and each factor on the right is clearly greater than 1. which is a contradiction, so n must be prime.

Transcription

1 MATH 324 Summer 200 Elemetary Number Theory Solutios to Assigmet 2 Due: Wedesday July 2, 200 Questio [p 74 #6] Show that o iteger of the form 3 + is a prime, other tha 2 = 3 + Solutio: If 3 + is a prime, sice 3 + = ( + )( 2 + ), the either + = or 2 + = The + = is impossible, sice, ad therefore we must have 2 + =, that is, ( ) = 0, so that = Questio 2 [p 74 #7] Show that if a ad are positive itegers with > ad a is prime, the a = 2 ad is prime Hit : Use the idetity a kl = (a k )(a k(l ) + a k(l 2) + + a k + ) Solutio: Suppose that a is prime, where > Sice a = (a )(a + a a + ) ad the secod factor is clearly greater tha, it follows that a =, that is, a = 2 Otherwise, the first factor would also be greater tha ad a would be composite Also, if is composite, so that = k l, with k > ad l >, the we ca factor 2 as i the hit: 2 kl = (2 k )(2 k(l ) + 2 k(l 2) k + ) ad each factor o the right is clearly greater tha which is a cotradictio, so must be prime Questio 3 [p 74 #0] Usig Euclid s proof that there are ifiitely may primes, show that the th prime p does ot exceed 2 2 wheever is a positive iteger Coclude that whe is a positive iteger, there are at least + primes less tha 2 2 Solutio: The proof is by strog iductio Base Case : If =, the p = = 2 Iductive Step : Now assume that p k 2 2k for k =, 2,, If M = p p 2 p +, sice M has a prime divisor p which is differet from each p i, with i, the p + p M = p p 2 p = = < = 2 2 By the priciple of mathematical iductio p 2 2 for all From the above, p + 2 2, ad sice 2 2 caot be prime if > 0, there must be + primes which are strictly less tha 2 2

2 Questio 4 [p 74 #2] Show that if p k is the k th prime, where k is a positive iteger, the p p p 2 p + for all itegers with 3 Solutio: Let M = p p 2 p +, where p k is the k th prime, from Euler s proof, some prime p differet from p, p 2,, p divides M, so that for all 3 Questio 5 [p 74 #3] p p M = p p 2 p + Show that if the smallest prime factor p of the positive iteger exceeds 3, the p must be prime or Solutio: Let p be the smallest prime factor of, ad assume that p > 3 Case : If is prime, the the smallest prime factor of is p =, ad i this case p = Case 2 : If > is ot prime, the must be composite, so that = p p, ad sice p > 3, the Now, if p is ot prime the p p < 3 = ( 3 ) 2 has a prime factor q with q < p < 3 < p ad this prime factor q is also a divisor of, which cotradicts the defiitio of p Therefore, p prime must be Questio 6 [p 87 #2] Show that every iteger greater tha is the sum of two composite itegers Solutio: If > ad is eve, the 4 is eve ad 4 > 7, so that 4 8 Therefore, = ( 4) + 4 is the sum of two composite itegers If > ad is odd, the 9 is eve ad 9 > 2, so that 9 4 Therefore, = ( 9) + 9 is the sum of two composite itegers

3 Questio 7 [p 87 #22] Let be a positive iteger greater tha ad let p, p 2,, p t be the primes ot exceedig Show that p p 2 p t < 4 ( ) Solutio: The proof is by strog iductio Base Case : if = 2, the p = 2 is the oly prime less tha or equal to 2, ad so that ( ) is true for = 2 2 < 4 2 = 6 Iductive Step : Now suppose that ( ) is true for 2, 3,, where 3 Note that we ca restrict our attetio to odd, sice if is eve, the p = p p p < 4 < 4 Settig = 2m +, we have ( ) 2m + (2m + )! = m m!(m + )! ad this is divisible by every prime p with m + 2 p 2m +, so that by the iductive hypothesis p 2m+ ( ) 2m + p m Now, the biomial coefficiets ( ) 2m + m p m+ ad ( ) 2m + p < 4 m+ m ( ) 2m + m + are equal ad both occur i the expasio of the biomial ( + ) 2m+, so that ad therefore ad ( ) is true for = 2m + also ( ) 2m + m 2 22m+ = 4 m, p 2m+ By the Priciple of Mathematical Iductio for all positive itegers 2 p < 4 m 4 m+ = 4 2m+ p < 4 p

4 Questio 8 [p 87 #23] Let be a positive iteger greater( tha) 3 ad let p be a prime such that 2/3 < p Show that p does 2 ot divide the biomial coefficiet Solutio: Note that the restrictios o are such that (a) p > 2 (b) p ad 2p are the oly multiples of p which are less tha or equal to 2, sice 3p > 2 (c) p itself is the oly multiple of p which is less tha or equal to From (a) ad (b) we have p 2 (2)! but p3 (2)!, while from (c), Therefore, that is, p 2 (!) 2 but p 3 (!) 2 p (2)! (!) 2, p ( ) 2 Questio 9 [p 87 #24] Use Exercises 22 ad 23 to show that if is a positive iteger, the there exists a prime p such that < p < 2 (This is Bertrad s cojecture) Solutio: You ca easily check usig the Sieve of Eratosthees that the result holds for 2 27 Now let 28, ad suppose that there is o prime betwee ad 2 Let be the prime power decompositio of ( ) 2 = p rp p 2 ( ) 2 By assumptio there are o primes betwee ad 2, so that ( ) 2 = p rp p If p is a prime with 2 3 < p, the < 4 3 < 2p < 2 ad 2 < 3p < 3, so that p divides! exactly oce ad p divides (2)! exactly twice, ad so p ( ) 2

5 Therefore, ( ) 2 = p rp p rp p 2 2<p 2/3 sice if ( ) 2 2 < p 2/3, the p divides exactly oce p 2 2 p 2/3 Now, the umber of primes less tha 2 is less tha the umber of odd itegers less tha 2, that is, less tha 2/2 = /2, therefore p 2 2 (2) /2 p From Questio 7, we have (replace by 2/3 ) which implies that p 2/3 p 2/3 p < 4 2/3, p < 4 2/3, so that ( ) 2 < (2) /2 4 2/3 ( ) 2 Sice is the largest of the 2 + terms i the biomial expasio of ( + ) 2, we have ( ) ( ) 2 2 (2 + ) > (2) > 2 2, so that which implies that 2 22 < ( ) 2 < (2) /2 4 2/3, 2 2/3 < (2) /2 Takig logarithms ad dividig by 2/6, we get 8 log 2 3 log(2) < 0 ( ) Now defie the fuctio f : N R by ad differetiate to get f() = 8 log 2 3 log(2), 2 log 2 3 f () = Note that f(28) = 8 log 2 > 0, ad f () > 0 for 28, so that f() is icreasig ad therefore positive for 28 However, this cotradicts the iequality ( ) Therefore, for ay positive iteger >, there is a prime p satisfyig < p < 2

6 Questio 0 [p 87 #27] Use Bertrad s postulate to show that m does ot equal a iteger whe ad m are positive itegers I particular, is ever a iteger for > Solutio: = Case : If m <, the m, so that + m 2, ad m m k= k ad the sum caot be a iteger i this case < }{{} times = =, Case 2 : If m, from Bertrad s postulate, there is a prime p such that < p < 2 + m Let p be the largest prime such that < p < + m, the + m < 2p, sice if ot, there would be a prime q with p < q < 2p + m, which cotradicts the choice of p Now suppose that m = N ( ) where N is a iteger Sice < p < + m < 2p, the p occurs as a factor i oly oe of the deomiators Defie M = ad multipy ( ) by M to get ad solvig for M p, we have +m k= k, ad M k = M k for k + m, M + M M +m = M N, ( ) m+ M p = M N M k Now ote that every term o the right is divisible by p, which implies that p Mp, which is a cotradictio k= k p Therefore, m is ever a iteger for ay positive itegers m ad

7 Questio Use the prime umber theorem lim x π(x) x/ log x = where π(x) is the umber of primes less tha or equal to x, to show that if p is the th prime, the so that p log for large lim p log =, Solutio: Suppose that the prime umber theorem is true, so that lim x if we let x = p for, the π(p ) =, so that π(x) x/ log x =, The atural logarithm is a cotiuous fuctio o R+, so that log p lim = (+) p lim {log + log log p log p } = 0, ad { log lim log p + log log p } = 0 log p log p Now, sice lim log p = +, this implies that { log lim + log log p } = 0, log p log p log log p ad sice lim = 0, we have log p that is, From (+), this implies that { } log lim = 0, log p lim log log p = ad therefore p log whe is large log log p log log p lim = lim = lim =, p p log p p Questio 2 [p 7 #4] Let p be a prime ad a positive iteger Show that the largest expoet ν() such that p ν()! is give by depoligac s formula: ν() = + p p 2 + p 3 +

8 Solutio: For k, the itegers i the sequece, 2, 3,, which are divisible by p k are where q = p k, 2 p k, 3 p k,, q p k p k So the umber of itegers i the sequece, 2, 3,, which are divisible by p k is p k O the other had, the expoet ν() of the largest power of p i the prime factorizatio of! is + p p p p r where r is determied by from p r < p r+ (eg the umber divisible by p 2 have to be couted twice, oce i ad oce i p p 2 )