Independence of the Miller-Rabin and Lucas Probable Prime Tests

Transcription

1 Idepedece of the Miller-Rabi ad Lucas Probable Prime Tests Alec Leg Metor: avid Corwi March 30, 017 1

2 Abstract I the moder age, public-key cryptography has become a vital compoet for secure olie commuicatio. To implemet these cryptosystems, rapid primality testig is ecessary i order to geerate keys. I particular, probabilistic tests are used for their speed, despite the potetial for pseudoprimes. So, we examie the commoly used Miller-Rabi ad Lucas tests, showig that umbers with may owitesses are usually Carmichael or Lucas-Carmichael umbers i a specific form. We the use these categorizatios, through a geeralizatio of Korselt s criterio, to prove that there are o umbers with may owitesses for both tests, affirmig the two tests relative idepedece. As Carmichael ad Lucas-Carmichael umbers are i geeral more difficult for the two tests to deal with, we ext search for umbers which are both Carmichael ad Lucas-Carmichael umbers, experimetally fidig oe less tha We thus cojecture that there are o such composites ad, usig multivariate calculus with symmetric polyomials, begi developig techiques to prove this.

3 1 Itroductio I the curret iformatio age, cryptographic systems to protect data have become a fudametal ecessity. With the quatity of data distributed over the iteret, the importace of ecryptio for protectig idividual privacy has greatly rise. Ideed, accordig to [EMC16], cryptography is allows for autheticatio ad protectio i olie commerce, eve whe workig with vital fiacial iformatio (e.g. i olie bakig ad shoppig). Now that more ad more trasactios are doe through the iteret, rather tha i perso, the importace of secure ecryptio schemes is oly growig. Thus, as some commoly used public key cryptography algorithms rely o large semiprimes that are extremely difficult to factor, fast prime geeratio is vital for key geeratio. For example, the RSA cryptosystem is oe of the most widely used systems for cryptography eve whe ot used directly, its ofte used to ecrypt the keys for other cryptographic algorithms ([CSE11]). To implemet it, RSA requires semiprimes that are aroud 048 bits (about 600 decimal digits) for its key ([EMC16]). These semiprimes are the product of two primes, so oe eeds to quickly geerate primes that are aroud 1048 bits (aroud 300 decimal digits). If we have a way to quickly determie if a umber is prime, we ca quickly geerate the ecessary primes by choosig cadidate umbers i the right size, util we fid oe that is prime. 1.1 Primality Tests Methods for determiig whether or ot a umber is prime are simply called primality tests. There are two distict types of primality tests. O oe had, determiistic tests always decide for certai whether or ot is prime. O the other had, probabilistic tests determie either that is composite, or that it might be prime. This is doe by checkig whether or ot satisfies some properties which are true of all primes ad few composites. While these probabilistic tests do ot always correctly determie if a umber is prime, they are geerally far faster tha determiistic tests. Ideed, whe compared with the fast (i the sese that it rus i polyomial-time), determiistic, Agrawal-Kayal-Saxea (AKS) primality test, effective probabilistic primality tests are still millios of times faster [AKS04]. Thus, probabilistic primality tests remai very relevat. Ideed, eve whe attemptig to prove that a umber is prime, probabilistic tests ca be rapidly applied to quickly exclude most composite umbers, reducig the overall time the primality test takes. 3

4 1. The Fermat Primality Test The Fermat test is costructed from Fermat s Little Theorem, which says that for a prime p, ad ay a relatively prime to p, we have a p 1 1 (mod p). So, to tur this ito a primality test, give a odd umber, we choose some positive iteger a < that is relatively prime to ad see if a 1 1 (mod ). If the cogruece holds, the might be prime. However, if the cogruece does ot hold, the we kow is a composite, ad we call a a witess for s compositeess. Similarly, if is actually composite, ad the cogruece holds for some a, the we call that a a owitess, ad a Fermat pseudoprime. However, the Fermat Primality test itself is ot particularly useful; there are some for which every a is a owitess. These umbers are called Carmichael umbers, ad they will play a vital role whe lookig at pseudoprimes to the Miller-Rabi test. 1.3 The Miller-Rabi Primality Test I the Miller-Rabi test, we express our odd iteger as = 1+d k, for d a odd iteger. The, as with the Fermat test, we choose a positive iteger a <, relatively prime to, as a potetial witess. However, with the Miller-Rabi test, we check that either a d 1 (mod ), or a r d 1 (mod ) for some r < k. If oe of these coditios hold, the is a probable prime. If is actually composite, the we say that it is a strog pseudoprime, ad that a is a owitess. Likewise, if either of the coditios hold, the a is a witess to ad is kow to be composite. As this paper will ot examie the Fermat test, witess ad owitess, whe used to describe a iteger a, will be used to mea a witess or owitess for the Miller-Rabi test. For a give, we will use N() to deote the umber of Miller-Rabi owitesses to. We will focus o this test, ad the Lucas Probable Prime test which we ow itroduce. 1.4 The Lucas Probable Prime Test To costruct the weak Lucas test, we first defie the Lucas Sequeces U ad V, whe give two itegers P ad Q, by U 0 = 0, U 1 = 1, U k+ = PU k+1 QU k, ad V 0 =, V 1 = ( P, V) k+ = PV k+1 QV k. The, let = P 4Q, ad let ε() deote the Jacobi symbol. We have the followig as a well-kow theorem: Theorem. [Ar97] Let p be a prime umber, relatively prime to Q. Set p ε(p) = k q for q a odd iteger. The, p U k q. 4

5 Note that this coditio ca be viewed as aalogous to the coditio for the Fermat test. So, just like with the Fermat test, there is a stroger form of the statemet; if p is prime, the oe of the followig coditios is satisfied: p U q or there exists some i, 0 i < k, where p V i q. Let O be the rig of algebraic itegers i Q( ). The, [Ar97] shows that each Lucas sequece is i a oe-to-oe correspodece with the orm-1 elemets τ i O where τ 1 is a uit i O/. Furthermore, [Ar97] also shows that for a give Lucas sequece, if is relatively prime to Q, the U k τ k 1 (mod ) ad V k τ k 1 (mod ). I light of these equivaleces, we will ofte study the Lucas test with algebraic itegers, ot with Lucas sequeces. Now, we ca tur the Lucas coditio ito a primality test, startig with some relatively prime to Q ad settig ε() = k q for q odd. If, for a composite, either U q or there is some i, 0 i < k, where p V i q, the we call a strog Lucas pseudoprime, sice it satisfies the strog form of the Lucas test, ad the pair of parameters (P,Q) (or the correspodig quadratic iteger τ), a owitess. We will ow itroduce some otatio that is useful whe talkig about Lucas pseudoprimes. Suppose > factors ito primes as = p e 1 1 pe m m. Let ε i deote ε(p i ). The, for 1 i m, let p i ε i = k i q i for q i odd. Fially, let SL(,) deote the umber of owitesses, aalogous to N() for Miller-Rabi owitesses. The Lucas test, whe used correctly, ca be highly idepedet of the Miller Rabi test, makig it particularly iterestig. Ideed, several algorithms exploit this idepedece to yield extremely powerful primality tests; of particular ote is the Baillie-PSW test, for which there are o kow composite which are probably prime. 1.5 This Research I Sectio, we begi by examiig prior results about umbers with may owitesses for both the Miller-Rabi ad Lucas Pseudoprime tests. We defie our cocept of Lucas- Carmichael umbers as a aalog of the Carmichael umbers, but for the Lucas Probable Prime test, ad give a categorizatio of umbers with may owitesses for both the Miller-Rabi ad Lucas tests i terms of Carmichael ad Lucas-Carmichael umbers. I order to prove that o umbers have may owitesses for both tests, we prove that if a composite has three prime factors, the it caot be a Carmichael umber ad a Lucas- Carmichael umber. The, i Sectio 3, we begi to geeralize this result to show that there are o composites that are both Carmichael ad Lucas-Carmichael umbers, developig 5

6 lemmas ad techiques before applyig them to a particular form of umbers with five prime factors to show that there are o umbers i that form that are both Carmichael ad Lucas-Carmichael umbers. We coclude i Sectio 4 by cosiderig possible future work. Fidig Composites with May Nowitesses.1 Prior Results I the case of the Miller-Rabi test, we have the below classificatio for the umbers with high amouts of owitesses, where ϕ() deotes the Euler totiet fuctio, ad v () is the greatest atural umber so that v (). v () is also kow as the -adic valuatio. Theorem 1. [Nar14] Suppose is a odd composite umber 81. The N() = ϕ() iff 4 is of the form (k + 1)(4k + 1), for k odd ad k + 1, 4k + 1 prime, or is a Carmichael umber of the form pqr, where p,q, ad r are distict primes 3 (mod 4). Furthermore, ϕ() < N() < ϕ() iff is of the form (k + 1)(6k + 1), for k odd ad k + 1, 6k prime, ad N() = ϕ() iff has the form (k + 1)(4k + 1) where k is eve. Fially, ϕ() 7 < N() < 5ϕ() 3 6 iff is a Carmichael umber of the form pqr, where v (p 1) = v (q 1) = v (r 1) > 1. Otherwise, N() ϕ() 8. [Ami15] provides a similar theorem for the Lucas test. Theorem. [Ami15] SL(,) 6 uless = 9 or 5, m = ad either = (k 1 q 1 1)( k 1 q 1 + 1), where is the product of twi primes, or = ( k 1 q 1 + ε 1 )( k 1+1 q 1 + ε ), or m = 3, ad = ( k 1 q 1 + ε 1 )( k 1 q + ε )( k 1 q 3 + ε 3 ), where q 1,q,q 3 q.. Carmichael ad Lucas-Carmichael Numbers I Theorem 1, several of the cases are Carmichael umbers. So, we state a well-kow theorem about Carmichael umbers that will allow us to talk about them more easily: Theorem 3 (Korselt s Criterio). A positive composite umber is a Carmichael umber if ad oly if it is square-free, ad for all its prime factors p i, p i 1 1 [Co]. Now, just like how Carmichael umbers completely defeat the weaker form of the Miller-Rabi test (the Fermat test), we cosider the umbers that completely defeat the weak Lucas test: 6

7 efiitio 1 (Lucas-Carmichael Number). A Lucas-Carmichael umber for a give value of is a composite umber, relatively prime to, such that for every Lucas sequece (P,Q), U ε(). Equivaletly, for every τ that is a orm-1 elemet i O where τ 1 is a uit i O/, τ ε() 1 (mod ). (A orm-1 elemet is the image of some x O uder the caoical map φ : O O/, where Norm(x) 1. We will deote the multiplicative group of such elemets by (O/).) We are the able to devise a aalog of Korselt s Criterio for these Lucas-Carmichael umbers, provig our theorem by workig with quadratic itegers. Theorem 4. A positive composite umber, relatively prime to, is a Lucas-Carmichael umber (with respect to ) if ad oly if it is square-free, ad for every prime p i, p i ε(p i ) ε(). Proof. We first prove that if is a Lucas-Carmichael umber, the it is square free, ad for every prime p i, p i ε(p i ) ε(). Suppose factors as p e 1 1 pe pe m m. Cosider the multiplicative group of orm-1 elemets i (O/p e i i ). Arault shows that this group is cyclic, with order pe i 1 i (p i ε(p i )). [Ar97] So suppose some g i is a geerator. The, by the Chiese Remaider Theorem, there is some orm-1 g O with g g i (mod p e i i ) for all i. Now, we claim g 1 must be ivertible i O for the sake of cotradictio, suppose otherwise. If 1 (mod 4), the O = Z[ ], so we ca express g = a + b. The, we kow N(g) = N(a + b ) = a b = 1. Note that N(g 1) = N(a 1 + b ) = (a 1) b = a b a+1 = N(g) a+1 = a. If this is relatively prime to i Z, the N(g 1) is ivertible, which would imply that g 1 is ivertible i O. So there is some p i with p i N(g 1). The a (mod p i ). We assumed that was relatively prime to, so p i. So a 1 (mod p i ). Now, N(g) = a b = 1, so a 1+b (mod p i ), takig the caoical mappig from Z to Z/p i. The, as a 1 (mod p i ), a 1 (mod p i ), so b 0 (mod p i ). So p i or p i b. But p i, ad we assumed that was relatively prime to. So p i b. Therefore i O/p i, g 1 (mod p i ). Alteratively, if 1 (mod 4), the O = Z[ 1+ ], ad we ca write g = a + b 1+. Substitute A = a + b,b = b, so that g = A+B. The, 1 = N(g) = N( A+B ) = A B 4. Also, N(g 1) = N( A +B ) = (A ) B 4 = N(g) A + 1 = A. If this is relatively prime to, the it is ivertible, which would make g 1 ivertible. So as before, there is some p i with N(g 1) = A 0 (mod p i ). The as p i, A 1 (mod p i). So A 4 1 (mod p i). The, as 1 = N(g) = A B 4, B 0 (mod p i ). Thus, as before, we 7

8 get that p i B, as ad are relatively prime. So, i O/p i, g 1 (mod p), regardless of the choice of. e Now, g is a geerator of (O/p i i ). But there are oly p e i 1 i < p e i 1 i (p i ε(p i )) elemets e i O/p i i which are cogruet to 1 (mod p i ), so g caot geerate the whole group. Thus, g 1 is ivertible i O. But the g is a orm-1 elemet i O where g 1 is ivertible. So by our iitial assumptio, g ε() 1 (mod ). So certaily, for ay p i, g ε() 1 (mod p e i i ). So, as the order of g is p e i 1 i (p i ε(p i )), as it is cogruet to some g i that is a geerator, p e i 1 i (p i ε(p i )) ε(). But as p e i i, pe i i is relatively prime to ε(). So, we must have that e i = 1. So is square-free. The, pluggig i e i = 1, we have that for ay p i, p i ε(p i ) ε(), as desired. Coversely, suppose that we ca factor ito distict primes = p 1 p p m where, for all p i, p i ε(p i ) ε(). The, for ay τ O with orm 1, τ O/p i. So, as the order of O/p i is p i ε(p i ), τ p i ε(p i ) 1 (mod p i ). So, as p i ε(p i ) ε(), τ ε() 1 (mod p i ). So, by the Chiese Remaider Theorem, τ ε() 1 (mod ), as desired. The followig two lemmas will allow us to restate Theorem. Lemma. For with two prime factors, is a Lucas-Carmichael umber if ad oly if it is the product of twi primes, p, p +, where ε(p) = 1,ε(p + ) = 1. Proof. Suppose is a Lucas-Carmichael umber with two prime factors. Usig the previously defied otatio, = k q + ε() = ( k 1 q 1 + ε 1 )( k q + ε ). Multiplyig out, = ( k 1+k q 1 q + k q ε 1 + k 1 q 1 ε ) + ε 1 ε. Note that ε() = ε 1 ε. Now, suppose without loss of geerality that k 1 k. is a Lucas-Carmichael umber, so k q ε = ( k 1+k q 1 q + k q ε 1 + k 1 q 1 ε ). So, k k 1 q 1 ε ). But q 1 is odd, so it is relatively prime to k. So k k 1. Thus, k k 1. But we assumed k 1 k. So k 1 = k. Furthermore, from k q ε = ( k 1+k q 1 q + k q ε 1 + k 1 q 1 ε ), we get that q k 1 q 1 ε. So as q is odd, q q 1. Oe ca similarly show that q 1 q. So, q 1 = q. The, = ( k 1 q 1 + ε 1 )( k 1 q 1 + ε ). ε 1 ε, or the two factors would be the same, cotradictig that is square-free. So, assume without loss of geerality that ε 1 = 1, ε = 1. Now, settig p = k 1 q 1 + ε 1 = k 1 q 1 1, we see that we have factored ito the product of twi primes, p (p+), where ε(p) = ε 1 = 1 ad ε(p + ) = ε = 1, as desired. Coversely, suppose = p(p+) where ε(p) = 1,ε(p+) = 1. The ε() = 1, so ε() = + 1 = p(p + ) + 1 = p + p + 1 = (p + 1). Note that p ε(p) = p + ε(p + ) = p + 1. So, p ε(p), p + ε(p + ) both divide ε(). So, as is clearly square-free, it is a Lucas-Carmichael umber. 8

9 Lemma. For with three prime factors, if ad oly if factors as ( k 1 q 1 + ε 1 )( k 1 q + ε )( k 1 q 3 +ε 3 ), where q 1,q,q 3 q, the is a Lucas-Carmichael umber with v (p 1 ε 1 ) = v (p ε ) = v (p 3 ε 3 ). Proof. Clearly, if factors i that way, the as the q i are odd, v (p 1 ε 1 ) = v (p ε ) = v (p 3 ε 3 ). Now, we ca multiply out the expressio for to get = 3 k 1 q 1 q q 3 + k 1 q q 3 ε 1 + k 1 q 1 q 3 ε + k 1 q 1 q ε 3 + k 1 q 1 ε ε 3 + k 1 q ε 1 ε 3 + k 1 q 3 ε 1 ε + ε 1 ε ε 3. So the, k 1 ε, as ε = ε 1 ε ε 3. So, as ε = k q, k 1 ε, ad q is odd, k1 k. So as q i q, k1 k, ad k1 ad q i are relatively prime, sice the q i are odd, p i ε i = k1 q i k q = ε. So is a Lucas-Carmichael umber with v (p 1 ε 1 ) = v (p ε ) = v (p 3 ε 3 ). Coversely, suppose is a Lucas-Carmichael umber with v (p 1 ε 1 ) = v (p ε ) = v (p 3 ε 3 ). Certaily, without loss of geerality, we ca factor = ( k 1 q 1 + ε 1 )( k q + ε )( k 3 q 3 + ε 3 ). The, the -adic valuatios are equal, so we must have k 1 = k = k 3. Fially, p i ε i ε, so k i q i k q. But q i is relatively prime to k, so q i q. So = ( k 1 q 1 + ε 1 )( k 1 q + ε )( k 1 q 3 + ε 3 ), where q 1,q,q 3 q. We the restate Theorem : Theorem 5 (Restatemeet of Result by [Ami15]). SL(,) uless = 9 or 5, m = 6 ad either is a Lucas-Carmichael umber or = ( k 1 q 1 + ε 1 )( k1+1 q 1 + ε ), or m = 3, ad is a Lucas-Carmichael umber where v (p 1 ε 1 ) = v (p ε ) = v (p 3 ε 3 )..3 Properties of Problematic Composites We ow wat to examie umbers that have N() > ϕ() ad SL(,) > ; call such 8 6 problematic. Assume that > 81, sice whe testig for primality, we would check that o small primes divide ito the umber, takig care of these smaller cases. Now, accordig to Theorem 1 ad Theorem, problematic umbers ca oly have two or three prime factors. The case for two prime factors ca be easily dealt with, so we will do that ow. ( ) Lemma 1. Suppose we choose a such that = 1. The there are o problematic umbers that have two prime factors. Proof. By Theorem 1, we kow that if N() > ϕ(), ad > 81 has two prime factors, 8 the = (k + 1)(4k + 1) for some iteger k, or = (k + 1)(6k + 1) for some odd k. By Theorem, we also kow that either is the product of twi primes, or = ( k 1 q 1 + ε 1 )( k 1+1 q 1 + ε ). Now, clearly, if = (k + 1)(4k + 1) or (k + 1)(6k + 1), the prime 9

10 factors of differ by more tha. So = ( k 1 q 1 + ε 1 )( k 1+1 q 1 + ε ). Now, so either ε 1 = 1,ε = 1 or ε 1 = 1,ε = 1. ( ) = 1, I the former case, = ( k 1 q 1 1)( k 1+1 q 1 + 1). Clearly the latter factor is larger. So if = (k + 1)(6k + 1) for some odd k, the 6k + 1 = k 1+1 q If 6k + 1 = k 1+1 q though, the as k is odd, v ( k 1+1 q 1 ) = v (6k) = 1. The k 1 = 0. But the the smaller prime factor of is k 1 q 1 1 = q 1 1. But q 1 is odd, so k 1 q 1 1 = q 1 1 is eve. But it also has form k + 1 for some k. Alteratively, if = (k + 1)(4k + 1) for some k, the 4k = k 1+1 q 1, k + 1 = k 1 q 1 1. So, as q 1 is odd, v (k) = k 1 1. The, v (k) = k 1. But ote that k + = k 1 q 1, so v (k + ) = k 1. However, k,k + are both odd, so oe of them is (mod 4) ad oe of them is 0 (mod 4) they certaily caot have equal -adic valuatios. I the latter case, = ( k 1 q 1 + 1)( k 1+1 q 1 1). Suppose that = (k + 1)(6k + 1), for k odd. We still have that the first factors are both the smaller oes, so 6k +1 = k 1+1 q 1 1, k + 1 = k 1 q The k = k 1 q 1, so as k is odd, k 1 = 1,k = q 1. So 6q = 6k + 1 = k 1+1 q 1 1 = 4q 1 1. So q 1 = 1. The = 3 7, ad we certaily have < 81, so is ot problematic. Fially, suppose that = (k + 1)(4k + 1). The, 4k + 1 = k 1+1 q 1 1, k + 1 = k 1 q So v (k) = k 1. The v (4k) = k But 4k + = k 1+1 q 1, so v (4k + ) = k So, we look oly at the case where has three prime factors. I this case, it has a large umber of Miller-Rabi owitesses, so it must be a Carmichael umber with v (p 1 1) = v (p 1) = v (p 3 1). Furthermore, sice it has a large umber of Lucas owitesses it is a Lucas-Carmichael umber, with v (p 1 ε 1 ) = v (p ε ) = v (p 3 ε 3 ). Usig some of these characteristics, we prove the followig lemma. Lemma. If has three distict prime factors, where p i = k 1 q i + ε i (so( v (p ) 1 ( ε 1 ) = ) v (p ε ) = v (p 3 ε 3 )), ad v (p 1 1) = v (p 1) = v (p 3 1), the = ( p i a for 1 i 3, where deotes the Jacobi symbol. b) Proof. ( ) Suppose ( ) without ( loss ) of geerality ( ) that for two of our prime factors, p 1 ad p,, where = 1, = 1. The, p 1 1 = k 1 q 1, so v (p 1 ) = k 1. p 1 p p 1 p Meawhile, p = k 1 q 1, so p 1 = k 1 q. But as v (p 1 1) = v (p 1), we thus get that v ( k 1 q ) = k 1. So, i particular, k 1 k 1 q. So k 1. So k 1 = 1. But the, p 1 = q = ( (q ) 1). ( But ) sice q is odd, (q ( 1), ) so( v (p)( ) k. )( So, for ) ay two prime factors, =. But we kow that =, by p i p j p 1 p p 3 10

11 the properties of the Jacobi symbol. So, The followig corollary is immediate. ( ) ( ) 3 ( ) = =, as desired. p 1 p 1 Corollary 1. All problematic umbers with three prime factors satisfy ( ) = ( ) ( ) So, we oly have two differet cases to cosider whe = 1 ad whe = ( ) 1. However, ote that i geeral, we are able to chose the value of based o our ( ) ( ) choice of. I fact, we wat to choose = 1; if = 1, the we have more problematic umbers, sice the costraits o are ot all idepedet. Lemma. If is a Carmichael umber with three prime factors, ( ad ) v (p 1 1) = v (p 1) = v (p 3 1), v (p 1 ε 1 ) = v (p ε ) = v (p 3 ε 3 ), ad = 1, the is a Lucas Carmichael umber. ( ) ( ) Proof. By Corollary 1, we have that = = 1 for all p i. Now, we earlier set ( ) p i ( ) ε = k q, where ε = = 1. So, as ε i is just, ad p i 1 1, by Korselt s p i Criterio (sice is a Carmichael umber), we have k 1 q i k q. So q i k q. But as q i is odd, it is relatively prime to k. So q i q for all i. The as v (p 1 ε 1 ) = v (p ε ) = v (p 3 ε 3 ), by our earlier lemma, is a Lucas-Carmichael umber. ( ) Ideed, [BW00] also otes that wat = 1, or the Miller-Rabi ad Lucas tests ( ) are ot as idepedet, though their justificatio is differet. They show how if = 1, ad is a strog pseudoprime to some as for the Miller-Rabi test, the must be a pseudoprime to certai ( ) pairs (P,Q) as well. So, we ow assume all are Lucas tests are performed where = 1, ad whe searchig for problematic umbers with three ( ) ( ) prime factors, we oly look at where = = 1..4 Results of Searchig We bega by testig the Carmichael umbers less tha 30 which have three prime factors such that v (p 1 1) = v (p 1) = v (p 3 1). I this search, o problematic umbers were foud. Ideed, we preset the followig result: p i ( p i ). 11

12 Theorem. There are o problematic umbers. To prove this, we first show the followig lemmas ( ) Lemma. If = 1, ad is both a Carmichael umber ad a Lucas-Carmichael ( ) umber, the = 1 for every prime p i. p i ( ) Proof. Suppose otherwise that for some p i, = 1. The, p i ε i ε, where ε = 1, ε i = 1, so p i But is also a Carmichael umber, so p i 1 1. The, p i 1. So ( as) p i is a prime, p i = 3. The, = 1, so there is some prime p j where ε j = 1. The p j 1 1, p j But p j 1 (mod 3), or 3 p j 1, implyig that 3 1, ad similarly, p j (mod 3), or 3 p j + 1, so these two are impossible, sice 3. But also, p j 3, uless p j = ( 3, ) cotradictig its distictess. So, there is o p i with εi = 1. Thus, for every p i, ε i = = 1, as desired. p i ( ) Corollary. If = 1, the there are o umbers with a eve umber of prime factors that are both Carmichael umbers ad Lucas-Carmichael umbers. ( ) Proof. Suppose has k prime factors. The, 1 = = ( 1) k by Lemma.4. So k must be odd. Lemma. If is both a Carmichael ad a Lucas-Carmichael umber, the for every prime p, p (mod p 1 ). Proof. As p, we ca write = ap for some iteger a. The, p 1 1 by Korselt s Criterio, so p 1 ap 1 ( = ) ap a+a 1 = a(p 1)+a 1. So p 1 a 1. Meawhile, from our last lemma, = 1, so from the aalog of Korselt s Criterio for Lucasp Carmichael umbers, we get p The, + 1 = ap + 1 = ap + a a + 1 = a(p + 1) a + 1, so p + 1 a + 1, or p + 1 a 1. Now, cosider gcd(p 1, p + 1). Certaily, this is either or 1 as p is a odd prime though, p 1, p + 1. So gcd(p 1, p + 1) =. Now, we kow p 1 = (p + 1)(p 1) = lcm(p + 1, p 1) gcd(p + 1, p 1) = lcm(p + 1, p 1). So lcm(p + 1, p 1) = p 1. The, as p 1 a 1, p + 1 a 1, we have lcm(p + 1, p 1) a 1. So p 1 p i a 1, or equivaletly, a 1 (mod p 1 ). 1

13 The, p is certaily relatively prime to p 1, so we ca multiply by p to get = ap p (mod p 1 ), as desired. Theorem 6. There are o composites with three prime factors that are both Carmichael Numbers ad Lucas-Carmichael umbers. Proof. Let = pqr be a Lucas-Carmichael umber, ad a Carmichael umber. The, without loss of geerality, suppose p < q < r. From our previous lemma, we get that pq 1 (mod r 1 ). So, pq = 1 + x r 1, for some iteger x. Now, if x, the we get pq r, cotradictig that p < q < r. Furthermore, we clearly kow that x 0. So, x = 1, ad pq = 1 + r 1. Note that 1 + r 1 = r + 1. So r + 1 = pq. I particular, r < pq. Note that 3 < q. So, 9(3 ) < q (3 ). I particular, 1 < 9(3 ). The 1 < q (3 ) = 3q q. So, q < 3q 3. As the right had side is positive, ( q 1) q < 3, ad as p < q, p pq < q. Sice r < (pr 1) pq, 1 q 1 < 3. The pr 1 ( q 1 ) < 3. By lemma 7, pqr q (mod q 1 ). So q 1 pr 1. The, pr 1 is a iteger, so ( q 1 ) it must be 1 or. But if pr 1 ( q 1 ) =, the pr = q, a cotradictio. So pr 1 = 1. Note ( q 1 ) that 1 = pq 1 ( r 1 ) < pr 1 ( q 1 ) = 1. After showig the above, we became aware of a equivalet result by [Wil77], usig a alterative proof. Theorem 7. There are o problematic umbers. Proof. Follows immediately from Lemma 1 ad Theorem 6, sice problematic umbers with three prime factors must be both Carmichael umbers ad Lucas-Carmichael umbers. 3 Numbers that are Carmichael ad Lucas-Carmichael I geeral, Carmichael ad Lucas-Carmichael umbers ted to have may owitesses for the Miller-Rabi ad Lucas Probable Prime tests. Ideed, this fact is evidet from the 13

14 formulas for the umber of owitesses stated i [Ar97]. Thus, umbers that are both Carmichael ad Lucas-Carmichael become of iterest. 3.1 Searchig for Carmichael ad Lucas-Carmichael Numbers I showig there were o problematic umbers, we were able to prove that there were o umbers with either a eve umber of prime factors, or three prime factors, that were both Carmichael ad Lucas-Carmichael. So, oe might expect that there are, i geeral, o umbers that are both Carmichael ad Lucas-Carmichael umbers. Ideed, after examiig all Carmichael umbers less tha 10 16, we foud o Lucas-Carmichael umbers ([Pi]). However, despite havig already show several lemmas about umbers that are both Carmichael ad Lucas-Carmichael umbers, tryig to geeralize the result to other odd umbers of factors proves much more difficult. The followig lemma provides a useful startig poit: Lemma. If is both a Carmichael umber ad a Lucas-Carmichael umber, the all primes dividig are equivalet mod 1. Proof. Clearly, must have multiple prime factors. If, there is aother odd prime p, so 1 is odd, but divisible by a eve umber. Similarly, if 3, but there is either aother odd prime p, p 1 (mod 3), i which case p 1 1 = 3 1, or there is a odd prime p, p (mod 3), so as p+1 +1 sice is a Lucas-Carmichael umber, Either way, we reach a cotradictio. So, gcd(,6) = 1. Furthermore, we caot have p,q, p 1 (mod 4),q 3 (mod 4) p 1 1, so 4 1, but q , so 4 + 1, a cotradictio. Similarly, we caot have p, q, p 1 (mod 3),q (mod 3) as p 1 1 = 3 1, ad q = So, as ay prime dividig must be 1,3 (mod 4), ad 1, (mod 3), we get that all primes dividig are equivalet mod 1 from the Chiese Remaider Theorem. ( ) Now, if is both a Carmichael ad a Lucas-Carmichael umber, whe = 1, the we have that lcm(p 1 1,..., p m 1) 1 ad lcm(p 1 +1,..., p m +1) +1. Furthermore, by defiitio lcm(p 1 1,..., p m 1) (p 1 1) (p m 1) ad lcm(p 1 + 1,..., p m + 1) (p 1 + 1) (p m + 1). 14

15 So both (p 1 1) (p m 1) lcm(p 1 1,..., p m 1) ad (p 1 + 1) (p m + 1) lcm(p 1 + 1,..., p m + 1) are itegers these umbers measure of much overlap there is betwee the differet p i 1s ad p i + 1s respectively. The smaller these umbers are, the stroger the coditios lcm(p 1 1,..., p m 1) 1 ad lcm(p 1 + 1,..., p m + 1) + 1 are, as more primes are required to divide 1 or Begiig to Geeralize We start developig techiques to geeralize the results of Theorem 6 by examiig the case where has five prime factors, 11 mod 1 (by lemma 8, these two coditios (p 1 1) (p 5 1) imply that for every prime p, p 11 mod 1), ad = 16. The lcm(p 1 1,..., p 5 1) last coditio comes from the fact that 4 is the miimum possible value for the ratio as p i 11 mod 1, divides p i 1 exactly oce for all p i, so divides the umerator five times ad the deomiator oce. So, the ratio must be a (positive iteger) multiple of 4 = 16. I the special case we are lookig at, we thus have that (p 1 1) (p 5 1) 1. So 16 without loss of geerality, there is some iteger r such that (p 1 1) (p 5 1) (16 + r) = As (p 1 1) (p 5 1) < 1, r must eve be positive. Now let s a,b deote the ath elemetary symmetric polyomial i b variables. (For example, s 1,b (x 1,...,x b ) = x i, 1 i b s,b (x 1,...,x b ) = x i x j, ad s b,b = x 1 x b.) The we ca express (p 1 1) (p 5 1) 1 i< j b as s 5,5 s 4,5 +s 3,5 s,5 +s 1,5 1, ad 1 as s 5,5 1, where the argumet for each of the symmetric fuctios is (p 1, p, p 3, p 4, p 5 ). So, substitutig ito (p 1 1) (p 5 1) ( r) = 1 ad rearragig, we have have that r(s 5,5 s 4,5 +s 3,5 s,5 +s 1,5 1) 16(s 4,5 s 3,5 + s,5 s 1,5 ) = 0. This motivates defiig the fuctio f r (x 1,x,x 3,x 4,x 5 ) = r(s 5,5 s 4,5 +s 3,5 s,5 +s 1,5 1) 16(s 4,5 s 3,5 +s,5 s 1,5 ) where here, each of the elemetary symmetric polyomials has (x 1,x,x 3,x 4,x 5 ) as its argumet. We ca ow employ calculus to show that f r has o positive iteger roots where for all i, p i 11 mod 1, thus showig that there are o umbers that are both Carmichael ad Lucas-Carmichael with the specific properties metioed earlier. Lemma. Cosider a poit i R 5 give by q = (q 1,q,q 3,q 4,q 5 ), where q 1 < q < q 3 < q 4 < q 5. The if f, f r f r 3 f r 4 f r,,, are all positive at q, the at x 5 x 4 x 5 x 3 x 4 x 5 x x 3 x 4 x 5 15

16 ay poit (q 1,q,q 3,q 4,q 5 ) where for all i, q i < q i, f is also positive. s a,b Proof. It is a well kow fact that, for a,b > 0, = s a 1,b 1, where s 0,b = 1 for x b ay b. Repeatably applyig this rule with the liearity of the derivative, we get that 5 f r 4 f r = r. I particular, this fuctio is always positive. Now, x 1 x x 3 x 4 x 5 x x 3 x 4 x 5 is oly a fuctio of x 1, so if its positive at q 1, the its positive at ay y 1 > q 1 sice its derivative is always positive. So is positive at ay pair (y 1,y ) if there is a permutatio 3 f r x 3 x 4 x 5 π with q 1 < y π(1),q < y π(), sice permutig the order of the argumets does ot chage the value of the fuctio (as the fuctio is a polyomial i terms of elemetary symmetric polyomials ad is thus symmetric), ad the derivative is positive for the relevat values. Repeatig this process, we fid that whe choosig ay poit (y 1,y,y 3,y 4,y 5 ) R 5, where some permutatio π of the y i s has q i < y π(i), f is positive at that poit. Usig the lemma, we the verify the required coditios at (63,75,87,99,111) for ay positive iteger r (oe ca easily show that icreasig r oly icreases the fuctio). So, as the five prime factors i a Carmichael umber that is also a Lucas-Carmichael umber must be uique, ad differ by at least 1, we ca rule out ay umbers whose least prime factor is greater tha 63, sice f r o the five prime factors caot be 0. Now, sice we are oly lookig at the case that is 11 mod 1, ote that the smallest prime factor must the either be 11,3,47, or 59. Checkig aroud (5,37,49,61,73), for r, we fid that the lemma is applicable. So for 47 ad 59, if f r has a root, we must have r = 1. Now, for 59, check that the lemma is applicable at (59,83,107,119,131), so for ay possible set of five primes, f 1 is too large if the secod smallest prime is ot 71. The use (59,71,107,119,131) to force the third smallest prime to be 83. The use (59,71,83,119,131) to force there to be 107. Fially its easy to check that for ay prime greater tha 119, f 1 has o zero. But we ca explicitly check f 1 (59,71,83,107,119) to see that it is ot 0. So there is o composite satisfyig the required properties with least factor 59. Oe ca use a similar series of tests to show that there are o umbers with the desired properties with least prime factor of 11,3, or 47, with the caveat that for 11 ad 3, oe first shows that f r is egative if r,1 respectively, ad always positive if r 4,3 respectively. Thus, there are (p 1 1) (p 5 1) o 11 mod 1, where = 16, which are both Carmichael ad lcm(p 1 1,..., p 5 1) Lucas-Carmichael umbers. 16

17 4 Future Plas Now that we have begu to develop techiques for showig the oexistece of composites that are both Carmichael ad Lucas-Carmichael umbers, these techiques could be geeralized to show that there are o umbers that are both Carmichael ad Lucas-Carmichael umbers. I particular, we cojecture that there is a explicit boud for the two ratios (p 1 1) (p m 1) lcm(p 1 1,..., p m 1) ad (p 1 + 1) (p m + 1), ad are curretly workig to fid lcm(p 1 + 1,..., p m + 1) such a boud. The, this would eable us to use a more geeral form of the techiques from Sectio 3 to show that there caot be umbers that are both Carmichael ad Lucas- Carmichael umbers i geeral. Furthermore, the Baillie-PSW test is a well kow probabilistic primality test that derives its stregth from the idepedece of the Miller-Rabi ad Lucas tests. However, as there are o kow composites that pass the test, there is ot much isight ito how to effectively tur it ito a determiistic test (ufortuately, heuristics suggest that some composites do pass the test). Our results could be used to develop a modified form of the Baillie-PSW algorithm where the composites that pass the test are easily characterized, opeig the way for a faster provably determiistic primality testig algorithm. After all, the Miller-Rabi ad Lucas Probable Prime tests are far faster tha the curret fastest determiistic tests. 5 Ackowledgemets We would foremost like to thak the PRIMES program ad faculty for makig this research possible, particularly Program irector r. Slava Gerovitch, Chief Research Adviser Professor Pavel Etigof, ad head metor r. Taya Khovaova. We also would like to express our utmost gratitude to our metor, avid Corwi, for guidig us through the research process. I additio, we recogize Stefa Wehmeir, from Mathworks, for suggestig the project. 17

18 Refereces [AKS04] Maidra Agrawal, Neeraj Kayal, ad Niti Saxea. PRIMES is i P. A. of Math. (), 160():78 793, 004. [Ami15] [Ar97] [BW00] avid Amirault. Better Bouds o the Rate of No-Witesses of Lucas Pseudoprimes materials/015/amirault.pdf. F. Arault. The Rabi-Moier theorem for Lucas pseudoprimes. Math. Comp., 66(18): , avid Bressoud ad Sta Wago. A course i computatioal umber theory. Key College Publishig, Emeryville, CA; i cooperatio with Spriger-Verlag, New York, 000. [Co] Keith Corad. Carmichael Numbers ad Korselt s Criterio. carmichaelkorselt.pdf. [CSE11] How ca I use asymmetric ecryptio, such as RSA, to ecrypt a arbitrary legth of plaitext?, 011. Accessed September [EMC16] ell EMC. Public-Key Cryptography Stadards, 016. Accessed September [Nar14] Shyam Narayaa. Improvig the Speed ad Accuracy of the Miller-Rabi Primality Test materials/014/narayaa.pdf. [Pi] Richard G.E. Pich. Tables relatig to Carmichael umbers. Accessed July cartable.html. [Wil77] H. C. Williams. O umbers aalogous to the Carmichael umbers. Caad. Math. Bull., 0(1): ,