is somewhat different than the text. In particular, the sections do not match up. We have seen issues with the failure of unique factorization already, e.g., Z[ 5] = O Q(

### Cyclotomic Extensions

Chapter 7 Cyclotomic Extensions A cyclotomic extension Q(ζ n ) of the rationals is formed by adjoining a primitive n th root of unity ζ n. In this chapter, we will find an integral basis and calculate

### Homework until Test #2

MATH31: Number Theory Homework until Test # Philipp BRAUN Section 3.1 page 43, 1. It has been conjectured that there are infinitely many primes of the form n. Exhibit five such primes. Solution. Five such

### Quotient Rings and Field Extensions

Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

### Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients

DOI: 10.2478/auom-2014-0007 An. Şt. Univ. Ovidius Constanţa Vol. 221),2014, 73 84 Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients Anca

### 3 1. Note that all cubes solve it; therefore, there are no more

Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if

### On the number-theoretic functions ν(n) and Ω(n)

ACTA ARITHMETICA LXXVIII.1 (1996) On the number-theoretic functions ν(n) and Ω(n) by Jiahai Kan (Nanjing) 1. Introduction. Let d(n) denote the divisor function, ν(n) the number of distinct prime factors,

### calculating the result modulo 3, as follows: p(0) = 0 3 + 0 + 1 = 1 0,

Homework #02, due 1/27/10 = 9.4.1, 9.4.2, 9.4.5, 9.4.6, 9.4.7. Additional problems recommended for study: (9.4.3), 9.4.4, 9.4.9, 9.4.11, 9.4.13, (9.4.14), 9.4.17 9.4.1 Determine whether the following polynomials

### Factoring Polynomials

Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent

### Settling a Question about Pythagorean Triples

Settling a Question about Pythagorean Triples TOM VERHOEFF Department of Mathematics and Computing Science Eindhoven University of Technology P.O. Box 513, 5600 MB Eindhoven, The Netherlands E-Mail address:

### k, then n = p2α 1 1 pα k

Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

### Alex, I will take congruent numbers for one million dollars please

Alex, I will take congruent numbers for one million dollars please Jim L. Brown The Ohio State University Columbus, OH 4310 jimlb@math.ohio-state.edu One of the most alluring aspectives of number theory

### Revised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m)

Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations.

### POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS

POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS RUSS WOODROOFE 1. Unique Factorization Domains Throughout the following, we think of R as sitting inside R[x] as the constant polynomials (of degree 0).

### Finding Carmichael numbers

Finding Carmichael numbers Notes by G.J.O. Jameson Introduction Recall that Fermat s little theorem says that if p is prime and a is not a multiple of p, then a p 1 1 mod p. This theorem gives a possible

### The Dirichlet Unit Theorem

Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if

### CONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12

CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.

### SOLVING POLYNOMIAL EQUATIONS

C SOLVING POLYNOMIAL EQUATIONS We will assume in this appendix that you know how to divide polynomials using long division and synthetic division. If you need to review those techniques, refer to an algebra

### a 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)

ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x

### ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS

ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for

### Factorization Algorithms for Polynomials over Finite Fields

Degree Project Factorization Algorithms for Polynomials over Finite Fields Sajid Hanif, Muhammad Imran 2011-05-03 Subject: Mathematics Level: Master Course code: 4MA11E Abstract Integer factorization is

### H/wk 13, Solutions to selected problems

H/wk 13, Solutions to selected problems Ch. 4.1, Problem 5 (a) Find the number of roots of x x in Z 4, Z Z, any integral domain, Z 6. (b) Find a commutative ring in which x x has infinitely many roots.

### Prime Numbers and Irreducible Polynomials

Prime Numbers and Irreducible Polynomials M. Ram Murty The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry.

### Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2)

Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Kevin Broughan University of Waikato, Hamilton, New Zealand May 13, 2010 Remainder and Factor Theorem 15 Definition of factor If f (x)

### On the largest prime factor of x 2 1

On the largest prime factor of x 2 1 Florian Luca and Filip Najman Abstract In this paper, we find all integers x such that x 2 1 has only prime factors smaller than 100. This gives some interesting numerical

### THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS

THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear

### QUADRATIC RECIPROCITY IN CHARACTERISTIC 2

QUADRATIC RECIPROCITY IN CHARACTERISTIC 2 KEITH CONRAD 1. Introduction Let F be a finite field. When F has odd characteristic, the quadratic reciprocity law in F[T ] (see [4, Section 3.2.2] or [5]) lets

### Lecture 13 - Basic Number Theory.

Lecture 13 - Basic Number Theory. Boaz Barak March 22, 2010 Divisibility and primes Unless mentioned otherwise throughout this lecture all numbers are non-negative integers. We say that A divides B, denoted

### Number Theory Hungarian Style. Cameron Byerley s interpretation of Csaba Szabó s lectures

Number Theory Hungarian Style Cameron Byerley s interpretation of Csaba Szabó s lectures August 20, 2005 2 0.1 introduction Number theory is a beautiful subject and even cooler when you learn about it

### Integer roots of quadratic and cubic polynomials with integer coefficients

Integer roots of quadratic and cubic polynomials with integer coefficients Konstantine Zelator Mathematics, Computer Science and Statistics 212 Ben Franklin Hall Bloomsburg University 400 East Second Street

### Some Polynomial Theorems. John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.

Some Polynomial Theorems by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.com This paper contains a collection of 31 theorems, lemmas,

### FACTORING IN QUADRATIC FIELDS. 1. Introduction. This is called a quadratic field and it has degree 2 over Q. Similarly, set

FACTORING IN QUADRATIC FIELDS KEITH CONRAD For a squarefree integer d other than 1, let 1. Introduction K = Q[ d] = {x + y d : x, y Q}. This is called a quadratic field and it has degree 2 over Q. Similarly,

### 2 Polynomials over a field

2 Polynomials over a field A polynomial over a field F is a sequence (a 0, a 1, a 2,, a n, ) where a i F i with a i = 0 from some point on a i is called the i th coefficient of f We define three special

### The van Hoeij Algorithm for Factoring Polynomials

The van Hoeij Algorithm for Factoring Polynomials Jürgen Klüners Abstract In this survey we report about a new algorithm for factoring polynomials due to Mark van Hoeij. The main idea is that the combinatorial

### SUM OF TWO SQUARES JAHNAVI BHASKAR

SUM OF TWO SQUARES JAHNAVI BHASKAR Abstract. I will investigate which numbers can be written as the sum of two squares and in how many ways, providing enough basic number theory so even the unacquainted

### The last three chapters introduced three major proof techniques: direct,

CHAPTER 7 Proving Non-Conditional Statements The last three chapters introduced three major proof techniques: direct, contrapositive and contradiction. These three techniques are used to prove statements

### 8 Divisibility and prime numbers

8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express

### I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called

### Congruence properties of binary partition functions

Congruence properties of binary partition functions Katherine Anders, Melissa Dennison, Jennifer Weber Lansing and Bruce Reznick Abstract. Let A be a finite subset of N containing 0, and let f(n) denote

### CONTRIBUTIONS TO ZERO SUM PROBLEMS

CONTRIBUTIONS TO ZERO SUM PROBLEMS S. D. ADHIKARI, Y. G. CHEN, J. B. FRIEDLANDER, S. V. KONYAGIN AND F. PAPPALARDI Abstract. A prototype of zero sum theorems, the well known theorem of Erdős, Ginzburg

### Factoring of Prime Ideals in Extensions

Chapter 4 Factoring of Prime Ideals in Extensions 4. Lifting of Prime Ideals Recall the basic AKLB setup: A is a Dedekind domain with fraction field K, L is a finite, separable extension of K of degree

### JUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson

JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### ALGEBRAIC NUMBER THEORY AND QUADRATIC RECIPROCITY

ALGEBRAIC NUMBER THEORY AND QUADRATIC RECIPROCITY HENRY COHN, JOSHUA GREENE, JONATHAN HANKE 1. Introduction These notes are from a series of lectures given by Henry Cohn during MIT s Independent Activities

### A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number

Number Fields Introduction A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number field K = Q(α) for some α K. The minimal polynomial Let K be a number field and

### FACTORING SPARSE POLYNOMIALS

FACTORING SPARSE POLYNOMIALS Theorem 1 (Schinzel): Let r be a positive integer, and fix non-zero integers a 0,..., a r. Let F (x 1,..., x r ) = a r x r + + a 1 x 1 + a 0. Then there exist finite sets S

### Applications of Fermat s Little Theorem and Congruences

Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4

### MATH 289 PROBLEM SET 4: NUMBER THEORY

MATH 289 PROBLEM SET 4: NUMBER THEORY 1. The greatest common divisor If d and n are integers, then we say that d divides n if and only if there exists an integer q such that n = qd. Notice that if d divides

### CHAPTER 5. Number Theory. 1. Integers and Division. Discussion

CHAPTER 5 Number Theory 1. Integers and Division 1.1. Divisibility. Definition 1.1.1. Given two integers a and b we say a divides b if there is an integer c such that b = ac. If a divides b, we write a

### GREATEST COMMON DIVISOR

DEFINITION: GREATEST COMMON DIVISOR The greatest common divisor (gcd) of a and b, denoted by (a, b), is the largest common divisor of integers a and b. THEOREM: If a and b are nonzero integers, then their

### MATH10040 Chapter 2: Prime and relatively prime numbers

MATH10040 Chapter 2: Prime and relatively prime numbers Recall the basic definition: 1. Prime numbers Definition 1.1. Recall that a positive integer is said to be prime if it has precisely two positive

### SUBGROUPS OF CYCLIC GROUPS. 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by

SUBGROUPS OF CYCLIC GROUPS KEITH CONRAD 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by g = {g k : k Z}. If G = g, then G itself is cyclic, with g as a generator. Examples

### r + s = i + j (q + t)n; 2 rs = ij (qj + ti)n + qtn.

Chapter 7 Introduction to finite fields This chapter provides an introduction to several kinds of abstract algebraic structures, particularly groups, fields, and polynomials. Our primary interest is in

### PROBLEM SET 6: POLYNOMIALS

PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other

### Mathematics Course 111: Algebra I Part IV: Vector Spaces

Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are

### Jacobi s four squares identity Martin Klazar

Jacobi s four squares identity Martin Klazar (lecture on the 7-th PhD conference) Ostrava, September 10, 013 C. Jacobi [] in 189 proved that for any integer n 1, r (n) = #{(x 1, x, x 3, x ) Z ( i=1 x i

### Putnam Notes Polynomials and palindromes

Putnam Notes Polynomials and palindromes Polynomials show up one way or another in just about every area of math. You will hardly ever see any math competition without at least one problem explicitly concerning

### 6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives

6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise

### The Prime Numbers. Definition. A prime number is a positive integer with exactly two positive divisors.

The Prime Numbers Before starting our study of primes, we record the following important lemma. Recall that integers a, b are said to be relatively prime if gcd(a, b) = 1. Lemma (Euclid s Lemma). If gcd(a,

### The Factor Theorem and a corollary of the Fundamental Theorem of Algebra

Math 421 Fall 2010 The Factor Theorem and a corollary of the Fundamental Theorem of Algebra 27 August 2010 Copyright 2006 2010 by Murray Eisenberg. All rights reserved. Prerequisites Mathematica Aside

### Mathematics for Computer Science/Software Engineering. Notes for the course MSM1F3 Dr. R. A. Wilson

Mathematics for Computer Science/Software Engineering Notes for the course MSM1F3 Dr. R. A. Wilson October 1996 Chapter 1 Logic Lecture no. 1. We introduce the concept of a proposition, which is a statement

### = 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that

Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without

### Basic Algorithms In Computer Algebra

Basic Algorithms In Computer Algebra Kaiserslautern SS 2011 Prof. Dr. Wolfram Decker 2. Mai 2011 References Cohen, H.: A Course in Computational Algebraic Number Theory. Springer, 1993. Cox, D.; Little,

### A Course on Number Theory. Peter J. Cameron

A Course on Number Theory Peter J. Cameron ii Preface These are the notes of the course MTH6128, Number Theory, which I taught at Queen Mary, University of London, in the spring semester of 2009. There

### A simple criterion on degree sequences of graphs

Discrete Applied Mathematics 156 (2008) 3513 3517 Contents lists available at ScienceDirect Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam Note A simple criterion on degree

### Introduction to Algebraic Geometry. Bézout s Theorem and Inflection Points

Introduction to Algebraic Geometry Bézout s Theorem and Inflection Points 1. The resultant. Let K be a field. Then the polynomial ring K[x] is a unique factorisation domain (UFD). Another example of a

### ON THE EMBEDDING OF BIRKHOFF-WITT RINGS IN QUOTIENT FIELDS

ON THE EMBEDDING OF BIRKHOFF-WITT RINGS IN QUOTIENT FIELDS DOV TAMARI 1. Introduction and general idea. In this note a natural problem arising in connection with so-called Birkhoff-Witt algebras of Lie

### An example of a computable

An example of a computable absolutely normal number Verónica Becher Santiago Figueira Abstract The first example of an absolutely normal number was given by Sierpinski in 96, twenty years before the concept

### 1 = (a 0 + b 0 α) 2 + + (a m 1 + b m 1 α) 2. for certain elements a 0,..., a m 1, b 0,..., b m 1 of F. Multiplying out, we obtain

Notes on real-closed fields These notes develop the algebraic background needed to understand the model theory of real-closed fields. To understand these notes, a standard graduate course in algebra is

### International Journal of Information Technology, Modeling and Computing (IJITMC) Vol.1, No.3,August 2013

FACTORING CRYPTOSYSTEM MODULI WHEN THE CO-FACTORS DIFFERENCE IS BOUNDED Omar Akchiche 1 and Omar Khadir 2 1,2 Laboratory of Mathematics, Cryptography and Mechanics, Fstm, University of Hassan II Mohammedia-Casablanca,

### Computing exponents modulo a number: Repeated squaring

Computing exponents modulo a number: Repeated squaring How do you compute (1415) 13 mod 2537 = 2182 using just a calculator? Or how do you check that 2 340 mod 341 = 1? You can do this using the method

### CONTINUED FRACTIONS AND FACTORING. Niels Lauritzen

CONTINUED FRACTIONS AND FACTORING Niels Lauritzen ii NIELS LAURITZEN DEPARTMENT OF MATHEMATICAL SCIENCES UNIVERSITY OF AARHUS, DENMARK EMAIL: niels@imf.au.dk URL: http://home.imf.au.dk/niels/ Contents

### Integer Factorization using the Quadratic Sieve

Integer Factorization using the Quadratic Sieve Chad Seibert* Division of Science and Mathematics University of Minnesota, Morris Morris, MN 56567 seib0060@morris.umn.edu March 16, 2011 Abstract We give

### Monogenic Fields and Power Bases Michael Decker 12/07/07

Monogenic Fields and Power Bases Michael Decker 12/07/07 1 Introduction Let K be a number field of degree k and O K its ring of integers Then considering O K as a Z-module, the nicest possible case is

PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient

### SOLUTIONS FOR PROBLEM SET 2

SOLUTIONS FOR PROBLEM SET 2 A: There exist primes p such that p+6k is also prime for k = 1,2 and 3. One such prime is p = 11. Another such prime is p = 41. Prove that there exists exactly one prime p such

### EXERCISES FOR THE COURSE MATH 570, FALL 2010

EXERCISES FOR THE COURSE MATH 570, FALL 2010 EYAL Z. GOREN (1) Let G be a group and H Z(G) a subgroup such that G/H is cyclic. Prove that G is abelian. Conclude that every group of order p 2 (p a prime

### RESULTANT AND DISCRIMINANT OF POLYNOMIALS

RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results