Primes of the form n 2 + 1

Transcription

1 Itroductio Ladau s Probles are four robles i Nuber Theory cocerig rie ubers: Goldbach s Cojecture: This cojecture states that every ositive eve iteger greater tha ca be exressed as the su of two (ot ecessarily differet) rie ubers. Twi Prie Cojecture: Are there ifiitely ay rie ubers such that + is also a rie uber? This roble was solved by Carlos Giraldo Osia (Lic. Mateáticas, USC, Cali, Colobia), who roved that if k is ay ositive eve iteger, the there are ifiitely ay rie ubers such that + k is also a rie uber. Legedre s Cojecture: Is there always at least oe rie uber betwee ad ( + ) for every ositive iteger? Pries of the for + : Are there ifiitely ay rie ubers of the for + (where is a ositive iteger)? Please see the article Prios Geelos, Deostració Kelliza, where C. G. Osia shows the ethod he used to solve the Twi Prie Cojecture. Pries of the for + I this docuet we are goig to rove that there are ifiitely ay rie ubers of the for +. I order to achieve our goal, we are goig to use the sae ethod that C. G. Osia used i his aer. Note: I this docuet, wheever we say that a uber b is betwee a uber a ad a uber c, it will ea that a < b < c, which eas that b will ever be equal to a or c (the sae rule will be alied to itervals). Moreover, the uber that we will use i this docuet will always be a ositive iteger. Page of 9

2 Theores,, ad 4 Let us suose that is a ositive iteger. We eed to kow what value eeds to have so that there is always a erfect square a such that < a <. Note: We say a uber is a erfect square if it is the square of a iteger. I other words, a uber x is a erfect square if x is a iteger. Perfect squares are also called square ubers. I geeral, if is ay ositive iteger, we eed to kow what value eeds to have so that there is always a ositive iteger a such that < a <. We have This eas that < a < < a ad < a ad a < a < To su u, < a < Page of 9 As we said before, the uber a is a ositive iteger. Now, the iteger iediately followig the uber will be called + d. I other words, + d is the sallest iteger that is greater tha : If is a iteger, the + d = +, because i this case we have d =. If is ot a iteger, the i the exressio + d we have 0 < d <, but we do ot have ay way of kowig the exact value of d if we do ot kow the value of first.

3 Now, let us ake the calculatio: + d < We eed to take the largest ossible value of d, which is d = (if + d < for all d such that 0 < d ): + <, the + < < < <.5 <.5 <.5.5 <.5 < >.5 > (.5 ) Page of 9

4 >.5 This eas that if is a ositive iteger, the for every ositive iteger there is at least oe ositive iteger a such that < a <. Now, if the >..5 aaaaaaaaaaaaaaaaaaaaaaaa > > Cosequetly, if is a ositive iteger, the for every ositive iteger 4.4 > there is at least oe ositive iteger a such that < a <. This.5 true stateet will be called Theore. Now we are goig to rove that if is ay ositive iteger, the ( ) Proof:.5 >..5 > >.5 >.5 >.5 >.5 + >.5 >.5 Page 4 of 9

5 >.5 It is very easy to verify that >.5 for every ositive iteger. Cosequetly, if is 4.4 > > > ay ositive iteger, the This eas that ( ). If 4.4 > 4.4 for every ositive iteger..5, the Note: I geeral, to rove that a iequality is correct, we ca solve that iequality ste by ste. If we get a result which is obviously correct, the we ca start with that correct result, work backwards fro there ad rove that the iitial stateet is true. As a cosequece, if is ay ositive iteger ad This true stateet will be called Theore. > 4.4.5, the > 4.4. I the docuet Ifiitely May Prie Nubers of the For a±b it was roved that for every ositive iteger > 4.4 there exist rie ubers r ad s such that < r < < s < (lease see that docuet for a roof). This true stateet will be called Theore. Accordig to Theores, ad, if is a ositive iteger, the for every ositive 4.4 iteger > there exist a ositive iteger a ad a rie uber s such.5 aaaaaaaaaaaaaaaaaaaaa that < a < < s <. This true stateet will be called Theore 4. Theores 5, 6 ad 7 Now, if is a ositive iteger, let us calculate what value eeds to have so that there is always a ositive iteger a such that < a < : We have Page 5 of 9

6 This eas that To su u, < a < < a ad a < < a ad a < < a < The uber a is a ositive iteger. Now, the iteger iediately followig the uber will be called greater tha : + d. I other words, + d is the sallest iteger that is If is a iteger, the d =. + d = +, because i this case we have If is ot a iteger, the i the exressio + d we have 0 < d <, but we do ot have ay way of kowig the exact value of d if we do ot kow the value of first. Let us ake the calculatio: + d < We eed to take the largest ossible value of d, which is d = (if + d < for all d such that 0 < d ): + <, the Page 6 of 9

7 + < < < <.5 <.5 <.5.5 <.5 (.5).5 < < < >.5 This eas that if is a ositive iteger, the for every ositive iteger > there is at least oe ositive iteger a such that < a <. Now,.5 if > the >..5 Page 7 of 9

8 Cosequetly, if is a ositive iteger, the for every ositive iteger 4.4 > there is at least oe ositive iteger a such that < a <..5 This true stateet will be called Theore 5. Now we are goig to rove that if is ay ositive iteger, the ( ) Proof:.5 >..5 > >.5 >.5 >.5 > >.5 > because.5 >, that is to say,.5 >. because. This eas that +.5 > which roves that +.5 > Page 8 of 9

9 >.5 Therefore, if is ay ositive iteger, the ( ) 4.4 > This eas that ( ) >, the.5. If ( ) 4.4 > 4.4 for every ositive iteger..5 As a cosequece, if is ay ositive iteger ad This true stateet will be called Theore 6. > 4.4.5, the > 4.4. Accordig to Theores, 5 ad 6, if is a ositive iteger, the for every ositive 4.4 iteger > there exist a rie uber r ad a ositive iteger a such.5 aaaaaaaaaaaaaaaaaaaaaaa that < r < < a <. This true stateet will be called Theore 7. Ifiitely ay rie ubers of the for + Accordig to Theore 4, for every ositive iteger > eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee ositive iteger a ad a rie uber s such that 4.4 (.5 ) < a < < s <. there exist a The ubers a ad s for what we will call air (erfect square, rie) of order k. This is because: We have a air of ubers: a erfect square a ad a rie uber s. The erfect square a is followed by the rie ubers s. I other words, a < s. We say that a + k = s. I other words, we say the air (erfect square, rie) is of order k because the differece betwee the ubers forig this air is k. Now we eed to defie soe other cocets: Page 9 of 9

10 The set ade u of all ositive itegers z such that Set A(). Exales: z > will be called Set A() is the set of all ositive itegers z such that z > 4.4 (.5 ) words, Set A() is ade u of all ositive itegers z such that z > Set A() is the set of all ositive itegers z such that z > 4.4 (.5 ) words, Set A() is ade u of all ositive itegers z > I other. I other The set ade u of all ositive itegers z such that Set B(). z > will be called Let us rove that there are ifiitely ay rie ubers of the for +. I order to achieve the goal, we are goig to use the sae ethod that C. G. Osia used i his article Prios Geelos, Deostració Kelliza.. Let us suose that i Set A() there are o airs (erfect square, rie) of order k < u startig fro = u. The ubers ad u are ositive itegers which belog to Set A().. Betwee u ad u, = u, there is at least oe air (erfect square, rie), accordig to Theore 4.. The differece betwee two itegers located betwee u ad u is k < u. 4. Betwee u ad u, = u, there is at least oe air (erfect square, rie) of order k < u, accordig to stateets. ad. 5. I Set A(), startig fro = u there is at least oe air (erfect square, rie) of order k < u, accordig to stateet Stateet 5. cotradicts stateet. 7. Therefore, o kid of air (erfect square, rie) of ay order k ca be fiite, accordig to stateet 6. Page 0 of 9

11 Note: It is already kow that for every ositive iteger k the olyoial + k is irreducible over R ad thus irreducible over Z, sice every secod-degree olyoial whose discriiat is a egative uber is irreducible over R. Accordig to stateet 7., for every ositive iteger k there are ifiitely ay airs (erfect square, rie) of order k. This eas that airs (erfect square, rie) of order ca ot be fiite. I other words, rie ubers of the for + ca ot be fiite. All this roves that there are ifiitely ay rie ubers of the for +. I geeral, if we use Set A(), Set B() ad Theores 4 ad 7 ad we use the sae ethod, we could rove that there are ifiitely ay rie ubers of the for + k ad ifiitely ay rie ubers of the for k for certai values of ad k (we oly have to take ito accout the cases where the olyoials + k ad k are irreducible over Z ). Coclusio We will restate the ost iortat theore that was roved i this docuet: There are ifiitely ay rie ubers of the for where is a ositive iteger. +, 0 Page of 9

12 New cojecture If is ay ositive iteger ad we take cosecutive itegers located betwee ( + ), the aog those itegers there is at least oe rie uber. ad I other words, if a, a, a, a 4,, a are cosecutive itegers such that < a < a < a < a4 <... < a < +, the at least oe of those itegers is a rie uber. This cojecture will be called Cojecture C. Legedre s Cojecture It is very easy to verify that the aout of itegers located betwee equal to. + is ad Proof: + = = + + = + We eed to exclude the uber ( + ) because we are takig ito cosideratio the itegers that are greater tha Accordig to this, betwee ad saller tha + = ad + : + there are two grous of cosecutive itegers each that do ot have ay iteger i coo. Exale for = : () ( + ) Grou A Grou B ( cosecutive itegers) ( cosecutive itegers) cosecutive itegers Page of 9

13 Grou A ad Grou B do ot have ay iteger i coo. Accordig to Cojecture C, Grou A cotais at least oe rie uber ad Grou B also cotais at least oe rie uber, which eas that betwee ad + there are at least two rie ubers. This is true because the ubers ad are both rie ubers. All this eas that if Cojecture C is true, the there are at least two rie ubers betwee ad + for every ositive iteger. As a result, if Cojecture C is true, the Legedre s Cojecture is also true. Brocard s Cojecture This cojecture states that if the betwee ( ) ad ad + are cosecutive rie ubers greater tha, + there are at least four rie ubers. Sice < < +, we have +. This eas that there is at least oe ositive iteger a such that < a < +. As a result, there is at least oe ositive iteger a aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa such that a + < <. Cojecture C states that betwee ( ) ad that betwee a ad a there are at least two rie ubers ad + there are also at least two rie ubers. I other words, if Cojecture C is true the there are at least four rie ubers betwee ( ) ad. As a cosequece, if Cojecture C is true the Brocard s Cojecture is also + true. Adrica s Cojecture This cojecture states that < for every air of cosecutive rie ubers ad + + (of course < + ). Obviously, every rie uber is located betwee two cosecutive erfect squares. If we take ay rie uber two thigs ay hae:, which is obviously located betwee ad +, Page of 9

14 Case : The uber is located aog the first cosecutive itegers that are located betwee + ad. These itegers for what we call Grou A, ad the followig itegers for what we call Grou B, as show below: < < ( + ) Grou A Grou B ( cosecutive itegers) ( cosecutive itegers) cosecutive itegers If is located i Grou A ad Cojecture C is true, the + is either located i Grou A or i Grou B. I both cases we have < +, because ( ) + = ad the ubers + ad are closer to each other tha ( + ) i relatio to. Case : The rie uber is located i Grou B. If is located i Grou B ad Cojecture C is true, it ay hae that + is also located i Grou B. I this case it is very easy to verify that < +, as exlaied before. Otherwise, if + is ot located i Grou B, the + is located i Grou C (see the grahic below). I this case the largest value + ca have is + = ( + ) + + = = + + ad the sallest value ca have is = + + (i order to ake the rocess easier, we are ot takig ito accout that i this case the ubers ca ot be both rie at the sae tie). ad + have differet arity, so they This eas that the largest ossible differece betwee + ad is + = Let us look at the grahic below: < < ( + ) <... Grou A Grou B Grou C ( cosecutive itegers) ( cosecutive itegers) ( + cosecutive itegers) cosecutive itegers axiu distace betwee ad + Page 4 of 9

15 = + + = = + + = + It is easy to rove that <. Proof: < + + < < < < < < + + < + + < + + < + + which is true for every ositive iteger. We ca see that eve whe the differece betwee + ad is the largest ossible differece, we have < +. If the differece betwee + ad were saller, the of course it would also hae that + <. Accordig to Cases ad, if Cojecture C is true the Adrica s Cojecture is also true. Page 5 of 9

16 To coclude, if Cojecture C is true, the Legedre s Cojecture, Brocard s Cojecture ad Adrica s Cojecture are all true. Possible ew iterval It is easy to verify that if Cojecture C is true, the i the iterval + +, + + (see the grahics o revious ages) there are at least two rie ubers for every ositive iteger. The uber + + will always be a odd iteger. Proof: If is eve, the is also eve. The we have ( eve iteger + eve iteger) + = eve iteger + odd iteger = odd iteger If is odd, the is also odd. The we have ( odd iteger + odd iteger) + = eve iteger + odd iteger = odd iteger Sice the uber + + will always be a odd iteger, the it ay be rie or ot. Now, the uber + + ca ever be rie because this uber will always be a eve iteger (ad it will be greater tha ). Proof: If = (sallest value ca have), the + + = + + = 6. If is eve, the ad are both eve itegers. The uber is also a eve iteger, ad we kow that If is odd, the eve iteger + eve iteger + eve iteger = eve iteger ad are both odd itegers, ad we kow that ( odd iteger + odd iteger) + eve iteger = eve iteger + eve iteger = eve iteger Page 6 of 9

17 Fro all this we deduce that if Cojecture C is true, the the axiu distace betwee two cosecutive rie ubers is the oe fro the uber + + to the uber + + = + +, which eas that i the iterval there are at least two rie ubers. I other words, i the [, ] iterval [ + +, + ] there is at least oe rie uber. The differece betwee the ubers + + ad + ( + + ) = + =. I additio to this, + is + + =. This eas that i the iterval + +, there is at least oe rie uber. I other words, if a = + + the the iterval a, a + a cotais at least oe rie uber. Note: The sybol reresets the floor fuctio. The floor fuctio of a give uber is the largest iteger that is ot greater tha that uber. For exale, x is the largest iteger that is ot greater tha x. Now, if Cojecture C is true, the the followig stateets are all true:. If a is a erfect square, the i the iterval a, a + a there is at least oe rie uber.. If a is a iteger such that a, a + a there is at least oe rie uber.. If a is a iteger such that a, a + a there is at least oe rie uber. < a + + < ( + ), the i the iterval < + + a < ( + ), the i the iterval We kow that a + a a + a. Proof: a + a a a a a a a + + a a a a + +, which is true for every ositive iteger a. Page 7 of 9

18 Ad we also kow that a + a a a > +. Proof: a + a > a + a a > a, which is obviously true for every aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa ositive iteger a. All this eas that the iterval a, a + a ca be alied to the uber a fro stateet., to the uber a fro stateet. ad to the uber a fro stateet. Therefore, if is ay ositive iteger ad Cojecture C is true, the i the iterval, + there is at least oe rie uber (we chage letter a for letter ). Accordig to this, we ca also say that if Cojecture C is true the i the iterval, + there is always a rie uber for every ositive iteger. Now how ca we rove Cojecture C? ===================0=================== See also: Eail: gera_a_az@hotail.co Núeros rios, fórula recisa. Origial aer: Cálculo de la catidad de úeros rios que hay or debajo de u úero dado (How to calculate the aout of rie ubers that are less tha a give uber) Núeros Prios de Sohie Gerai, Deostració de su Ifiitud (There Are Ifiitely May Sohie Gerai Prie Nubers) Ifiitely May Prie Nubers of the For a±b Ifi itos Núeros Prios de la Fora a±b Page 8 of 9

19 A Paers by Carlos Giraldo Osia I recoed (these aers are available at htt://uerosrios.8.co/docuetos.ht ): Núeros Prios (Prie Nubers) Prios, Disersió Parabólica (Pries, Parabolic Disersio) Prios de Mersee, Disersió Parabólica (Mersee Pries, Parabolic Disersio) Proof of Legedre s ad Brocard s Cojectures Page 9 of 9