Lecture 1. January 8, 2018

Transcription

1 Lecture 1 Jauary 8, Primes A prime umber p is a positive iteger which caot be writte as ab for some positive itegers a, b > 1. A prime p also have the property that if p ab, the p a or p b. This is a special property of the itegers about which you will lear i Algebraic Number Theory course. Oe of the cetral questios i aalytic umber theory cocers coutig primes. Euclid proved that there are ifiitely may primes. Do you kow that proof? He assumes that this is ot so, therefore all the primes i the world form a fiite set. Let us assume that this set is {p 1, p,..., p k }. The Euclid forms the umber N = p 1 p p k + 1. Because we have added 1 to the product p 1 p k, the umber N caot be a multiple of p 1. Or of p. Or of ay of p 1,..., However, sice N is a positive iteger, it must be a product of some primes. This is the fudametal theorem of arithmetic. Ayoe of those primes dividig N is a prime umber ot i the set {p 1, p,..., p k }. This shows that the set of primes is ifiite. Problem 1. Let p k be the kth prime. Deduce from Euclid s proof that p k k 1 holds for all k 1. So, ow that we kow that there are ifiitely may primes, how do we cout them? Oe way is to let be ay positive umber, ad cout the umber of primes p. This umber is called π(). Thus, π() = p 1. Problem. What is π(100)? 1

2 Chebyshev s estimates I 1896, de la Vallée Poussi ad Hadamard proved idepedetly the Prime Number Theorem. That is, they showed that lim π() / log = 1. The result had bee cojectured by Gauss. Earlier, Chebyshev had established that c 1 log π() c log for all 10, where c 1 = log( 1/ 3 1/3 5 1/5 /30 1/30 ) ad c = 6c 1 /5. Here, we prove somethig weaker by a method which is easier tha Chebyshev s. The followig proof is due to Erdős. Theorem 1. For, ( ) 3 log 8 log We shall eed the followig lemma. < π() < (6 log ) log. Lemma 1. Let p be a prime ad e p (!) be the epoet at which p appears i!. The e p (!) = k 0 Proof. Iductio o. The formula is clearly true if = 1 (for all p). Assume that it holds for ad write + 1 = p u m, where p m. The, by the iductio hypothesis, e p (( + 1)!) = e p (!) + u = u k=1 ( ) p k k>u All is left to ote is that + 1 p k = p k + 1 if 1 k u ad + 1 = p k p k if k > u.

3 Proof of Theorem 1. We first prove the lower boud o π(). We start with the observatio that = ()! (!) p rp (1) p< if > 1, where r p is that positive iteger such that p rp < p rp+1. Ideed, to prove divisibility ((1)) ote that the epoet at which p appears i the biomial coefficiet equals e p (()!) e p ((!) ) = e p (()!) e p (!) = ( ) p k k 1 Divisibility relatio (1) follows ow by observig that y y {0, 1} holds for all real umbers y (whe is it zero ad whe is it 1?) together with the fact that whe k > r p, we have p k >, therefore /p k = 0 for such values of k. Divisibility relatio (1) certaily gives that () π(). O the other had, sice ad sice we get that k By iductio, oe checks that Hece, if 3, we have (1 + 1) = k=0 k holds for all k = 0, 1,...,, > > holds for all 3. < + 1 < < () π(), 3

4 which after takig logarithms becomes π() > log( ) log = log log(). Assume ow that 8. Let be that positive iteger such that < +. Note that 3. Further, we have > 3/4 (because 8). The fuctio y y/ log y is icreasig for y > e (to see this, study the sig of its derivative!) ad certaily 3/4 6 > e for 8. Puttig together all the above we get that if 8, the π() π() log log() log which is the desired lower boud for 8. yourself that it also holds for [, 8). Thus, We ow tur to the upper boud. Note that ( ) p. <p <p which after takig logarithms leads to Hece, p < (1 + 1) =, 3/4 log(3/4) > 3 log 8 π() log π()(log(/) + log ) < log. log, You should ow check by π() log π() log(/) < log + π() log (3 log ), () where we used the obvious fact that π(). Put f() = π() log ad otice that the iequality () above is f() f(/) < (3 log ). Let = i for i = k, k 1,...,. We the have f( k ) f( k 1 ) < (3 log ) k f( k 1 ) f( k ) < (3 log ) k 1. f(4) f() < (3 log ) 4 (3) 4

5 Summig up iequalities (3), we get so π( k+1 ) log( k ) = f( k+1 ) = (3 log )( k ) + f() = (3 log )( k ) + π(4) log < (3 log )( k ) < (3 log ) k+1, ( ) π( k+1 k ) < (6 log ) log( k. ) Now give, choose k 1 such that k < k+1. If 4, the k so k 4 > e. Thus, k / log( k ) / log wheever 4. We thus get that ( ) π() π( k+1 k ) < (6 log ) log( k < (6 log ) ) log for 4. You should also check that the claimed iequality holds for all [, 4). 3 Homework Problem 3. Show that if + 1 is prime the is a power of. Problem 4. Show that if 1 is prime the is prime. Problem 5. A positive iteger is pseudoprime to base if is composite ad the cogruece 1 1 (mod ) holds. Let F = + 1 be the th Fermat umber. Show that if k 1 < <... < s k, the F 1 F s is either a prime or a base pseudoprime. Deduce that there are ifiitely may base pseudoprimes. Problem 6. Show that ( 1)! 1 (mod ) if is prime ad ( 1)! if > 4 is composite. Use this to prove that 1/ p = 1 + m= m j= (j 1)!+1 j (j 1)! j Problem 7. Prove that holds for all 9. lcm[1,,,..., ] 5