Theorems Related To Mersenne Primes Mathematics Essay

Theorems Related To Mersenne Primes maths EssayIntroductionIn the past m either use to pass on that the itemizes of the type 2p-1 were bloom of youth for in completedly(prenominal) strands poem which is p, but when Hudalricus Regius (1536) distinctly established that 211-1 = 2047 was not inflorescence because it was divisible by 23 and 83 and later on Pietro Cataldi (1603) had properly substantiate nearly 217-1 and 219-1 as deuce give prime teleph cardinal sums but also inaccurately stated that 2p-1 for 23, 29, 31 and 37 gave prime rime. Then Fermat (1640) proved Cataldi was wrong about 23 and 37 and Euler (1738) showed Cataldi was also incorrect regarding 29 but make an accurate conjecture about 31.Then subsequently this extensive report of this dilemma with no accurate result we saw the entry of Martin Mersenne who decl ard in the introduction of his Cogitata Physica-Mathematica (1644) that the functions 2p-1 were prime for-p= 2, 3, 5, 7, 13, 17, 19, 31, 67, 1 27 and 257 and forother peremptory integers where p So simply the definition is when 2p-1 forms a prime issue forth it is recognized to be a Mersenne prime.Many years later with unseasoned moments being discovered belonging to Mersenne Primes in that respect ar compose many fundamental questions about Mersenne primes which remain unresolved. It is still not set whether Mersenne primes is infinite or finite. There ar still many aspects, functions it performs and applications of Mersenne primes that are still unfamiliarWith this concept in mind the focus of my panoptic essay would beWhat are Mersenne Primes and it cerebrate functions?The reason I learn this topic was because while researching on my extended essay topics and I came across this part which from the beginning intrigued me and it gave me the opportunity to fill this gap as very(prenominal) little was taught about these aspects in our school and at the same measure my enthusiasm to learn something new throug h research on this topic. through this paper I will explain what are Mersenne primes and certain theorems, related to other aspects and its application that are related with it.Theorems Related to Mersenne Primesp is prime only if 2p1 is prime. certainty If p is composite so it bottomland be written as p=x*y with x, y 1.2xy-1= (2x-1)*(1+2x+22x+23x+..+2(b-1)a) and then we fall in got 2xy 1 as a product of integers 1.If n is an odd prime, then any prime m that divides 2n 1 essential(prenominal) be 1 plus a septuple of 2n. This holds even when 2n 1 is prime.Examples Example I 25 1 = 31 is prime, and 31 is multiple of (2-5) +1Example II 211 1 = 23-89, where 23 = 1 + 2-11, and 89 = 1 + 8-11.Proof If m divides 2n 1 then 2n 1 (mod m). By Fermats Theorem we bed that 2(m 1) 1 (mod m). Assume n and m 1 are comparatively prime which is similar to Fermats Theorem that states that (m 1)(n 1) 1 (mod n). Hence there is a number x (m 1)(n 2) for which (m 1)x 1 (mod n), an d thus a number k for which (m 1)x 1 = kn. Since 2(m 1) 1 (mod m), rhytidoplasty both(prenominal) sides of the congruence to the power x gives 2(m 1)x 1, and since 2n 1 (mod m), raising both sides of the congruence to the power k gives 2kn 1. Thus 2(m 1)x/2kn = 2(m 1)x kn 1 (mod m). still by means, (m 1)x kn = 1 which implies that 21 1 (mod m) which means that m divides 1. Thus the first gear conjecture that n and m 1 are comparatively prime is unsustainable. Since n is prime m 1 guide to be a multiple of n.Note This information provides a confirmation of the infiniteness of primes different from Euclids Theorem which states that if there were finitely many primes, with n being the largest, we give up a contradiction because every prime dividing 2n 1 must be larger than n.If n is an odd prime, then any prime m that divides 2n 1 must be congruous to +/-1 (mod 8).Proof 2n + 1 = 2(mod m), so 2(n + 1) / 2 is a square root of 2 modulo m. By quadratic equali ty reciprocity, any prime modulo which 2 has a square root is congruent to +/-1 (mod 8).A Mersenne prime canisternot be a Wieferich prime.Proof We show if p = 2m 1 is a Mersenne prime, then the congruence does not satisfy. By Fermats diminutive theorem, m p 1. right off write, p 1 = m. If the granted congruence satisfies, then p2 2m 1, therefore Hence 2m 1 , and therefore .This leads to , which is unsurmountable since .The Lucas-Lehmer TestMersenne prime are found using the interest theoremFor n an odd prime, the Mersenne number 2n-1 is a prime if and only if 2n -1 divides S(p-1) where S(p+1) = S(p)2-2, and S(1) = 4.The assumption for this psychometric test was initiated by Lucas (1870) and then made into this straightforward experiment by Lehmer (1930). The onward motion S(n) is calculated modulo 2n-1 to conserve time. This test is undefiled for binary computers since the division by 2n-1 (in binary) can only be completed using rotation and addition.Lists of know M ersenne PrimesAfter the discovery of the first few Mersenne Primes it took more than devil centuries with taut verification to obtain 47 Mersenne primes. The sideline table under lists all recognized Mersenne primes-It is not well-known whether any undiscovered Mersenne primes present betwixt the 39th and the 47th from the above table the position is consequently short-lived as these be werent always discovered in their increasing order.The sideline graph shows the number of digits of the largest known Mersenne primes year wise.Note The vertical exfoliation is logarithmic.FactorizationThe factoring of a prime number is by meaning itself the prime number itself. Now if talk about composite verse. Mersenne numbers are excellent investigation cases for the particular number field cover algorithm, so frequently that the largest figure they have factorized with this has been a Mersenne number. 21039 1 (2007) is the record-holder after estimating took with the help of a couple of hundred computers, mostly at NTT in Japan and at EPFL in Switzerland and yet the time consequence for calculation was about a year. The special number field screen door can factorize figures with more than one large factor. If a number has one huge factor then other algorithms can factorize larger figures by initially finding the serve up of small factors and after that making a primality test on the cofactor. In 2008 the largest Mersenne number with confirmed prime factors is 217029 1 = 418879343 - p, where p was prime which was confirmed with ECPP. The largest with possible prime factors allowed is 2684127 1 = 23765203727 - q, where q is a likely prime.GeneralizationThe binary depiction of 2p 1 is the digit 1 copyed p times. A Mersenne prime is the grounding 2 repunit primes. The base 2 depiction of a Mersenne number demonstrates the factorization congressman for composite exponent.Examples in binary notation of the Mersenne prime would be251 = 1111122351 = (111111111 111111111111111111111)2Mersenne Primes and Perfect NumbersMany were intense with the relationship of a ii sets of different numbers as two how they can be interconnected. One such connection that many multitude are concerned still today is Mersenne primes and Perfect Numbers.When a arbitrary integer that is the sum of its proper confident(p) divisors, that is, the sum of the positive divisors excluding the number itself then is it said to be known as Perfect Numbers.Equivalently, a unblemished number is a number that is one-half the sum of all of its positive divisors. There are said to be two types of better numbers1) Even utter(a) numbers- Euclid revealed that the first four meliorate numbers are generated by the recipe 2n1(2n1)n = 2 2(4 1) = 6n = 3 4(8 1) = 28n = 5 16(32 1) = 496n = 7 64(128 1) = 8128.Noticing that 2n1 is a prime number in each instance, Euclid proved that the formula 2n1(2n1) gives an even finished number whenever 2p1 is prime2) Odd perfect nu mbers- It is unidentified if there might be any odd perfect numbers. Various results have been obtained, but none that has helped to locate one or otherwise resolve the question of their existence.An example would be the first perfect number that is 6. The reason for this is so since 1, 2, and 3 are its proper positive divisors, and 1+2+3=6. Equivalently, the number 6 is equal to half the sum of all its positive divisors (1+2+3+6)/2=6.few Theorems related with Perfect numbers and Mersenne primesTheorem One z is an even perfect number if and only if it has the form 2n-1(2n-1) and 2n-1 is a prime.Suppose first that p = 2n-1 is a prime number, and set l = 2n-1(2n-1). To show l is perfect we need only show sigma(l) = 2l. Since sigma is multiplicative and sigma(p) = p+1 = 2n, we knowsigma(n) = sigma(2n-1).sigma(p) =(2n-1)2n = 2l.This shows that l is a perfect number.On the other hand, suppose l is any even perfect number and write l as 2n-1m where m is an odd integer and n2. Again sigma is multiplicative sosigma(2n-1m) = sigma(2n-1).sigma(m) = (2n-1).sigma(m).Since l is perfect we also know thatsigma(l) = 2l = 2nm.Together these two criteria give2nm = (2n-1).sigma(m),so 2n-1 divides 2nm hence 2n-1 divides m, say m = (2n-1)M. Now substitute this back into the equation above and divide by 2n-1 to get 2nM = sigma(m). Since m and M are both divisors of m we know that2nM = sigma(m) m + M = 2nM,so sigma(m) = m + M. This means that m is prime and its only two divisors are itself (m) and one (M). Thus m = 2n-1 is a prime and we have prove that the number l has the prescribed form.Theorem Two n will also be a prime if 2n-1 is a prime.Proof Let r and s be positive integers, then the polynomial xrs-1 is xs-1 times xs(r-1) + xs(r-2) + + xs + 1. So if n is composite (say r.s with 1Theorem Three Let n and m be primes. If q divides Mn = 2n-1, then q = +/-1 (mod 8) andq = 2kn + 1 for some integer k.Proof If p divides Mq, then 2q=1 (mod p) and the order of 2 (mod p) divides the p rime q, so it must be q. By Fermats Little Theorem the order of 2 also divides p-1, so p-1=2kq. This gives 2(p-1)/2 = 2qk = 1 (mod p) so 2 is a quadratic balance wheel mod p and it follows p = +/-1 (mod 8), which completes the proof.Theorem Four If p = 3 (mod 4) be prime and then 2p+1 is also prime only if 2p+1 divides 2p-1.Proof Suppose q = 2p+1 is prime. q=7 (mod8) so 2 is a quadratic residue modulo q and it follows that there is an integer n such that n2=2 (modq). This shows2p = 2(q-1)/2 = nq-1 = 1 (mod q),showing q divides Mp. Conversely, let 2p+1 be a factor of Mp. Suppose, for proof by contradiction, that 2p+1 is composite and let q be its least prime factor. Then 2p=1 (modq) and the order of 2 modulo q divides both p and q-1, hence p divides q-1. This shows qp and it follows(2p+1) + 1 q2 p2which is a contradiction since p 2.Theorem Five When we add the digits of any even perfect number with the exception of 6 and then sum the digits of the resulting number and keep doing it once more until we get a unmarried digit which will be one.Examples.28 10 1,496 19 10 1, and8128 19 10 1Proof Let s(n) be the sum of the digits of n. It is easy to see that s(n) = n (mod 9). So to prove the theorem, we need only show that perfect numbers are congruent to one modulo nine. If n is a perfect number, then n has the form 2p-1(2p-1) where p is prime which see in the above theorem one. So p is either 2, 3, or is congruent to 1 or 5 modulo 6. Note that we have excluded the case p=2 (n=6). Finally, modulo nine, the powers of 2 repeat with period 6 (that is, 26 = 1 (mod 9)), so modulo nine n is congruent to one of the three numbers 21-1(21-1), 23-1(23-1), or 25-1(25-1), which are all 1 (mod 9).Conjectures and Unsolved ProblemsDoes an odd perfect number exist?We have so far known that even perfect numbers are 2n-1(2n-1)from the Theorem One above, but what about odd perfect numbers? If there is an odd perfect number, then it has to follow certain conditions-To be a p erfect square times an odd power of a single primeIt is divisible by at least eight primes and has to have at least 75 prime factors with at least 9 distinctIt has at least 300 decimal digits and it has a prime divisor greater that 1020.Are there infinite numbers of Mersenne primes?The set is probably yes because of the harmonic sequence deviation.The New Mersenne ConjectureP. T. Bateman, J. L. Selfridge and Wagstaff, Jr., S. S., have conjectured the following-Let n be any odd natural number. If two of the following statements hold, subsequently so does the thirdn = 2p+/-1 or n = 4p+/-32n-1 is a prime(2n+1)/3 is a prime.Are all Mersenne number 2n-1 square free?This is kind of like an open question to which the answer is still not known and hence it cannot be called a conjecture. It is simplex to illustrate that if the square of a prime n divides a Mersenne, then p is a Wieferich prime which are uncommon Only two are acknowledged lower than 4,000,000,000,000 and none of these squar ed divide a Mersenne. If C0 = 2, then let C1 = 2C0-1, C2 = 2C1-1, C3 = 2C2-1 then are all of these prime numbers?Dickson Catalan (1876) responded to Lucas stating 2127-1 (which is C4) being a prime with this sequenceC0 = 2 (which is a prime)C1 = 3 (which is a prime)C2 = 7 (which is a prime)C3 = 127 (which is a prime)C4 = 170141183460469231731687303715884105727 (which is a prime)C5 1051217599719369681875006054625051616349 (is C5 a prime or not?)It looks as if it will not be very likely that C5 or further larger terms would be prime number. If there is a single composite term in this series, then by theorem one each and every one of the following terms would be composite.Are there more double-Mersenne primes?Another world(a) misunderstanding was that if n=Mp is prime, then so is MnLets endure this number Mn to be MMp which would be a double-Mersenne.As we dedicate this to the first four such numbers we get prime numbersMM2 = 2(4-1) -1= 23-1 =7MM3 =2(8-1)-1 =127MM5 =2(32-1)-1=21474 83647,MM7 =2(128-1)-1 =170141183460469231731687303715884105727.Application of Mersenne PrimeIn computer science, unspecified p-bit integers can be utilized to express numbers up to Mp.In the numeric problem loom of capital of Vietnam is where the Mersenne primes are used. It is a mathematical cause consisting of three rods, and a number of phonograph records of different sizes, which can slide onto any rod. The puzzle begins with the disks in ascending order of size on the first rod, the largest at the bottom to the smallest at the top. A diagram given below illustrates the Tower of Hanoi.The objective of the puzzle is to move the entire stack to other rod, obeying the following rulesOnly one disk whitethorn be travel at a time.Each move consists of taking the upper disk from one of the rods and sliding it onto another rod, on top of the other disks that whitethorn already be present on that rod.No disk may be placed on top of a smaller disk.Now to solve this punt with a p-dis c tower needs the minimum of Mp no of steps, where p is the no of disc used in the Tower of Hanoi and if we use the formula of Mersenne then we get the required result.An example of this would be if there were 5 discs involved in this Tower of Hanoi then the least number of steps required to finish this game would be 31 steps minimum.ConclusionAfter investigating the entire aspects, functions, and few applications of Mersenne Primes I believe that there is still many dissonant theories when it comes to Mersenne primes. These primes are also useful to investigates much further and deeper into the number governing body and help us to understand more sets of numbers such as Fermat prime, Wieferich prime, Wagstaff prime, Solinas prime etc.