Math 5330 Spring Notes Prime Numbers

Transcription

1 Math 5330 Sring 208 Notes Prime Numbers The study of rime numbers is as old as mathematics itself. This set of notes has a bunch of facts about rimes, or related to rimes. Much of this stuff is old dating back years. We start with ossibly the most imortant question about rimes: how many are there? We have an ancient method for calculating lists of rimes, which is still unbelievable good. It is called the Sieve of Erotosthenes. The idea is as follows: Form an array of numbers, 2, 3,.... Ignore because it is a unit. Circle 2, and cross out every subsequent multile of 2. Suose that the last number in your array is n. Reeat the following until you circle a number greater than? n : Find the next number not crossed off, circle it, and cross off every subsequent multile. Once a number greater than? n is circled, dont cross off any additional numbers, just circle the remaining numbers not crossed off, they will all be rime. For examle, to find all rimes less than 00 takes 4 stes:

2 This algorithm is very fast. Ive heard it said that if you are writing a rogram that makes use of all rimes less than,000,000 it is faster to use the sieve of Erotosthenes than to read in a reexisting file of rimes. One thing that makes the algorithm so fast is that we can sto sieving once we reach? n. The reason for this is the following theorem. Theorem If n is not rime, then n has a rime divisor ď? n. Proof: Let n be comosite and let q be a rime divisor of n, so n qn{qq. If q ď? n, then let q, in the theorem. Otherwise, n{q ă? n, so let be any rime divisor of n{q. As a consequence of this theorem, we have the following factoring technique: To factor a number n, try dividing n in turn by the rimes 2, 3, 5,.... We continue until a rime gets larger than the square root of the unfactored art. For examle, suose we wish to factor We divide by 2 until an odd number results: The unfactored art is not divisible by 3, but it is by 5, so we have Continuing, we have Note that after dividing by 3 to leave an unfactored art of 0, we can sto because 0 ă 3 2. This means that 0 must be rime. At this oint, we have a good way to find all small rimes, and a reasonable algorithm for factoring small numbers. We still have not answered the question of how many rimes there are. Consider following table: range ` ` rime count This table gives the number of rimes in ranges of 00 consecutive integers. Based on the table, one might exect that the number of rimes is finite and that there is some largest rime. However, this is not the case. The following is a result due to Euclid: Theorem 2 There are infinitely many rimes. Proof: Suose not, and let t2, 3,..., k u be a comlete list of rimes. Let M be the number M k `. (M is one greater than the roduct of all rimes.) When M is divided by 2 or 3 or... or k, the remainder will be. Thus, M is not divisible by any of the rimes Page 2

3 in our list. But the Fundamental Theorem says M is divisible by some rime. This contradicts the assumtion that our list was comlete, which comletes the roof of the theorem. Given the first k rimes, we can define the number M k 2 3 k `. The first several are M 3, M 2 7, M 3 3, M 4 2. These first ones are rimes and it is a common conjecture by students that M k is always rime. But this is not the case: M It turns out that rimes are fairly common. We give a name to the number of rimes u to some bound: πnq the number of rimes ď n. For examle, π00q 25 because there are 25 rimes less than 00. In 793 at the age of 5, Gauss made the following conjecture about how common rime numbers are: πxq «ż x 2 lntq dt. This integral has a name, it is called lixq, the logarithmic integral. In 896, this conjecture was roved roughly simultaneously by Hadamard and de la Vallée Poussin. The exact statement of the result is this: Theorem 3 (The Prime Number Theorem or PNT) πxq lim xñ8 lixq. It turns out that lixq «x, so we usually say πxq «x lnxq htt://mathworld.wolfram.com/primenumbertheorem.html. Here is a grah from lnxq Page 3

4 The following table also gives a feel for how good these aroximations are. n n πnq lnnq linq ,000,000 78,498 72,382 78,626,000,000,000 50,847,478 48,254,942 50,849,234 It took about 00 years to rove the Prime Number Theorem because it took that long to develo the necessary results in Comlex Analysis. What does the field of analysis have to do with rime numbers? To give a feel, here is a second roof that there are infinitely many rimes, making use of analysis. This roof will also show how logarithms might enter into the icture, and how many different areas of mathematics seem to be related to number theory. Our actual theorem is the following: The sum ÿ 2 ` 3 ` 5 ` 7 ` ` diverges. This shows there are infinitely many rimes because any finite sum would converge. Moreover, it shows that rimes must be fairly common, much more common than squares, 8ÿ because the sum of the recirocals of the squares, n converges. In fact, ÿ 8 2 n π n n This isn t relevant, just an interesting fact. Our roof is indirect, and is essentially due to Euler. What he did was to first consider the roduct ź ` ` ` 2 `. 3 ďn For examle, when n 0, the roduct is ` 2 ` 2 ` ` 2 3 ` 3 ` ` 2 5 ` 5 ` ` 2 7 ` 7 `. 2 If we were to multily this out, the sum would be ` 2 ` 3 ` 4 ` 5 ` 6 ` 7 ` 8 ` 9 ` 0 ` 2 ` 4 ` 5 ` 6 ` 8 `. The denominators are those divisible by 2 s, 3 s, 5 s, and 7 s. For examle, to get 8 we multily the terms 2 from the first roduct, from the second, and the s from the third 32 and fourth roducts. In articular, ź ďn ` ` ` 2 ` ą 3 nÿ k k ą ln n. Page 4

5 The reason nÿ k k ż ą ln n is that the sum can be aroximated by the integral dx. In x ż n dx, we get the x 2 ` 3 ` 4 ` 5 ` 6 ` 7 ` ż 9 8 ą dx ln9q. That is, the x articular, if we use uer rectangles to estimate the area of lnnq sum. For examle, below, ` sum is actually larger than lnn ` q, but lnnq looks a little simler. Now the sum is a geometric series with sum have ź ` ` ` 2 ` 3 ďn. Putting the ieces together so far, we ą ln n. Next, we take the logarithm of this to convert the roduct to a sum giving ÿ ln ą ln ln n. ďn To get to the sum of the recirocals of the rimes, we could use the MacLauran exansion: ln xq x ` x2 2 ` x3 3 ` but it works out better to use the formula ln xq ď x ` x2 x 2, for 0 ď x ă. A grah will show this is true. Alternatively, if you let fxq x ` x2 ` ln xq, x2 Page 5

6 then f0q 0 and f xq ą 0 for all 0 ă x ă, so fxq starts at 0 and increases. We have ln ln n ă ÿ ln ď ÿ f ÿ ` ÿ 2. ďn ďn ďn ďn One last trick: ÿ ďn 8 2 ď ÿ k 2 k ÿ It isn t imortant that k 2 3, only that it is some finite number, but it is hard to 4 k 2 ass u a nice formula. To see this summation is correct, we use artial fractions: k 2 2 k, k ` so 8ÿ 8 k 2 ÿ 2 k k ` k ` 2 4 ` 3 5 ` 4 6 ` 5 ` ` k 2 Putting everything together, n ln n ă ÿ ďn ` 3 4 or ÿ ďn ą ln ln n 3 4. Since ln ln n Ñ 8 as n Ñ 8, the sum of the recirocals of the rimes diverges. Perfect numbers A number is called a erfect number if it equals the sum of its roer divisors (or if the sum of all the divisors is twice the number.) The first several erfect numbers are 6 ` 2 ` 3 28 ` 2 ` 4 ` 7 ` ` 2 ` 4 ` 8 ` 6 ` 3 ` 62 ` 24 ` ` 2 ` 4 ` 8 ` 6 ` 32 ` 64 ` 27 ` 254 ` 508 ` 06 ` 2032 ` Looking at the rime factorizations of these numbers, we have 6 2 3, , , This might lead us to guess that erfect numbers always have the Page 6

7 form 2 k for some rime number. We can say more: If n 2 k, then the factors of n are, 2, 2 2,..., 2 k,, 2, 2 2,..., 2 k. These consist of two geometric sequences, and their sum is 2 k` q ` 2 k` q. On the other hand, we want the sum to be 2n 2 k`. For this to be the case, we need 2 k` q ` 2 k` q 2 k` Ñ 2 k` 0, or 2 k`. This means we can t use any k and, they must be linked. In articular, we can only use rimes that are one less than a ower of 2. The mathematician Mersenne looked into this in some deth. Let Mnq 2 n. We call such numbers Mersenne numbers. If Mnq is rime, we refer to it as a Mersenne rime. We have roven the following. Theorem 4 If 2 n is rime, then m 2 n is a erfect number. Do all erfect numbers have the form 2 n where 2 n? We don t know. Here is a artial answer, however. Theorem 5 (Euler) If m is a erfect number and m is even, then m 2 n 2 n q where 2 n is rime. Proof: Let m be an even erfect number, and suose that m 2 k Q where Q is odd. Suose the sum of the divisors of Q is S. Then the sum of the divisors of m is S ` 2 ` 2 2 ` ` 2 k q S2 k` q. You should convince yourself that the sum, indeed, looks like this. It is not quite obvious. As above, we want the sum to be 2m so 2m S2 k` q. Solving for S, S 2m 2k` 2 k` Q 2 k` Q ` Q 2 k`. Q But S is the sum of all the divisor of Q and Q and are both divisors of Q. This 2 k` means that Q cannot have any other divisors and the only way this could haen is if Q is Q rime, and 2 k` giving Q 2k`. This comletes the roof. Some obvious questions: Question : Are there any odd erfect numbers? Question 2: Are there infinitely many erfect numbers? No one knows the answers to these questions, though there is strong circumstantial evidence that there are infinitely many erfect numbers. The second question could be osed in terms of Mersenne rimes: are there infinitely many Mersenne rimes? Again, it is thought that the answer is yes. What does it take for a Mersenne number to be a Mersenne rime? Here is a table of the first several Mersenne numbers. n n Page 7

8 A attern resents itself: only rime numbers n can give rise to rimes (in articular, ) In fact, Mersenne numbers have a multilicative roerty: If m n then Mmq Mnq. For examle, 7 M3q divides M9q. This is not too hard to rove: If n km then 2 n 2 km 2 m q k 2 m q2 mk q ` 2 mk 2q ` ` 2 m ` q, so 2 n has 2 m as a factor. Unfortunately, even for rimes,, 2 need not be rime. For examle, As of February 208 there are 50 known rimes for which 2 is rime. these are: 2, 3, 5, 7, 3 the ones known to the Greeks, 7, 9, 3, 6 discovered before the 20th century, 89, 07, 27, 52, 607, 279, 2203, 228, 327 discovered before 960, 4253, 4423, 9689, 994, 23, 9937, 270, 23209, discovered before 980, 86243, 0503, 32049, 2609 found in the 980 s, , , , , , , found in the 990 s, 3,466,97 (200), 20,996,0 (2003), 24,036,583 (2004), 25,964,95, 30,402,457 (2005), 32,582,657 (2006), 43,2,609, 37,56,667 found in August, Setember, 2008, 42,643,80 found in Aril, 2009, 57,885,6 found January 25, 203, 74,207,28 found in January, ,232,97 found in January, 208. It is known that 2 42,643,80 is the 46 th Merseene rime. Also, all exonents u to 76,872,9 have been checked at least once so it is highly likely that 2 74,207,28 is the 49 th Mersenne rime and that 2 77,232,97 is the 50 th Mersenne rime Page 8

9 The following is an algorithm to check if a number 2 is a Mersenne rime. It is called the Lucas-Lehmer test. Set U 4. for i from 3 to do relace U by U 2 2 mod 2 q at the end of the loo, if U 0, then 2 is rime, otherwise, 2 is comosite. For examle, if 9, we have U 4 Ñ 4 Ñ 94 Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ 0, so 2 9 is a Mersenne rime. However, when 23, we have U 4 Ñ 4 Ñ 94 Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ so 2 23 is not a Mersenne rime. Page 9