MATH 324 Summer 2006 Elementary Number Theory Solutions to Assignment 2 Due: Thursday July 27, 2006

Transcription

1 MATH 34 Summer 006 Elemetary Number Theory Solutios to Assigmet Due: Thursday July 7, 006 Departmet of Mathematical ad Statistical Scieces Uiversity of Alberta Questio [p 74 #6] Show that o iteger of the form 3 + is a prime, other tha = 3 + Solutio: If 3 + is a prime, sice 3 + = ( + )( + ), the either + = or + = The + = is impossible, sice, ad therefore we must have + =, that is, ( ) = 0, = Questio [p 74 #7] Show that if a ad are positive itegers with > ad a is prime, the a = ad is prime Hit : Use the idetity a kl = (a k )(a k(l ) + a k(l ) + + 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 = Otherwise, the first factor would also be greater tha ad a would be composite Also, if is composite, = k l, with k > ad l >, the we ca factor as i the hit: kl = ( k )( k(l ) + k(l ) + + 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 wheever is a positive iteger Coclude that whe is a positive iteger, there are at least + primes less tha Solutio: The proof is by strog iductio Base Case : If =, the p = 0 = Iductive Step : Now assume that p k k for k =,,, If M = p p p +, sice M has a prime divisor p which is differet from each p i, with i, the p + p M = p p p = = + < + = By the priciple of mathematical iductio p for all From the above, p +, ad sice caot be prime if > 0, there must be + primes which are strictly less tha

2 Questio 4 [p 74 #] Show that if p k is the k th prime, where k is a positive iteger, the p p p p + for all itegers with 3 Solutio: Let M = p p p +, where p k is the k th prime, from Euler s proof, some prime p differet from p, p,, p divides M, for all 3 Questio 5 [p 74 #3] p p M = p p 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 : If > is ot prime, the must be composite, = p p, ad sice p > 3, the Now, if p is ot prime the p p < 3 = ( 3 ) 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 #] 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, 4 8 Therefore, = ( 4) + 4 is the sum of two composite itegers If > ad is odd, the 9 is eve ad 9 >, 9 4 Therefore, = ( 9) + 9 is the sum of two composite itegers

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

4 Questio 8 [p 87 #3] Let be a positive iteger greater( tha) 3 ad let p be a prime such that /3 < p Show that p does ot divide the biomial coefficiet Solutio: Note that the restrictios o are such that (a) p > (b) p ad p are the oly multiples of p which are less tha or equal to, sice 3p > (c) p itself is the oly multiple of p which is less tha or equal to From (a) ad (b) we have p ()! but p3 ()!, while from (c), Therefore, that is, p (!) but p 3 (!) p ()! (!), p ( ) Questio 9 [p 87 #4] Use Exercises ad 3 to show that if is a positive iteger, the there exists a prime p such that < p < (This is Bertrad s cojecture) Solutio: You ca easily check usig the Sieve of Eratosthees that the result holds for 7 Now let 8, ad suppose that there is o prime betwee ad Let be the prime power decompositio of ( ) = p rp p ( ) By assumptio there are o primes betwee ad, ( ) = p rp p If p is a prime with 3 < p, the < 4 3 < p < ad < 3p < 3, p divides! exactly oce ad p divides ()! exactly twice, ad so p ( )

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

6 Questio 0 [p 87 #6] (Extra Credit) Use Bertrad s postulate to show that every positive iteger with 7 is the sum of distict primes Solutio: First ote that the result is true for all positive itegers with 7 6 : 7 = + 5 = = = + 9 = = = = = 6 = Now let be a positive iteger with 7, ad let =, from Bertrad s postulate, there is a prime p with 7 7 < p, 5 7 or sice = +, 5 7 p Now let = p, ad if 7, we repeat the above procedure By Bertrad s postulate there exists a prime p with 5 7 p Now let 3 = p, ad if 3 7, we repeat the above procedure By Bertrad s postulate there exists a prime p 3 with p 3 Cotiuig i this way, at each stage if j 7, Bertrad s postulate guaratees the existece of a prime p j such that j 5 j 7 p j This process will stop whe j = k ad k+ 6 Now, j 5 ad p j implies that k+ = k p k = k p k p k = = j j 5 j 5 < j p j < j j 5 + p j +, k p i < j p j, i=j = j + 5

7 Therefore, k+ j p j j + 5, j + 5 j + 6 k+ < Also, sice for j k, the Therefore, the fial value k+ will satisfy j 7 p j j 7 k+ = k p k 7 7 k+ 6 = j 7, Note that the above argumet has show that for each j with j k, we have ( ) j 7 j + 6 j 8 j 5 p j+ 7 / < p j, ad the sequece {p j } will be decreasig with o duplicates Now we ote that j p j + 6 I fact, sice j 5 p j the there exists a real umber θ with 0 θ < such that j 5 j 5 = + θ, that is, ad sice j is a iteger, the j 5 θ p j, j p j θ < p j + 7, j p j + 6 Therefore, sice j > 6 for j k, the p j > 5, p k, the smallest of the p j s will be at least 7 Also ote that: if p j = 7, the j 0 ad j+ 3 if p j =, the j 8 ad j+ 7 if p j = 3, the j 3 ad j+ 9 Now all that is left is to show that ca be writte as a sum of distict primes whe k+ = 6, 5,, 9, 8, 7 ad we do this by cosiderig each of these cases i tur

8 case : If k+ = 6, the 6 = k p k p k + 6 p k = p k + 6, p k 0, ad sice p k is prime, this implies p k Sice 6 = = + 5, we oly eed to be cocered with the case that p k = ad p k = 3 If this happes, the k p k + 6 = = 3, ad k+ = k p k p k 3 3 = 8, which cotradicts the fact that k+ = 6, so we caot have both p k = ad p k = 3 Therefore, usig either 6 = or 6 = + 5 we have a partitio of ito distict primes case : If k+ = 5, the agai 5 = k p k p k + 6 p k, p k 9, ad sice p k is prime, this implies that p k Sice 5 = , we have a partitio of ito distict primes case 3: If k+ = 4, the agai 4 = k p k p k + p k, p k 8, ad sice p k is prime, this implies that p k Sice 4 = , we have a partitio of ito distict primes case 4: If k+ = 3, the 3 = k p k p k + 6 p k, p k 7 We caot have both p k = 3 ad p k =, sice this implies k p k + 6 = = 3, k+ = k p k p k 3 3 = 8 which is a cotradictio Usig either 3 = 3 or 3 = + we have a partitio of ito distict primes case 5: If k+ =, the = k p k p k + 6 p k, p k 6, ad so p k 7 if p k > 7, sice = 7 + 5, we have a partitio of ito distict primes if p k = 7, the k+ = k p k implies that k = + 7 = 9 We caot also have p k =, sice this implies that k + 6 = 8, k+ 8 7 = 0, which is a cotradictio Therefore, 9 = , ad we have a partitio of ito distict primes case 6: If k+ =, the = k p k p k + 6 p k implies that p k 5, so i fact we must have p k 7 We caot have both p k = 7 ad p k =, otherwise, k p k + 6 = 8, k+ = k p k p k 8 7 = 0, which is a cotradictio So if p k = 7 or p k >, we have a partitio of ito distict primes If p k =, the k+ = k p k imples that k = + =, ad we caot also have p k = 3, sice the = k p k p k 3 3 = 8, which is a cotradictio Thus, with = 3+7+ we have a partitio of ito distict primes case 7: If k+ = 0, sice p k 7 ad 0 = , we have a partitio of ito distict primes case 8: If k+ 9, ad p k = 7, the k = k+ + p k 6, which is a cotradictio, sice we costructed the sequece k 7 Therefore, we must have p k > 7, ad with 9 = 7 +, 8 = 5 + 3, or 7 = 7, we have a partitio of ito distict primes Note: This result was first prove by H E Richert i 950, a proof by iductio ca be foud i the book Elemetary Theory of Numbers by Sierpiński

9 Questio [p 87 #7] 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, + m, ad m m + + k= k ad the sum caot be a iteger i this case < }{{} times = =, Case : If m, from Bertrad s postulate, there is a prime p such that < p < + m Let p be the largest prime such that < p < + m, the + m < p, sice if ot, there would be a prime q with p < q < p + m, which cotradicts the choice of p Now suppose that m = N ( ) where N is a iteger Sice < p < + m < p, 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

10 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 p log for large lim p log =, Solutio: Suppose that the prime umber theorem is true, lim x if we let x = p for, the π(p ) =, π(x) x/ log x =, The atural logarithm is a cotiuous fuctio o R+, ad Now, sice lim log p = +, this implies that log p lim = (+) p lim {log + log log p log p } = 0, { log lim log p + log log p } = 0 log p log p { 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