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

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

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

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.

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.

Further linear algebra. Chapter I. Integers.

Further linear algebra. Chapter I. Integers. Andrei Yafaev Number theory is the theory of Z = {0, ±1, ±2,...}. 1 Euclid s algorithm, Bézout s identity and the greatest common divisor. We say that a Z divides

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,

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

(x + a) n = x n + a Z n [x]. Proof. If n is prime then the map

22. A quick primality test Prime numbers are one of the most basic objects in mathematics and one of the most basic questions is to decide which numbers are prime (a clearly related problem is to find

Solutions to Practice Problems

Solutions to Practice Problems March 205. Given n = pq and φ(n = (p (q, we find p and q as the roots of the quadratic equation (x p(x q = x 2 (n φ(n + x + n = 0. The roots are p, q = 2[ n φ(n+ ± (n φ(n+2

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)).

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

b) Find smallest a > 0 such that 2 a 1 (mod 341). Solution: a) Use succesive squarings. We have 85 =

Problem 1. Prove that a b (mod c) if and only if a and b give the same remainders upon division by c. Solution: Let r a, r b be the remainders of a, b upon division by c respectively. Thus a r a (mod c)

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

CONTINUED FRACTIONS, PELL S EQUATION, AND TRANSCENDENTAL NUMBERS

CONTINUED FRACTIONS, PELL S EQUATION, AND TRANSCENDENTAL NUMBERS JEREMY BOOHER Continued fractions usually get short-changed at PROMYS, but they are interesting in their own right and useful in other areas

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

MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES

MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES 2016 47 4. Diophantine Equations A Diophantine Equation is simply an equation in one or more variables for which integer (or sometimes rational) solutions

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

Finite Fields and Error-Correcting Codes

Lecture Notes in Mathematics Finite Fields and Error-Correcting Codes Karl-Gustav Andersson (Lund University) (version 1.013-16 September 2015) Translated from Swedish by Sigmundur Gudmundsson Contents

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

Orders Modulo A Prime

Orders Modulo A Prime Evan Chen March 6, 2015 In this article I develop the notion of the order of an element modulo n, and use it to prove the famous n 2 + 1 lemma as well as a generalization to arbitrary

MA2C03 Mathematics School of Mathematics, Trinity College Hilary Term 2016 Lecture 59 (April 1, 2016) David R. Wilkins

MA2C03 Mathematics School of Mathematics, Trinity College Hilary Term 2016 Lecture 59 (April 1, 2016) David R. Wilkins The RSA encryption scheme works as follows. In order to establish the necessary public

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

Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

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(

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

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.

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

Is this number prime? Berkeley Math Circle Kiran Kedlaya

Is this number prime? Berkeley Math Circle 2002 2003 Kiran Kedlaya Given a positive integer, how do you check whether it is prime (has itself and 1 as its only two positive divisors) or composite (not

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

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:

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

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

3. QUADRATIC CONGRUENCES 3.1. Quadratics Over a Finite Field We re all familiar with the quadratic equation in the context of real or complex numbers. The formula for the solutions to ax + bx + c = 0 (where

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,

Practice Problems for First Test

Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the human mind will never penetrate.-

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

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).

EULER S THEOREM. 1. Introduction Fermat s little theorem is an important property of integers to a prime modulus. a p 1 1 mod p.

EULER S THEOREM KEITH CONRAD. Introduction Fermat s little theorem is an important property of integers to a prime modulus. Theorem. (Fermat). For prime p and any a Z such that a 0 mod p, a p mod p. If

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.

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

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

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

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

NUMBER THEORY AMIN WITNO

NUMBER THEORY AMIN WITNO ii Number Theory Amin Witno Department of Basic Sciences Philadelphia University JORDAN 19392 Originally written for Math 313 students at Philadelphia University in Jordan, this

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

Algebraic Systems, Fall 2013, September 1, 2013 Edition. Todd Cochrane

Algebraic Systems, Fall 2013, September 1, 2013 Edition Todd Cochrane Contents Notation 5 Chapter 0. Axioms for the set of Integers Z. 7 Chapter 1. Algebraic Properties of the Integers 9 1.1. Background

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.

CHAPTER 5: MODULAR ARITHMETIC

CHAPTER 5: MODULAR ARITHMETIC LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we will consider congruence modulo m, and explore the associated arithmetic called

Primitive Prime Divisors of First-Order Polynomial Recurrence Sequences

Primitive Prime Divisors of First-Order Polynomial Recurrence Sequences Brian Rice brice@hmc.edu July 19, 2006 Abstract The question of which terms of a recurrence sequence fail to have primitive prime

Homework 5 Solutions

Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which

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

Introduction to finite fields

Introduction to finite fields Topics in Finite Fields (Fall 2013) Rutgers University Swastik Kopparty Last modified: Monday 16 th September, 2013 Welcome to the course on finite fields! This is aimed at

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 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

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

PERIMETERS OF PRIMITIVE PYTHAGOREAN TRIANGLES. 1. Introduction

PERIMETERS OF PRIMITIVE PYTHAGOREAN TRIANGLES LINDSEY WITCOSKY ADVISOR: DR. RUSS GORDON Abstract. This paper examines two methods of determining whether a positive integer can correspond to the semiperimeter

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.

ON INTEGERS EXPRESSIBLE BY SOME SPECIAL LINEAR FORM

Acta Math. Univ. Comenianae Vol. LXXXI, (01), pp. 03 09 03 ON INTEGERS EXPRESSIBLE BY SOME SPECIAL LINEAR FORM A. DUBICKAS and A. NOVIKAS Abstract. Let E(4) be the set of positive integers expressible

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

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

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

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

Prime Numbers. Chapter Primes and Composites

Chapter 2 Prime Numbers The term factoring or factorization refers to the process of expressing an integer as the product of two or more integers in a nontrivial way, e.g., 42 = 6 7. Prime numbers are

Congruences. Robert Friedman

Congruences Robert Friedman Definition of congruence mod n Congruences are a very handy way to work with the information of divisibility and remainders, and their use permeates number theory. Definition

PYTHAGOREAN TRIPLES PETE L. CLARK

PYTHAGOREAN TRIPLES PETE L. CLARK 1. Parameterization of Pythagorean Triples 1.1. Introduction to Pythagorean triples. By a Pythagorean triple we mean an ordered triple (x, y, z) Z 3 such that x + y =

MATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear:

MATH 89 PROBLEM SET : INDUCTION The induction Principle The following property of the natural numbers is intuitively clear: Axiom Every nonempty subset of the set of nonnegative integers Z 0 = {0,,, 3,

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,

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

Solutions to Homework Set 3 (Solutions to Homework Problems from Chapter 2)

Solutions to Homework Set 3 (Solutions to Homework Problems from Chapter 2) Problems from 21 211 Prove that a b (mod n) if and only if a and b leave the same remainder when divided by n Proof Suppose a

Algebra 2. Rings and fields. Finite fields. A.M. Cohen, H. Cuypers, H. Sterk. Algebra Interactive

2 Rings and fields A.M. Cohen, H. Cuypers, H. Sterk A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 1 / 20 For p a prime number and f an irreducible polynomial of degree n in (Z/pZ)[X ], the quotient

3. Applications of Number Theory

3. APPLICATIONS OF NUMBER THEORY 163 3. Applications of Number Theory 3.1. Representation of Integers. Theorem 3.1.1. Given an integer b > 1, every positive integer n can be expresses uniquely as n = a

Consequently, for the remainder of this discussion we will assume that a is a quadratic residue mod p.

Computing square roots mod p We now have very effective ways to determine whether the quadratic congruence x a (mod p), p an odd prime, is solvable. What we need to complete this discussion is an effective

Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof. Harper Langston New York University

Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof Harper Langston New York University Proof and Counterexample Discovery and proof Even and odd numbers number n from Z is called

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

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

Proofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate.

Advice for homework: Proofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate. Even if a problem is a simple exercise that doesn t

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 square-free kernel of x 2n a 2n

ACTA ARITHMETICA 101.2 (2002) The square-free kernel of x 2n a 2n by Paulo Ribenboim (Kingston, Ont.) Dedicated to my long-time friend and collaborator Wayne McDaniel, at the occasion of his retirement

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

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

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,

Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00

18.781 Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00 Throughout this assignment, f(x) always denotes a polynomial with integer coefficients. 1. (a) Show that e 32 (3) = 8, and write down a list

1. LINEAR EQUATIONS. A linear equation in n unknowns x 1, x 2,, x n is an equation of the form

1. LINEAR EQUATIONS A linear equation in n unknowns x 1, x 2,, x n is an equation of the form a 1 x 1 + a 2 x 2 + + a n x n = b, where a 1, a 2,..., a n, b are given real numbers. For example, with x and

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

Chapter 6 Finite sets and infinite sets. Copyright 2013, 2005, 2001 Pearson Education, Inc. Section 3.1, Slide 1

Chapter 6 Finite sets and infinite sets Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 1 Section 6. PROPERTIES OF THE NATURE NUMBERS 013 Pearson Education, Inc.1 Slide Recall that denotes

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

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

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

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