PRIME NUMBERS YANKI LEKILI

Transcription

1 PRIME NUMBERS YANKI LEKILI We denote by N the set of natural numbers:,2,..., These are constructed using Peano axioms. We will not get into the hilosohical questions related to this and simly assume the usual roerties of natural numbers: There is an addition and multilication law on numbers. These satisfy the commutative, associative and distributive laws. There is an order on N so that either a < b or b < a for distinct natural numbers. Furthermore, every non-emty set in N has a smallest element, i.e. the order on N is a well-ordering. Finally, we shall aeal to the rincile of mathematical induction. We write Z for all integers: {..., 2,, 0,, 2,...} and Z 0 for non-negative integers.. Divisibility Definition. An integer a is divisible by b if there is a third integer c such that a = bc We write b a if b divides a and b a if b does not divide a. The relation is reflexive, a a; transitive, b a and c b imlies c a, but not symmetric, if b a then it is not usually the case that a b. In fact, if b and a are ositive integers and b a, then we have b a. Definition 2. A ositive integer is said to be a rime number (or simly a rime) if > and has no ositive divisors excet and. The first few rime numbers are: 2, 3, 5, 7,, 3, 7, 9, 23,... The rimes are the building blocks of numbers. The following theorem makes this recise: Theorem 3. Every ositive integer excet is a roduct of rimes. Proof. Let n N be a number. Either n is rime, when there is nothing to rove or n has divisors between and n. Let S be the set of divisors of n greater than. Then this set has a smallest element m. We claim that m is a rime. Otherwise, there would be natural number l with < l < m such that l m but since m n, by transitivity, we have that l n. We obtained an element of S, namely l, that is smaller than the smallest element of S. This is a contradiction. Hence, m must have been a rime. Therefore, n is either rime or divisible by a rime less than n, say in which case, n = n, < n < n

2 2 YANKI LEKILI Here n is either a rime, in which case the roof is comleted or it is divisible by a rime 2, in which case, we have n = n = 2 n 2, < n 2 < n < n Reeating the argument, we obtain a sequence of decreasing numbers n, n,..., n k,... The sequence stos when n k is rime for some k, and then we have: n = 2... k Note that i s in the above roof do not have to be distinct, we can grou them together and write: n = e e es s to get the rime factorisation of the integer n. For examle, we have: 666 = We will see later that the factors e i i are unique aart from rearrangement of factors. But, first we need to develo our understanding of division a little more. Lemma 4. (Division Algorithm) Given a Z and b > 0, there exists a unique q, r Z such that a = qb + r and 0 r < b. Proof. Consider the arithmetic rogression..., a 3b, a 2b, a b, a, a + b, a + 2b, a + 3b... extending indefinitely in both directions. Let S be the set of nonnegative elements in this list. S is non-emty: Either a is nonnegative then a S, or if a is negative,then a ab = a( b) 0 hence a ab S. Now, S is non-emty, hence has a smallest element r. Thus, by definition, we have r = a qb for some q and r 0. Also r < b because otherwise r b = a (q + )b would be an element of S that is smaller than r. Next, to rove uniqueness, let a = q b + r = q 2 b + r 2 satisfying the same conditions. Then (q q 2 )b = r 2 r. Taking absolute values, we get q q 2 b = r 2 r, hence b r 2 r, but 0 r, r 2 < b hence r 2 r < b. Hence, it must be that r 2 r = 0 and q 2 q = 0. In other words, r = r 2 and q = q 2. Definition 5. Let a, b N. The greatest common divisor of a and b is the greatest number d N such that d a and d b. We write (a, b) (or gcd(a, b)) for the greatest common divisor of a and b. The numbers a and b are said to be corime (or relatively rime) if (a, b) =. For examle, by listing all divisors of 2 and all divisors of 8, one can easily comute (2, 8) = 4 but soon we will do this in a much more efficient way. Theorem 6. If d = (a, b) then there exists integers x 0 and y 0 such that d = (a, b) = ax 0 + by 0

3 PRIME NUMBERS 3 Another way to state this fundamental result is that the greatest common divisor of a and b is a Z-linear combination of a and b. Proof. Consider the set S of all natural numbers of the form ax + by with x and y in Z. The set is non-emty, for examle it contains a and b. Hence, S has a smallest element m. So m is a natural number of the form m = ax 0 + by 0 for some integers x 0 and y 0. Every common divisor of a and b divides m, hence in articular d m. To conclude that d = m, we shall show that m a and m b. Using the division algorithm, write a = qm + r for 0 r < m. Let x = ( qx 0 ) and y = qy 0. Then, we have ax + by = a qax 0 qby 0 = a qm = r Hence, by the minimality roerty of m it follows that r = 0. This shows that m a, similarly we show that m b. Note that the integers x 0, y 0 are not uniquely determined. Indeed, given one solution (x 0, y 0 ) to d = ax 0 + by 0, we can obtain infinitely many other solutions as: d = a(x 0 + nb) + b(y 0 na) for n Z. The revious theorem gives a characterization of the greatest common divisor of a and b. Namely, it is least ositive integer value of ax + by where x and y ranges over all integers. But how to comute this value? (and the integers x 0, y 0? ) We shall use Euclid s algorithm. The crucial observation is the following lemma: Lemma 7. If a = qb + r then (a, b) = (b, r). Proof. If d is a common divisor of a and b, then d a qb = r, hence d is a common divisor of b and r. Conversely, if d is a common divisor of b and r, then d qb + r = a, hence d is a common divisor of a and b. Therefore, the set of common divisors of a and b agree with the set of common divisors of b and r, hence the greatest common divisors are the same. Now, given a, b N, Euclid s algorithm works as follows to determine (a, b). Without loss of generality, suose b < a, then we aly division algorithm to write a = q b+r with 0 r < b. If r 0, then we aly division algorithm to b and r to write b = q 2 r + r 2 with 0 r 2 < r. We reeat this until we find a remainder which is zero. (This must haen at some finite ste, since b > r > r Thus, we have a system of equations: a = q b + r, 0 < r < b b = q 2 r + r 2, 0 < r 2 < r r = q 3 r 2 + r 3, 0 < r 3 < r 2. r n 3 = q n r n 2 + r n, r n 2 = q n r n + r n, r n = q n r n < r n < r n 2 0 < r n < r n

4 4 YANKI LEKILI We aly the above lemma reeatedly to deduce Thus, we roved: (a, b) = (b, r ) = (r, r 2 ) =... = (r n 2, r n ) = (r n, r n ) = (r n, 0) = r n Theorem 8. The last non-zero remainder r n of Euclid s algorithm is the greatest common divisor of a and b. Here is an examle. Let us comute (87, 35). We have 87 = = =. + =. Thus, we see that (87, 35) =. Thus, we should be able to find integers x 0 and y 0 such that 87x y 0 = Euclid s algorithm also gives a way to do this. Namely, we have = 2. = 2.(35 2.2) = (35 2.( )) = Indeed, one can use Euclid s algorithm to give another roof of Theorem 6. We will now return back to factorisations of natural numbers into rimes. We start with the following: Theorem 9. (Euclid s first theorem) Let be a rime number and let a, a 2 N. If a a 2 then a or a 2. More generally, if a rime divides a roduct a a 2... a n, then divides a i for some i. Proof. The case of a roduct with n factors follows easily from the case with two factors. Suose is a rimes and a a 2. If a then (a, ) = and therefore, by Theorem 6, there are an x 0 and a y 0 for which Multilying this by a 2 gives x 0 a + y 0 = x 0 a a 2 + y 0 a 2 = a 2 Now, x 0 a a 2 and y 0 a 2, hence a 2. Let s use this to rove a theorem due to Pythagoras. Theorem 0. 2 is irrational.

5 PRIME NUMBERS 5 Proof. If 2 is rational, we can write it as 2 = a for integers a, b such that (a, b) =. b Then, a and b satisfy the equation: a 2 = 2b 2 Hence, b a 2. Therefore, a 2 for any rime factor of b. It follows from Theorem 9 that a. But, this is contrary to the assumtion that (a, b) =. Hence b =, and this also is clearly false. We now come to one of the main tools of elementary number theory: Theorem. (Fundamental theorem of arithmetic) Every ositive integer a > has a factorisation into rime factors as a = e e es s, and aart from rearrangement of factors, this factorisation is unique. Proof. We have already seen the existence of a factorisation in Theorem 3. Now, we show uniqueness. Suose that a =... k = q... q j are two rime factorisations of a. Then, by Theorem 9, q i for some i. Since q i is a rime, this imlies that = q i. We can then divide out and q i from both sides to get two rime factorisations of a/ = 2... k = q... q i q i +... q j. We can then match 2 with q i2 for some i 2 by the same argument. Continuing this way, we get that for all s, s = q is for some i s. After cancelling out all of, 2,... k, the remaining roduct must equal. Hence, there are no remaining factors on the right hand side either. Hence k = j and the matching = q i, 2 = q i2,... k = q ik shows that the two factorisations differ by only a rearrangement of factors. It is now clear why should not be counted as a rime. If it were, then Theorem would be false, since we could insert any number of unit factors. If we know the rime factorisation of ositive integer a then we can immediately write down all ositive divisors: if a = e e e k k then b a if and only if b has a rime factori- with 0 f i e i for all i. This observation gives the sation of the form b = f f f k k following lemma: Lemma 2. If m, n N are corime then every natural number d with d mn can be written uniquely as d = d d 2 where d, d 2 N and d m and d 2 n. Proof. Since m and n are corime, they don t have any rime factors in common. So, m = k... kr r and n = k r+ r+... k r+s r+s where all the i are distinct. If d mn then d = l... l r+s r+s with 0 l i k i for i r +s. Let d = l... lr r and d 2 = l r+ r+... l r+s r+s. Then, obviously d = d d 2, d m and d r n. Conversely, if d = d d 2, d m and d 2 n then we must have d = l... l r r and d 2 = l r+ r+... l r+s r+s. From d = d d 2 it follows that l i = l i for i r + s and therefore d = d and d 2 = d 2. This shows uniqueness. We will also need the following lemma. Lemma 3. If, 2... r be distinct rime numbers and let n be any integer. If i n for all i then 2... r n.

6 6 YANKI LEKILI Proof. For every r, we must show the following statement: If, 2..., r are distinct rimes, n is any integer and i n for i =, 2,..., r, then 2... r n. To do this, we use induction on r. The case r = is obviously true. Now assume r 2 and that we have shown the statement for r. Let..., r be distinct rimes and n an integer such that i n for all i. By induction hyothesis, we get that 2... r n. So, we can write n = 2... r m for some m Z. Now, since r n and r i for i r, it follows, by Theorem 9, that r m. Hence, it follows that 2... r 2... r m = n as desired. Exercise: Give an alternative roof of Lemma 3 using FTA. Finally, let us discuss factorials and their rime factorisations. natural number N, we have the notation: Recall that given a N! := N i =.2... (N ).N i= This number is equal to the number of ermutations of a set of N elements. Let us observe that if N!, then must divide one of the numbers, 2,..., N and therefore N. On the other hand, every rime number N is a rime factor of N!. So, to find a rime factorisation of N!, we need to determine the exonent of each rime N which divides N!. Let us write this first as: N! = N e where the roduct is over all N and e are non-negative integers. To state the next result, we find convenient to introduce the following notation: Definition 4. For any real number x R, one signifies by [x] the largest integer x, that is, the unique integer such that x < [x] x. This function is called the integral art of x. Lemma 5. N! = N e where e = m= [ N m ] Note that the sum m= [ N m ] has only finitely many non-zero terms because if m > N, then [ N m ] = 0. Proof. Consider a rime number N. We must count how often aears in the roduct N! =.2..N. Clearly, [ N ] of the factors, 2,..., N are multiles of ; [ N ] 2 factors are multiles of 2 etc. Hence, in e = [ N ] + [ N ] +... we have counted once the 2 number of factors which are divisible by but not 2 (as art of [ N ]), we have counted twice the number of factors which are divisible by 2 but not 3 (as art of [ N ] + [ N ]) etc. 2 This comletes the roof.

7 PRIME NUMBERS 7 Note that it follows easily from above that N e [ ] for the sum m= [ N m ] < m= N m = N( ) = N. As an examle, let us find the largest integer k such that 7 k 50!. We can comute this as: k = [ 50 7 ] + [50 7 ] = 7 + = Comutational roblems. Lemma 6. A ositive integer n is comosite if and only if n has a rime divisor n Proof. If n is comosite then n = ab with < a < n and < b < n. We can assume that a b. Let be a rime factor of a. Then is also a rime factor of n and a = a.a a.b = n. A rimality test is an algorithm that determines whether an integer n > is a rime or comosite. The above lemma gives the following test. For every rime n test whether n is divisible by or not. We know that if n for some n then n is comosite, otherwise n is rime. The sieve of Eratosthenes is the method of comuting list of rimes u to a number n by using this algorithm. Write down all the integers n and cross out multiles of 2, then cross out multiles of 3 and continue until we cross out multiles of all rimes n. Then the remaining numbers are rime. This method is useful for small numbers but, of course, it is clear that it is not very efficient for large numbers. In 2002, Agrawal, Kayal and Saxena develoed the first olynomial time rimality test (known as the AKS rimality test). Polynomial time means that there exists constants C, k such that for every integer n > the algorithm needs at most C.(log n) k many stes to decide whether n is rime or not. Note that AKS test determines whether n is rime or not without finding a rime factor. 2. Basic distribution results Recall that fundamental theorem arithmetic leads us to think that rime numbers are building blocks of all numbers. We begin with the famous result of Euclid that says that there are infinitely many rimes: Theorem 7. (Euclid s second theorem) The number of rimes is infinite. We will give two roofs of this result. Here is Euclid s own roof: Proof. Let 2, 3, 5,..., be the list of rimes u to, and let q = Then q is not divisible by any of the numbers 2, 3, 5,...,. It is therefore either rime, or divisible by a rime between and q. In either case, there is a rime greater than, which roves the theorem.

8 8 YANKI LEKILI To study the distribution of rime numbers among all natural numbers, we give the following definition: Definition 8. For a real number x R, we let π(x) = the number of rimes that are not greater than x For examle, π(7) = 4, π(0) = 4, π(π) = 2, π() = 0. Note that if we know the function π then we also know all the rime numbers: an integer n is rime if and only if π(n) > π(n ). Obviously, π(x) x for all x 0. Euclid s second theorem imlies that π(x) as x. Can we show that π(x) is given by a formula in terms of more familiar functions? Let us try to deduce a bit more from Euclid s argument. Let n denote the n th rime. So = 2, 2 = 3,... etc.. Let us define q = n + Then, since i < n for i =,..., n, we deduce that for n >, and so that q < n n + n+ < n n + as either q = n+ or there is a rime bigger than,..., n that divides q. In fact, we can do a bit better. Suose that n < 2 2n for n =, 2,... N, then Euclid s argument shows that N N + < N + < 2 2N+ Since, = 2 < 2 2 = 4, by induction we conclude that n < 2 2n, for all n. Therefore, we have that π(2 2n ) n. Now, given any ositive real number x, we can sandwich it as e en < x e en, for some n since the series e em for m =, 2,..., is monotonically increasing. In other words, we have: n < log log x n Let us suose n 3, so that e n > 2 n and so e en > 2 2n, then we see π(x) π(e en ) π(2 2n ) n log log x Note that the only lace where we used the assumtion n 3 was in deducing π(e en ) n. This can easily be checked directly for n = and n = 2 as well. Hence, we roved that π(x) log log x for all x > e. We also note that log log x 0 for < x e, hence we get that: Theorem 9. π(x) log log x, for x >.

9 PRIME NUMBERS 9 However, log log x is a rather weak bound. For examle, for x = 0 9, it gives π(x) 3, whereas the value of π(x) is over 50 million. Figure shows the grah of π(x) for 0 x 00. Figure. Grah of π(x) for 0 x 00 To draw this for yourselves, go to Mathematica and tye Plot[PrimePi[x],{x,0,00}]. Around 800, several mathematicians conjectured aroximations for π(x) as x. Legendre suggested in 798 that π(x) is aroximately equal to the function x. Figure log x 2 shows how π(x) and x comares for x 400. log x It is close but does not quite match. Of course, we are able to see this much easily today with the hel of comuters. A better aroximation to π(x) is rovided by the logarithmic integral, defined by: Li(x) := x 2 dt log t Here is the Mathematica code for laying with these functions: = P lot[p rimep i[x], {x, 2, 00000}, ImageSize Large] 2 = P lot[x/log[x], {x, 2, 00000}, P lotstyle {Green}, ImageSize Large] 3 = P lot[logintegral[x], {x, 2, 00000}, P lotstyle {Red}, ImageSize Large] Show[, 2, 3] We will return to this interesting subject of asymtotic aroximation of π(x) later in the next section.

10 0 YANKI LEKILI Figure 2. Grah of π(x) vs. x/ log x for 2 x 400 Let s discuss a second roof of Euclid s theorem. The roof uses a secial sequence of rather fast growing numbers called Fermat s numbers. They are defined by: so that F n = 2 2n + F = 5, F 2 = 7, F 3 = 257, F 4 = It can be checked easily that F, F 2, F 3, F 4 are rime numbers. Fermat conjectured that all F n are rimes but in fact this was disroved by Euler in 732 when he showed that F 5 = = = An easy roof was given by Coxeter. Indeed, 64 = = hence divides each of = 2 28 ( ) and (since x 4 = (x + )(x )(x 2 + )). So, it divides their difference. In fact, there is no known rime number of the form F n for n > 4. Peole are still searching for them (why?). Here is a website for it: htt:// However, we have a more theoretical use of Fermat numbers in mind. Namely, let us rove the following: Theorem 20. (Goldbach) No two Fermat numbers have a common divisor greater than.

11 PRIME NUMBERS Proof. Suose that F n and F n+k where k > 0 are two Fermat numbers and that m F n and m F n+k. Letting x = 2 2n we can write: F n+k 2 F n 2 2n + = x2k x + = x2k x 2k 2 + x 2k 3... = 22n+k So, we see that F n F n+k 2, but now since m F n+k and m F n+k 2, we deduce that m 2. So m = or 2 but it can t be 2 since all the Fermat numbers are odd. Hence, the theorem follows. Corollary 2. There are infinitely many rimes. Proof. Each F i is either rime or have an odd rime divisor which does not divide any other. So, there are at least as many rimes as there are Fermat numbers. There are infinitely many Fermat numbers by definition. This roof also gives n+ F n = 2 2n +, which is slightly better than what we have seen before (but not much). Can one find a simle function f : N N such that for each natural number n, f(n) is a rime number? Clearly, Fermat s sequence could not do this job. There is no known satisfactory answer to this. Of course, one could define f(n) = n, the n th rime number but this is by no means a simle function. There is a remarkable olynomial function given by f(n) = n 2 n + 4 It turns out this olynomial takes rime values for all n with 0 n 40, but obviously f(4) is comosite, since 4 f(4). The following theorem says that olynomial functions with integer coefficients are no good for answering the above question. Theorem 22. No olynomial f(n) with integral coefficients, not a constant, can be a rime for all n, or for all sufficiently large n. Proof. Consider a olynomial given by f(x) = a 0 x k + a x k +... a k We may assume that the leading coefficient a 0 > 0 so that f(n) as n since otherwise f(x) will become negative when x is big. Thus, f(x) > for all x sufficiently large, so say f(x) > for all x > N. Let x 0 > N be such a number, and let us take Then, for all integers r Z, we see that y = f(x 0 ) >. f(ry + x 0 ) = a 0 (ry + x 0 ) k +... is divisible y by the binomial exansion theorem. Now, f(ry +x 0 ) as r. Hence, there are infinitely many comosite values of f(n).

12 2 YANKI LEKILI What about simle functions f : N N such that f(n) is rime for infinitely many n? If f(n) is of the form f(n) = an + b for some a, b with a > 0 and (a, b) = then there is a nice answer to this. Let s first ractice in a few examles. Theorem 23. There are infinitely many rimes of the form 4n + 3. Proof. Let,..., n be the first n rimes, then consider: q = n Then q is of the form 4n + 3, and is not divisible by any of the rimes,..., n, hence it should have rime divisors > n. Furthermore, it cannot be that all the rime divisors of q are of the form 4n + since the roduct of such numbers is of the same form, hence there must be at least one rime divisor of q of the form 4n + 3 and greater than n. By letting n, we can construct infinitely many rimes of this form. Here is a similar result: Theorem 24. There are infinitely many rimes of the form 6n + 5. Proof. The roof is similar. We define q by q = n and observe that any rime number, excet 2 or 3 is of the form 6n + or 6n + 5 (why?), and the roduct of two numbers of the form 6n + is again of the same form. All these theorems are articular cases of a famous theorem of Dirichlet: Theorem 25. (Dirichlet 837) If a > 0 and b are integers such that (a, b) =, then there are infinitely many rimes of the form an + b. The roof of this theorem uses analytical methods too difficult to discuss here. We shall not cover its roof in this course. That deals with the linear functions. What about quadratic olynomials? The question becomes much harder and we don t even know whether the following conjecture is true or not. Conjecture 26. There are infinitely many rimes of the form n Chebyshev s theorem We now return back to our study of the rime distribution function π(x). We will give a roof of a theorem due to Chebyshev (also selled Tchebychef): Theorem 27. There exists constants c, c 2 > 0 such that x c log x < π(x) < c x 2 log x

13 PRIME NUMBERS 3 The closer the constants c j are to, the more technical the roof becomes. Here we will show that c = log 2 and c 2 2 = 6 log(2). The roof will use the following elementary facts: () 22n ( ) 2n 2n n 2 2n (2) ( ) 2n n is not divisible by any > 2n. (3) ( ) 2n n is divisible by all rimes n < 2n. The first one follows from ( + ) 2n = 2n ( 2n m=0 m), the second and third follows from our formula for the rime factorizations of factorials (2n)! and n!. Proof. (Proof of Chebyshev s theorem) Uer bound: Any with n < 2n divides ( ) 2n n so by Lemma 3, the roduct divides ( 2n n ). Therefore, we have n π(2n) π(n) n< 2n n< 2n ( ) 2n 2 2n n Taking the log gives n π(2n) π(n) 2 log(2) log n Using induction, we now easily see that π(2 k ) 3 2k k In fact, this is checked directly for k 5; k > 5, we argue by induction: π(2 k+ ) π(2 k ) + 2k+ 3.2k k k + 2.2k k 5.2k k 3.2k+ k + Next, since the function f(x) = x is monotonically increasing for x e, we have that log x if 4 2 k < x 2 k+, then Since π(x) 6 log 2 x log x bound for all x. Lower bound π(x) π(2 k+ 2 k ) 6. k + 6 log 2 2k log 2 6 log 2 x k log x for x 4 as well, we have now established the claimed uer Put N = ( ) 2n n, let v (N) denote the highest ower of that divides N. By the formula from Lemma 5, we now that v (N) = m Now, we use the following lemma: ([ ] 2n m [ ]) n 2 m

14 4 YANKI LEKILI Lemma 28. For all x R, we have [2x] 2[x] {0, }. Proof. Let us write x = [x] + {x} where {x} is called the fractional art of x. Now, if {x} < /2, then 2x = [2x] + {2x}, hence [2x] 2[x] = 0. Otherwise, if {x} /2, then we get [2x] 2[x] =. [ ] [ ] If m log 2n > 2n or equivalently m >, then we have that 2n n 2 = 0. Thus, we log m m find that [ ] log 2n v (N) log Now, log ( ) 2n n 2n 2n log 2 log 2n log ( 2n ), because 22n n 2n ( ) 2n n [ ] log 2n log, because N = v(n), and v (N) log 2n 2n This yields the lower bound We claim that this imlies that [ ] log 2n log log 2n = π(2n) log 2n log 2n π(2n) log 2 2n log 2n π(x) log 2 2 x, for all x 2 log x [ ] log 2n log This inequality can be checked directly for x 6, hence it suffices to rove it for x > 6. Pick an integer n with 6 2n < x 2n + 2. Then, hence, as required. π(x) π(2n) log 2 2n log 2n 2n log 2n n + log 2n = n log 2n 7 4 log 2 > log 2 (n + ) log 2 log(2n) (n + ) log 2 log(2n + 2) log 2 x 2 log x We will next rove Bertrand s ostulate. It was conjectured by Bertrand in 845 and roved by Chebyshev in 850. Theorem 29. For every integer n, there is a rime satisfying n < 2n.

15 PRIME NUMBERS 5 Chebyshev introduced an auxiliary function, the θ-function. It is defined by θ(x) = x log for real numbers x (summation over all rime numbers x). For examle, θ(0) = log 2 + log 3 + log 5 + log 7 Chebyshev roved uer and lower estimates for the function θ, and then deduced uer and lower estimates for the function π. We have the following uer estimate for θ. Lemma 30. If x > 0, then θ(x) < log(4) x. Proof. The statement is clearly true for 0 < x <, so we can assume that x. Since θ(x) = θ([x]), it is enough to rove the statement θ(n) < log(4) n for n N. For this, we use induction on n. The cases n = and n = 2 are obviously true. Now, assume that n 3 and that θ(m) < log(4) m for m < n. We must distinguish the cases n even and n odd. If n is even then θ(n) = θ(n ) < log(4) (n ) < log(4) n, as required. If n is odd, let n = 2m + for m. We will show below that θ(2m + ) θ(m + ) < log(4) m. It then follows that θ(n) = θ(2m+) θ(m+)+θ(m+) < log(4) m+log(4) (m+) = log(4)(2m+) = log(4) n as required. So, it remains to show that θ(2m + ) θ(m) < log(4) m for every m. Consider M = ( ) 2m+ m = (2m+)2m (m+2). If is a rime number with m+2 2m+, m! then divides M (because divides the numerator but not the denominator). Hence, by Lemma 3, the roduct m+2 2m+ divides M, in articular, is less than or equal to M. On the other hand, M < 2 2m because ( ) ( ) 2m + 2m + 2M = + < ( + ) 2m+ m m + It follows that θ(2m+) θ(m+) = m+2 2m+ log = log ( m+2 2m+ ) log M < log 2 2m = m log 4 Proof. (Proof of Bertrand s ostulate) Recall first that any rime number that divides N = ( 2n n ) has to satisfy 2n and if there is any rime n < 2n then it divides N.

16 6 YANKI LEKILI Next, let us observe that for n 3 if 2n < n, then does not divide N = ( ) 2n 3 n. Indeed, 2n < 3 2, hence 2 2n < 3. Thus, [ ] [ ] 2n n v (N) = 2 = 2 2 = 0 Now, we rove Bertrand s ostulate by contradiction. Suose that there is an integer n and there is no rime in the interval (n, 2n]. By the discussion above, this imlies that there is no rime > 2 n that divides N. 3 Next, consider rimes N, such that v (N) >. They satisfy 2 v(n) 2n, hence we must have 2n for such rimes. The number of such rimes is clearly bounded by 2n. Now, we have 2 2n 2n Taking logs, we get: ( ) 2n = v(n) n v (N)> 2n(log(2)) log(2n) 2n log(2n)+log( v (N)= v (N)= (2n) 2n v (N)= ) 2n log(2n)+θ( 2n 3 ) 2n log(2n)+log(4)( 2n 3 ) Reorganizing, we get 2n log(2) 3( + 2n) log(2n) x Now, since is monotonically increasing for x > 3, this inequality cannot hold log x for large n. In fact, it is false for n 52 and we arrive at a contradiction if n 52. For n < 52, Bertrand s ostulate is roved by looking at the sequence of rimes 7, 3, 23, 43, 83, 63, 37, 63. Corollary 3. Let n denote the n-th rime number. Then n 2 n. Proof. By Bertrand s ostulate we know that each of the intervals (, 2], (2, 4], (4, 8], etc. contains at least one rime number. Hence the interval (, 2 n ] contains at least n rime numbers. Thus the n-th rime number must be contained in this interval, i.e. n 2 n. 4. The rime number theorem We have mentioned before that π(x) is aroximately equal to the function will now make this statement more recise. x. We log x Definition 32. Let f and g be functions which are defined for all sufficiently large real numbers, and assume that f(x) and g(x) are ositive for all large x. We say that f and g are asymtotically equal, and write this as f g if f(x) g(x) as x, i.e. the limit lim x f(x)/g(x) exists and is equal to. We can now state one of the landmark theorems in elementary number theory:

17 PRIME NUMBERS 7 Theorem 33. (Prime number theorem) π(x) x log x The rime number theorem was roved in 896 indeendently by Hadamard and de la Vallée Poussin using methods of comlex analysis. An elementary roof was given in 948 by Selberg and also by Erdös (based on Selberg s lemma). We shall not cover the roof of this theorem in this class but if you are seriously interested in number theory you should learn its roof. Corollary 34. Prime number theorem imlies Chebyshev s theorem. Proof. lim x f(x)/g(x) = means for any numbers c < < c 2 and sufficiently large x, we have that c < f(x)/g(x) < c 2 Hence, c g(x) < f(x) < c 2 g(x) as x Here is another corollary of the rime number theorem Corollary 35. Let n denote the n-th rime number. Then (Here n n log n means that lim n n n log n as n n n log n =.) Proof. If y = x, then log y = log x log log x. Therefore, log x log y lim x log x = lim log log x x log x = Thus, x = y log x y log y as x. Since, by the rime number theorem, π(x) y, it follows that x π(x) log π(x). In other words, π(x) log π(x) lim = x x Now, let x = n, then π(x) = n, so we get n log n lim = n n as required. In fact, it is not too hard to show that the statement given in the above corollary is equivalent to the rime number theorem. Finally, we shall give a corollary of the Prime Number Theorem that recovers an asymtotic version of Bertrand s ostulate. Corollary 36. Let δ >, then for all sufficiently large x, the interval (x, δx] contains a rime number.

18 8 YANKI LEKILI (Note that this is not necessarily true for all x, for examle (, δ ] does not contain a rime for δ < 2.) Proof. The interval (x, δx] contains a rime number if and only if π(δx) π(x). Thus, we must show that π(δx) π(x) for all sufficiently large x. We have π(δx) π(x) = π(δx) x/ log(x) δx/ log(δx) δx/ log(δx) π(x) x/ log(x), By the rime number theorem, we see that the first two factors go to as x. For the third factor, we find δx/ log(δx) x/ log(x) log(x) = δ log(δ) + log(x) δ, as x. Hence, lim π(δx)π(x) = δ. x Now, fix any γ with < γ < δ. From lim x π(δx) π(x) = δ it follows that π(δx) γπ(x) for all sufficiently large x, and from π(x) as x, it follows that (γ )π(x) for all sufficiently large x. Thus for all large enough x, we obtain π(δx) π(x) (γ )π(x) as required. A refinement of the aroximation given in the rime number theorem can be obtained if instead of x/ log x one uses the following function: Definition 37. For x 2, define li(x) = x 2 log t dt The function li(x) is called the logarithmic integral. One can then show that li(x) π(x) The bound on error terms of these aroximations is still an imortant area of research in number theory. We mention the following imortant conjecture: Conjecture 38. (Riemann hyothesis) Let ɛ > 0. Then, π(x) li(x) < x /2+ɛ

19 PRIME NUMBERS 9 5. Arithmetic functions and Dirichlet series Definition 39. A real- or comlex-valued function defined on the ositive integers is called an arithmetic function. Definition 40. Given an arithmetic function f(n) = α n C, we define its Dirichlet series: α n F (s) = n s The most imortant examle of a Dirichlet series is the Riemann zeta function associated to the arithmetic function u defined by u(n) = for all n N. Definition 4. The Riemann zeta function, denoted by ζ(s), is the function of a real variable s > defined by the series ζ(s) = n s We must check that the series converges for s >. Since all summands in the series are ositive, it suffices to check that the artial sums N n s are bounded above. These artial sums can be estimated as follows: N N N n s = + n s < + x s dx < + x s dx = + 0 (as s ). s n=2 Here the inequality N n=2 n s < N x s dx can be seen by comaring the area under the curve x s for x N to the sum of the areas of the rectangles of width and height 2 s, 3 s,... N s under this curve. The following imortant result shows how Riemann zeta function is related to rime numbers: Theorem 42. (Euler roduct formula) If s > then Proof. Since 2, we have ζ(s) = s s = + s + 2s +... for s > (indeed for s > 0). If we take = 2, 3,... q, and multily the series together, the general term resulting is of the tye where 2 a 2s 3 a 3s... q aqs = n s, n = 2 a 2 3 a 3... q aq (a 2 0, a 3 0,..., a q 0)

20 20 YANKI LEKILI A number n will occur if and only if it has no rime factors greater than q, and then by Fundamental theorem of arithmetic, once only. Hence, = n s s q (q) the summation on the right-hand side extending over the numbers formed from the rimes u to q. These numbers include all the numbers u to q, so that we have 0 < n s < n s n s (q) and the last sum tends to 0 when q. Hence, n s = lim n s = lim q q s (q) q q+ Note that the essential ste in the roof was the existence of a unique rime factorisation for every ositive integer. Therefore, the Euler roduct can be considered as an analytic exression of the fundamental theorem of arithmetic. Proosition 43. ζ(s) as s and s >. Proof. For every s > the integral x s dx is smaller than the sum n s. Thus for every s >, we have ζ(s) = n s > x s dx = s Since s as s, s >, it follows that ζ(s) as s, s >. Note that from this it follows easily that there are infinitely many rimes (This roof is due to Euler). Suose there were finitely many rimes, then we have: lim ζ(s) = lim ( s ) = lim ( s s s s ) = ( ) contradicting ζ(s) as s. Corollary 44. The series is divergent. Proof. Recall that log x = x + x2 2 + x for x <. Alying this to s for s >, we have: log s = s + 2s 2 + 3s

21 PRIME NUMBERS 2 Taking the logarithm of ζ(s), we get: = ( log ) s log ζ(s) = log = s ) ( s + 2s 2 + 3s = s + Now, we observe that the second sum is bounded above by. Namely, k 2 ks k < < n=2 k 2 k = 2 k k = n 2 n = ( n ) =. n n=2 k 2 ks k 2 Hence, s > log ζ(s) for all s >. Since ζ(s) as s with s >, this imlies that s as s. Now, we have > s for all s >, hence it follows that is divergent. We have seen that the trivial arithmetic function u with u(n) = for n N gives rise to the imortant Riemann zeta function. Here are some imortant arithmetic functions: For n N, () u(n) =. (2) d(n) = #{d N : d n} = d n. (3) σ(n) = d n d. (4) σ k (n) = d n dk. µ() =, (5) (Möbius function) µ(n) = µ(n) = 0, if 2 n for some rime µ(n) = ( ) s, if n =... s where i are distinct. (6) (Euler s totient function) φ(n) = #{d N : d < n, (d, n) = }. Definition 45. An arithmetic function f is multilicative if f(mn) = f(m)f(n) whenever (m, n) =. Observe that a multilicative function is uniquely determined by its values on owers of rime numbers. Indeed, if n = e e e k k and f is multilicative, then f(n) = i f( e i i ) Clearly, the function n u(n) = is multilicative. We shall next see that d(n) is also multilicative by the following lemma:

22 22 YANKI LEKILI Lemma 46. Let f : N C an arithmetic function. g : N C by the formula g(n) = f(d). d n Define the arithmetic function If f is multilicative, then so is g. Proof. Let m, n N, then g(mn) = d mn f(d) = d m,d 2 n f(d d 2 ) = d m f(d ) d 2 m f(d 2 ) = g(m)g(n). Corollary 47. n d(n) = d n is multilicative. n σ k(n) = d n dk is multilicative. It is easy to check that µ(n) is multilicative. As art of the homework, you will see that φ(n) is also multilicative. Maniulations of Dirichlet series. Dirichlet series allows one to study an arithmetic function via comlex analysis. We shall not delve into analysis here but discuss basic maniulations of Dirichlet series. We shall not care about convergence questions. To justify the rearrangement of terms one often requires absolute convergence of the series. Lemma 48. Let F (s) = a(n) n s where c(n) = d n and G(s) = F (s)g(s) = a(d)b(n/d) = d n c(n) n s b(n), then their roduct is given by n s a(n/d)b(d) Proof. The roduct is F (s)g(s) = where we set n = dk. d= k= Lemma 49. Let F (s) = a(d)b(k) d s k s = d n a(d)b(n/d) n s = a(n). Then F (s) = b(n) n s n s b(n) = log(n) a(n). Proof. Since d ds (n s ) = log(n) n s, we have F (s) = d a(n) = ds n s where log(n) a(n) n s. c(n) n s

23 PRIME NUMBERS 23 Examles: The Dirichlet series associated to the identity function N N, n n is given by ζ(s) ζ(s) = σ(n) n s d(n) n s n n = s. Indeed, ζ(s) ζ(s) = n s = ζ(s )ζ(s). Indeed, n ζ(s )ζ(s) = n s ζ (s) = log(n). n s = ζ(s ). ns n = s n = s d n n s = d n d. n s = d(n) n s. σ(n) n s Lemma 50. Let ɛ be the arithmetic function defined by ɛ() = and ɛ(n) = 0 for all n >. We have: µ(d) = ɛ(n) d n Proof. It is easy to see that by definition µ is multilicative. Therefore, the function n d n µ(d) is also multilicative. Thus, it suffices to comute its value on rime owers. We have { µ( k, if k = 0 ) = , if k > 0 d k Corollary 5. µ(n) =. n s ζ(s) Proof. We have µ(n) n s n = s d n µ(d) n s =. In a similar way, using d n φ(n) = n, one shows that We end with Möbius inversion formula: φ(n) n s = ζ(s ) ζ(s) Theorem 52. For arithmetic functions f and g, the following statements are equivalent: () g(n) = d n f(d) for all n.

24 24 YANKI LEKILI (2) f(n) = d n µ(d)g(n/d) = d n µ(n/d)g(d) for all n. Proof. Assume that () holds, then µ(d)g(n/d) = d n d n µ(d) c n d f(c) = c n f(c) d n c µ(d) = f(n) where we used the fact that a air of ositive integers (c, d) satisfies d n and c n if and d only if c n and d n and that c d µ(d) = 0 unless c = n. n c Assume that (2) holds, then f(d) = µ(c)g(d/c) = µ(d )g(c) = g(n) d n d n c d c n where we used the fact that a air (c, d) of ositive integers satisfies d n and c d if and only if the air (c, d ) = (c, d/c) satisfies c n and d n c. d n c