4 A Survey of Congruent Results 12

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1 4 A urvey of Congruent Results 1 ECTION 4.5 Perfect Nubers and the iga Function By the end of this section you will be able to test whether a given Mersenne nuber is rie understand what is eant be a erfect nuber use the roerties of the siga function The largest nown rie at resent, July 014, is the Mersenne rie M This rie was found using the Great Internet Mersenne Prie earch GIMP which was briefly described in the last section. But how did GIMP find this rie? In essence it uses what we call the Lucas Leher Test which we characterise below. The Electronic Frontier Foundation EFF is offering the following rewards for finding large rie nubers: $150,000 to the first individual or grou who discovers a rie nuber with at least 100,000,000 decial digits. $50,000 to the first individual or grou who discovers a rie nuber with at least 1,000,000,000 decial digits Lucas and Leher Édouard Lucas was born in Aiens which is located in North West France near the Belgiu border. He was educated at the École Norale in Aiens. His first aointent was to tae u the ost of assistant astronoer at the observatory in Paris. After this he becae an artillery officer during the Franco Prussian war (180 1). Later he becae rofessor of atheatics in Paris. Lucas built u a reutation for delivering entertaining lectures and stretching his students with atheatical uzzles. Lucas is nown for forulating an algorith in 1856 which tested given Mersenne nubers for riality without checing all the ries less than the square root of the given nuber. Figure 5 Lucas 184 to 1891 In Father Mersenne list of Mersenne ries he included 6 1. However in 186, Lucas was 6 1 the first to show that is not rie. Lucas s algorith does not give the factors for coosite Mersenne nubers. Figure 6 Derric Leher 1905 to 1991 Derric Leher was born in Berley California and was one of five children. He too an interest in nuber theory in articular nuber theory calculations because of the influence of his father who was rofessor of atheatics at University of California, Berley. Derric read hysics at University of California, Berley (19) and then undertoo a PhD (190) at Brown University in Rhode Island UA.

2 4 A urvey of Congruent Results 1 In the 190 s Derric wored at various institutions finally settling at the University of California, Berley in 1940 where he reained for the rest of his acadeic career, retiring in 19. Derric Leher refined the Lucas algorith in the 190 s for testing Mersenne nubers and that is why the test is nown as the Lucas - Leher test. Leher is nown as the father of coutational nuber theory Lucas - Leher Test We will just state this test but not rove it. Lucas Leher Test (4.5). The Mersenne nuber 1 M is rie 0 od M as the least non negative residue such that od M for integer 1 where is defined Exale 4.19 how that M 1 is rie. olution Alying the Lucas Leher test with M 1 1, 0 4 and 1,,, 4 and 5 have od 1 [ubstituting 1] od 1 [ubstituting ] Fro the last calculation we have 5 0 od od 1 [ubstituting ] od 1 [ubstituting 4 ] od 1 [ubstituting 5 ] Hence by the Lucas Leher test we conclude that Exale 4.0 By alying the Lucas Leher test, show that olution Using the Lucas Leher test with because. so 0 od M M 1 is rie. M 1 is coosite. M 1 04, od 04 and 1,,,, 9 : od od od od od od od 04 we

3 4 A urvey of Congruent Results 14 We sto here because we have evaluated ince od 04 0 od 04 so 0 od M 1 is not rie which eans it is coosite. We have shown in the last two sections that M You ight thin that this, Lucas Leher test is a retty cubersoe test. Yes it is but couter rograes norally use this test to find whether a given Mersenne nuber is rie or not Perfect Nubers We will show in this subsection that erfect nubers are closely related to Mersenne ries. Perfect nubers have been nown for over two thousand years. However after the ancient Grees, erfect nubers were forgotten about until the 1500s when Cataldi, Ferat, Descartes and Mersenne studied these. Actually Ferat derived his well nown Little Theore by exaining erfect nubers. Before we define erfect nubers we need to define roer factors or roer divisors: Definition (4.6). A roer factor of a nuber n is any ositive factor of n aart fro the nuber n itself. For exale the roer factors of 1 are 1,,, 4 and 6. Factors of 1 are 1,,, 4, 6 and 1. What are the roer factors of 0? 1,,, 5, 6, 10 and 15 Note that the roer factors of 0 does not include the factor 0. What are the factors of 6? 1,, and 6 Only 1, and are roer factors of 6 and if we add these we get 1 6 What are the factors of 8? 1,, 4,, 14 and 8 Again if we add all the roer factors of 8 which are 1,, 4, and 14 we get What do you notice about the su of roer factors in these two exales? Well the su of all the roer factors of 6 gives 6. The su of all the roer factors of 8 gives 8. uch nubers are called erfect nubers. Definition (4.). Let n be a ositive integer. Then n is called a erfect nuber if the su of all the roer factors (divisors) of n is equal to n. Equivalently if d1, d, d,, d are roer factors (divisors) of n and d1 d d d n Then we say n is a erfect nuber. Another exale of a erfect nuber is 496 because the roer divisors (factors) of 496 are 1,, 4, 8,16, 1, 6,14 and 48 Adding all these gives

4 4 A urvey of Congruent Results 15 Why call these nubers erfect? Nubers lie 1 are called abundant nubers because their roer factors 1,,, 4 and 6 su to and 16 1 The Grees ade analogies of these abundant nubers with anials having ore than 5 fingers on one hand. Nubers lie 10 are called deficient nubers because their roer factors 1, and 5 su to and 8 10 Again the Grees viewed these deficient nubers as reresenting anials having less than 5 fingers on one of its hand. This is why nubers lie 6 whose roer factors su to exactly 6 are called erfect nubers. A reasonable set of questions to as are: How can we locate these erfect nubers? Is there a forula we can use to generate these or do we have to chec each nuber and add its roer factors? Yes there is a forula which we describe next. The einent Gree atheatician Euclid stated and roved the following result: Theore (4.8). Let be a rie nuber. If the Mersenne nuber 1 is rie then 1 N 1 is a erfect nuber. rie Before atteting a roof, let us loo at the first few erfect nubers N for various ries: with rie with rie with rie with rie with rie The first 4 erfect nubers in this list was nown to the Grees but the fifth erfect nuber was not discovered until the 1500 s. Why did we iss the rie in this list? Reeber fro the revious Exale (4.0) we had [coosite] 1 N 1 we ust have 1 is rie but for we have In the given forula 1 is not rie. Note that there are large gas between consecutive erfect nubers; 6, 8, 496, 818, 550 6, Peter Barlow in his boo Theory of Nubers ublished in 18 claied the following : about the erfect nuber is the greatest that ever will be discovered; for as they are erely curious, without being useful, it is not liely that any erson will ever attet to find one beyond it. How do we rove the stated theore? By using the su of the geoetric series:

5 4 A urvey of Congruent Results 16 a 1 r n1 (4.9) a ar ar ar 1 r We can assue that 1 starting with 0 1 to is rie. The roer factors of 1 N 1 1 and then their ultiles with the rie 1 roer factors (divisors) is given by: 1,,,,, 1, 1, 1, 1,, 1 First art 1 Note that we do not include the factor 1 1 N econd art 1 n are the owers of. Hence the list of because it is not a roer factor of Dividing this list into two arts by first suing all the owers of and then all the ultiles of 1. uing the owers of which is the first art of the nubers in the list (*): 1 1 This is a geoetric series with a 1, r and n. Alying the su forula (4.9) yields ( ) uing the ultiles of 1 with owers of which is the second art of the nubers in the list (*): (**) Factorizing If we exaine the square bracets su then we see that this is a geoetric series with a 1 and r. Using the geoetric su series forula (4.9) with a 1, r and n 1 we get Putting this into the revious calculation (**) gives ( ) Adding equations ( ) and ( ) which gives the su of all the roer factors of N: ince the su of all the roer factors is This coletes our roof Factorizing iilifying 1 N so N is a erfect nuber. 000 years after Euclid, one of greatest atheatician of all tie, Euler roved that the converse of the above theore is also true: N (*)

6 4 A urvey of Congruent Results 1 Theore (4.0). Every even erfect nuber N is of the for 1 N 1 where 1 is rie Exercises 4.5. This theore gives rise to the question: Are there any odd erfect nubers? We don t now because it is an oen roble. No one has been able to find one but this does not ean there aren t any. This is one of the oldest conjecture in atheatics The iga Function Definition (4.1). The siga function n of a ositive integer n is defined as the su of all the ositive divisors (factors) of n. Let d1, d,, d be the divisors of n then n d d d This siga function Exale 4.1 n 1 is soeties called the su of divisors function. Deterine (a) 10 (b) 1 (c) 8 (d) 1 olution (a) The ositive divisors of 10 are 1,, 5 and 10 so (b) iilarly we have the ositive divisors of 1 are 1,,, 4, 6 and 1. Therefore we have (c) The ositive divisors of 8 are 1,, 4,, 14 and 8. Adding these nubers gives (d) Note that 1 is a rie so the only divisors are 1 and 1. Adding these two nubers gives Proerties of the iga Function 1. What do you redict the siga function Notice fro the revious exale that of a rie nuber will be? As the only factors of a rie are 1 and itself therefore we should have Proosition (4.). We have is a rie nuber 1 Exercises Fro the revious exale art (c) we have Reeber 8 is a erfect nuber. What do you thin the siga function will be for a erfect nuber n? n n

7 4 A urvey of Congruent Results 18 Proosition (4.). Let n be a erfect nuber then n n. Let n be a erfect nuber and d 1, d,, d 1 and d n be the ositive factors (divisors) of the given ositive integer n. Note that the last factor is n itself. Adding these gives n d1 d d 1 n (*) These are roer divisors of n Fro the definition of erfect nubers we have d d d n 1 1 ubstituting this d1 d d 1 n into (*) gives n d1 d d 1 n n n n This is our required result. Therefore and nubers. The siga function is ultilicative. What does this ean? Definition (4.4). A general function n because 6 and 496 are erfect f n f ab f a f b whenever a of a ositive integer n is called ultilicative if gcd, b 1 Can you thin of any exales of ultilicative functions? For exale if f is the square root function then n n where and n are ositive integers This eans that the square root function is ultilicative. However there are any functions that are not ultilicative such as the logarithic function. For exale, let and n be ositive integers then ln n ln ln n [Not Equal] In the next subsection we rove that the siga function will show: 1 If n 1 n where s are distinct ries then is ultilicative. Actually we 1 1 n 1 1 How does ultilicative hel in evaluating n for a given ositive integer n? Let us find 561. We really don t want to find all the factors of 561 as this will be a rather dull tas and tae soe tie to find. Exale Deterine olution Let us first decoose 561 into its rie factors (you can easily chec that 561 is divisible by and by using aroriate tests): 561 1

8 4 A urvey of Congruent Results 19 Clearly, and 1 are distinct ries. o assuing have n But now we have to find, and 1. How? ince, and 1 are all rie so by Proosition (4.): We have is ultilicative, therefore we ( ) 1 4, 1 and 1 18 ubstituting these into ( ) gives What does ean? If we add all the ositive factors of 561 then we get without listing all the factors of 561 and then In this exale we easily evaluated suing the. However we do need to find the rie decoosition of a given n. where 1. ay we want to find Proosition (4.5). Let be rie and be a ositive integer then Listing the ositive factors of where is rie we have 1,,,,, Adding these factors gives so 1 This is a geoetric series so alying the suation forula: a 1 r n1 (4.9) a ar ar ar 1 r With a 1, r and n 1 we have We have shown which is our required result. 1 n Multilying nuerator and denoinator by 1 Exale Deterine olution The rie decoosition of 500 is

9 4 A urvey of Congruent Results 0 ince the siga function is ultilicative so ( ) Alying the above roosition (4.5) we have Alying Alying ubstituting these and into ( ) yields Hence [Adding all the ositive factors of 500 gives 109.] Multilicatively of the iga Function In this subsection we rove that the revious defined siga function Proosition (4.6). Let n q where and q are different ries. Then n q q n is ultilicative. Listing the factors of n q where and q are distinct ries in a table and suing these using the geoetric series forula (4.9) we have: Factors List of Factors uation of factors Factors of 1,,,, 1, Factors of q,,, q q q, q q 1 q q 1 1 Factors of q,,, q q q, q q 1 q q 1 1 Factors of q,,, q q q, q q 1 q q 1 1 Factors of q q, q,, q, q q 1 q q 1 Adding the nubers in the highlighted last colun gives q 1 q 1 q 1 q 1 q q q q q 1 q 1 q 1 q 1 q 1 q 1 q 1 Factorizing q q 1 q 1 1 q 1 1 q q (*)

10 4 A urvey of Congruent Results 1 Adding the first entry in last colun in the table to this suation in (*) gives 1 1 q q q q q 1 1 q 1 Factorizing 1 q q q q 1 1 q 1 1 q 1 1 q 1 By the revious Proosition (4.5): Using this to evaluate and q ubstituting this gives and q and q q 1 q 1 1 q 1 into the above gives q 1 1 q 1 1 q 1 This coletes our roof. q q Proosition (4.). Let the rie decoosition of a ositive integer n be given by 1 n where s are distinct ries Then n We say the siga function Exercises n is ultilicative. References: A fantastic article on the history of erfect nubers is at the following url: htt://www-history.cs.st-and.ac.u/histtoics/perfect_nubers.htl UMMARY A ositive integer n is a called erfect nuber if the su of its roer factors is equal to n. We can use the roerties of the siga function n n which tells us the su of divisors of n. to evaluate