ON LARGE ZSIGMONDY PRIMES

Transcription

1 PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 02, Number, January 988 ON LARGE ZSIGMONDY PRIMES WALTER FEIT (Communicated by Donald Passman) Abstract. If a and m are integers greater than, then a large Zsigmondy prime is a prime / such that l\am, / a' for < i < m and either I2 am - or / > m +. The main result of this paper lists all the pairs (a, m) for which no large Zsigmondy prime exists.. Introduction. Let a and m be integers greater than. A Zsigmondy prime for (a, m) is a prime / such that l\am - but / a' - for l.< j < m. A well-known theorem of Zsigmondy asserts that Zsigmondy primes exist except if {a, m) = (2,6) or m = 2 and a = 2k I (see [5]). This paper contains a refinement of this result. Observe that if / is a Zsigmondy prime for (a, m), then a has order m modulo / and so / = (mod m). Thus / > m +'.- A prime / is a large Zsigmondy prime for (a,m) if / is a Zsigmondy prime for (a, m) and either / > m + or I \am. Theorem A. // a and m are integers greater than, then there exists a large Zsigmondy prime for (a,m) except in the following cases. (i) m = 2 and a = 2*3' for some natural number s, and t = 0 or. (ii) a = 2 andm = 4,6,0,2 or 8. (iii) a = 3 and m = 4 or 6. (iv)(a, m) = (5,6). If «is a natural number and / is a prime, let \n\ denote the /-part of n. In other words \n\i = lk where /* \n but lk + n. Thus a Zsigmondy prime / for (a, m) is large if and only if \am, > m +. An argument similar to that used to prove Theorem A also yields the following result. Theorem B. Let N be a positive integer. Then for all but a finite number of pairs of integers (a,m) with a>\ and m > 2, there exists a Zsigmondy prime I with \am - \\,> mn +. The proof of Theorem B actually yields a bound which can be used to find the exceptions in Theorem B. This method is used to prove Theorem A. Received by the editors August 27, 986 and, in revised form, September 22, Mathematics Subject Classification (985 Revision). Primary A05; Secondary A07. The work on this paper was partly supported by NSF grant #DMS American Mathematical Society /88 $.00 + $.25 per page

2 30 WALTER FEIT Zsigmondy's Theorem was independently, but later, discovered by Birkhoff and Vandiver [2]. Of course it follows from Theorem A by checking that in cases (ii), (iii), and (iv) a Zsigmondy prime exists except when (a, m) - (2,6). Artin gave an elegant proof of the original result in []. The proof of Theorem A given here uses his approach and is based on his work. The special case of Theorem A for a ^ 3 was proved in [3]. However there is no appreciable simplification in handling the general case rather than the case a = 2, so the proof given here is independent of earlier work and is self-contained. The motivation for Theorem A comes from the theory of finite groups and is needed for some forthcoming work [4]. The relevance of Theorem A and Zsigmondy's Theorem for the theory of finite groups can also be seen for instance in [, 3]. I am indebted to the referee for suggesting Lemma 2.. This is stronger than my earlier result and shortens some of the arguments below. 2. Some preliminary results. For any natural number m let m(x) denote the wth cyclotomic polynomial and let P(m) denote the largest prime which divides m. Let <p(m) = degree of $m(x) denote the Euler function. Lemma 2.. Let a, m be integers with a > and m > 2. Let S be the set of all Zsigmondy primes for (a,m). Let Z = T\,eS\$m(a)\i. Then (2.) 4rm(a)\ZP(m). Proof. This is a direct consequence of [, Lemma ]. D Lemma 2.2. Let a, m be integers with a > and m > 2. Let N be a positive integer. Suppose that if I is a Zsigmondy prime for (a,m) then \am \ < Nm +. Then (2.2) \^m(a)\<(nm + l)np(m). Proof. There are at most N Zsigmondy primes / < Nm +. Thus if Z is defined as in Lemma 2. then Z < (Nm + l)n. The result follows from Lemma 2.. D Lemma 2.3. Let a, m be integers with a > and m > 2. Suppose there exists no large Zsigmondy prime for (a, m). Then (2.3) *m(a) <(m + l)p(»i). Proof. Let N = in Lemma 2.2. D The proofs of Theorems A and B are based on the fact that (2.2) and (2.3) only hold for a finite number of cases. For any positive real number r define (2.4) Lm(r)= min *w(«). \u\-r Here u ranges over all complex numbers with \u\ = r. Since this set is compact, the minimum exists. Lemma 2.4. Let m = nt, where n is relatively prime to t. Let r be a positive real number. Then Lm(r) > (Ln(r))"*'\ Proof. Let «be a complex number with \u\ = r. Then! «(«) i-in.mí,

3 ON LARGE ZSIGMONDY PRIMES 3 where e ranges over all the primitive rth roots of. As \ue\ = \u\, this implies that l^m(m)l > \Ln(r)\vW. The result follows as u was arbitrary with \u\ = r. D The next two elementary results will be needed later. Lemma 2.5. Let m be a natural number. Then <p(m)^ \4m. Proof. If p is a prime then <p(pb) = (p - l)/>fc_ for any natural number b. Thus (p(ph) > ph/2 if p > 2 and <p(2h) > 2b/2'\ This proves the result as <p(nt) = <p(n)q>(t) for n, t relatively prime. D Lemma 2.6. Let 0 < n < m. Thenf(x) = (xm - l)/(x" + ) is monotone increasing for x > 0. Proof. If x > 0 then,,,. (m - n)xm+"~l + mxm~l + nx"' /(*) = --;-~2->0. D (x" + ) 3. The proof of Theorem B. Lemma 3.. Let r > 2 be a real number. Let m, N be natural numbers, suppose that Lm(r) < (Nm + l)np(m). Then (r - ) < (2N + )2(*+>. Proof. By Lemma 2.4 {Nm + l)np(m) > Ljr) > L(r),'(m) = (r - \y(m). If (r - ) > (2N + )2<A'+) this imphes that (Nm + l)n+ > (Nm + l)nm > (Nm + l)np(m) > (2N + ifn+w"\ Hence by Lemma 2.5 which is impossible for m >. D {Nm + ) > (27V + l)2v(m) > (2N + )* Lemma 3.2. Let r > 2 be a real number. Let m, N be natural numbers such that m > and m + 2n with n odd. Suppose that Lm(r) < 2N(Nm + l)*p(m). Then there exists C = C(N), depending only on N, with m < C. Proof. Suppose first that m = pb is a prime power. By assumption, pb > 2. There exists a complex number u with \u\ = r such that 4N{Np» + l)np> u?'- - rp _ i i rp _ i > -Sï rp" + " 3 ri > \rpb-p"- > lrpba > I?/-6^ This imphes the existence of C0 = C0(N) > 020, depending only on N, such that Pb < Q.

4 32 WALTER FEIT We will now prove the result with C = C, where k = ir(c0) is the number of primes p < C0. Suppose m> C. Then m= pht for some prime p with ph > C0 and /> t. Therefore V(r)>2*(V + l)v Hence Lemma 2.4 implies that 2»(Np»t + ) V(0 > Lm(.) > { V(r)}"(" > {2«>(V + l)"p. By assumption, Hence í > 2 and so <p(t) > 2. Thus 2Npt(2Npbt)N'> 2aV(V)W/>- *(2Arp»i)'f>2w/»(Ar/»*)w')> {2(/V>*)v(0) ". After taking Ath roots, Lemma 2.5 implies that (3.) A//2 > (TV/)*)*0 > {Npbf/2. Let v = A//>*. By (3.) vf2 > y^/2. As v > /»* > C0 > 020 this implies that if r > 5 then 25 v > v^/2. Hence 25 > v(^-)/2> v/0> 00 which is not the case. Since t # 2i0 with f0 odd, (3.) implies that one of the following holds: / = 4 and 6y > y2, r = 3 and 9v > y2. None of these can hold as y > 020. D Lemma 3.3. Let a > 2 e a natural number. Let m, N be natural numbers with m > 2. Suppose that \4>Ja)\*i(Nm + l)np(m). Then there exists C = C(N), depending only on N, with m < C. Proof. If m =t 2«with n odd this follows directly from Lemma 3.2. Suppose that m = 2/ with n odd. Then $m(«) = $ (-«) for any complex number u and P(m) = P(n). Thus Lm(a) = Ln(a). Hence Lemma 3.2 implies that if no such C = C(N) exists then \*m(a)\>l (a)>2n(nn + l)np(») > (2Nn + l)np(n) = (Nm + l)np(m). D Proof of Theorem B. This is an immediate consequence of Lemmas 2.2, 3., and The proof of Theorem A. The proof of Theorem A is analogous to that of Theorem B but the relevant inequalities need to be considered more precisely.

5 ON LARGE ZSIGMONDY PRIMES 33 Lemma 4.. Suppose that m = nt with n > and n relatively prime to t. Assume that m # 2 (mod4). Let r be a real number such that Ln(r) > 2(n + l)p(n). Then Lm(r) > 2(m + l)p(m). Proof. By induction on the number of primes dividing t it may be assumed that / = pb for some prime p. By assumption n, t # 2 (mod 4) and so t > 2 and n > 2. Suppose the result is false. Then by Lemma 2.4 Thus 2(npb + l)pp(n) > Lm(r) > LB(r)^^" > {2(n + l)p(n)}^-^\ (4.) 4np2h > 2(np" + \)p > (2(«+ l))'*-'é~>(ii)'*~'*"l~ > {2(n + l)}"h-^'2ph-phi-. Let v = \ph. As ph - pb~ > f/»6, (4.) implies that I6ny2 > 22v"(«+ l)-v. Hence I6y2 > 22-"-(«+ l)y~l > 22-v-4-v'- = 24v"3. Thus pb = y2 <. Hence ph = 3, 4 or 5 contrary to (4.). Lemma 4.2. Let r > 2 be a real number. Let m be an integer such that m # 2«/or n odd and m >. 77ie«(4.2) Lm(r)>2(m + l)p(m) except possibly in the following cases. (i) m = 3 awav < 6. (ii) w = 4 and r < 5. (iii) w = 2 and r < 4. (iv) r < 3 and m = 5,7,8,9,5 or 20. Proof. Assume that m does not satisfy the conclusions of the lemma. In particular (4.2) does not hold. Suppose first that m = ph > 2 for some prime p. Then for suitable u with \u\ = r, Lemma 2.6 implies that (4.3) 2(p"+l)p> «'*" - D r^ - 2p - * r"h ' + " 2"h ' + Suppose that m = p is a prime. Then (4.3) implies that 6p(p + ) > 2'. Hence p <. Thus /? = 3,5 or 7 and (4.3) implies that rp 2/>(/> + l)> T^y- If p = 7 this yields that 2 > (r7 - l)/(r + ) and so r < 3. If /> = 5 then 60 ^ (r5 - l)/(r + ) and so r < 3. If p = 3 then 24 > (r3 - l)/(r + ) and so r < 6. Suppose that m = pb with o >. Then r'" ' > 4 and so rp" ' - > (/ '* ' + ). Hence (4.3) implies that /-l _, (4.4) 2(/>» +!)/»> 2 -» ',p

6 34 WALTER FEIT Let y = \pb. Then y «/>* - ph~l. Thus (4.4) yields that 2(2 v+ l)y>2y~l. Hence v < 9 and so p*> < 8. Thus pb = 4,8,9 or 6. If pb = 6 then (4.3) implies that 68 > (26- l)/(28 + ) = 28 - which is not the case. If pb = 9 then (4.4) imphes that 20 > r6 and so r < 3. If pb = 8 then (4.3) implies that 36 > (r8 - l)/(r4 + ) = r4 - and so r < 3. If pb = 4 then (4.3) imphes that 20 > (r4 - l)/(r2 + ) = r2 - and so r < 5. In summary, we have shown that if m = pb, then m satisfies the conclusion of the lemma. In particular (4.2) holds unless m = 4,8,3,9,5 or 7. Hence Lemma 4. implies that w Suppose that m = 7í. Then As (4.2) does not hold, Lemma 2.4 implies that for suitable u with \u\ = r. (4.5) 4(7? + ) > - v(0 r7-l r + This yields the following inequalities. lit = 5 then 4 36 > ((r7 - l)/(r + l))4. If t = 9 then 4 64 > ((r7 - l)/(r + l))6. If í = 8 then 4 57 ^ ((r7 - l)/(r + l))4. None of these can hold for r > 2. Hence 5 + r, 9 + t, and 8 + t by Lemma 4.. Thus i 2. As í # 2 (mod 4) either / = 3 or 4. Suppose t = 4; then (4.5) becomes 4 29 > ((r7 - l)/(r + l))2, which is impossible for r > 2. Thus 4 / by Lemma 4. and so t = 3. Now (4.5) becomes 4 22 > ((r7 - l)/(r + l))2, which is impossible for r > 2. Therefore it may be assumed that w Suppose that m = 5?. Then t As (4.2) does not hold, Lemma 4. imphes that for suitable u with lui = r. (4.6) 0(5? + ) > u5-l u+ <p(0 3s - r+ This yields the following inequalities. If í = 9 then 460 > ((r5 - l)/(r + l))6. If í = 8 then 40 > ((r5 - l)/(r + l))4. Neither of these can hold for r > 2. Thus Lemma 4. implies that 9 f t and Sit. Hence t \ 2 and so t = 3,4, or 2. If t = 2 then (4.6) becomes 60 > ((r5 - l)/(r + l))4, which is impossible for r > 2. If t = 4 then (4.6) becomes 20 > ((r5 - l)/(r + l))2. Hence r < 3. If t = 3 then (4.6) becomes 60 > ((r5 - l)/(r + l))2. Hence r < 3. 9>(0?(0

7 ON LARGE ZSIGMONDY PRIMES 35 Therefore it may be assumed that Suppose that m = 9t with / = 4 or 8. As (4.2) does not hold, Lemma 4. implies that for suitable u with \u\ = r (4.7) 6(9i + )» u9-l u3-l 9(0 - This yields the following inequalities. If / = 8 then 6 73 > ((29 - l)/9)4. If t - 4 then 6 37 > ((29- l)/9)2. Neither of these holds and so 9 * w. Thus m 24. Hence m = 2 or 24. If m = 24 and (4.2) does not hold, Lemma 4. implies that for suitable u with M = r, - a 50 > + l %(r4-l)2. This is impossible for r ^ 2. If m = 2 and (4.2) does not hold, Lemma 4. implies that for suitable u with \u\ = r, Thus r < ^ - + 9(0 2 = IT +» (r2 D2- Lemma 4.3. Suppose that a, m are integers with a > 2 a/w m > 3. 77ien (4.8) *w(a) >(m + l)/>(m) except possibly in the following cases. (i) m = 3 or 6 anda < 6. (ii) m = 4 a/ida < 5. (iii) m = 2 0/0* a < 4. (iv) a < 3 and m = 5,7,8,9,0,4,5,8,20 or 30. Proof. If m = 2«with n odd, then the result follows directly from Lemma 4.2. Suppose that m = 2n with n odd. Then $,(«) = $ (-") for any complex number u and P(m) = P(n). Thus Lm(a) = L (a). Hence if (4.2) holds for n then «Dm(a) > L (a) > 2(«+ l)p(n) = (2«+ 2)P(m) > (m + l)p(m). Suppose that (4.2) does not hold for n. By Lemma 4.2 one of the following cases must occur: n = 3, m = 6, and a < 6. a<3 and «= 5,7,9 or 5. Thus w = 0,4,8 or 30. D Proof of Theorem A. Suppose that m = 2. Since 2 is the greatest common divisor of a and a + it follows that any odd prime dividing a + is a Zsigmondy prime. This implies the result for m - 2. Suppose now that m > 3.

8 36 WALTER FEIT Assume that there exists no large Zsigmondy prime for (a, m). By Lemmas 2.3 and 4.3 (a, m) is one of the cases listed in Lemma 4.3. We will exhibit a large Zsigmondy prime / in each case that is not listed in the statement of Theorem A. This will conclude the proof. m = 3, m = 6, m = 4, m = 2, a = 2, / a l m / D REFERENCES. E. Artin, The orders of the linear groups, Comm. Pure Appl. Math. 8 (955), G. D. Birkhoff and H. S. Vandiver, On the integral divisors of a" - b", Ann. of Math. (2) 5 (904), W. Feit, Extensions of cuspidal characters of G\*m(q) (to appear). 4. W. Feit and G. Seitz (to appear). 5. K. Zsigmondy, Zur Theorie der Potenzreste, Monatsch. Math. Phys. 3 (892), Department of Mathematics, Yale University, New Haven, Connecticut 06520