GENERALIZED LUCAS-LEHMER TESTS USING PELL CONICS 1. Introduction Like Elliptic Curves, There Is a Group Law on the Pell Conics

Home , Pseudoprime

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 140, Number 8, August 2012, Pages 2653–2661 S 0002-9939(2011)11196-1 Article electronically published on December 20, 2011

GENERALIZED LUCAS-LEHMER TESTS USING PELL CONICS

SAMUEL A. HAMBLETON

(Communicated by Ted Chinburg)

Abstract. Pell conics are used to write a Proth-Riesel twin-primality test. We discuss easy-to-ﬁnd primality certiﬁcates for integers of the form mnh ± 1. The known primality test for 3nh ± 1 is associated with X2 +3Y 2 =4.

1. Introduction Like elliptic curves, there is a group law on the Pell conics [4]. These are aﬃne curves of genus 0 of the form C : X2 − ΔY 2 = 4 where Δ is a fundamental discriminant. Geometrically, points P and Q on Pell conics are added by taking the line parallel to PQ, passing through the point O =(2, 0), and intersecting with C at P + Q. Algebraically this is x x +Δy y x y + x y (1) (x ,y )+(x ,y )= 1 2 1 2 , 1 2 2 1 . 1 1 2 2 2 2 Multiplication by 2 and 3 is given by

2(x, y)=(x2 − 2,xy), (2) 3(x, y)=(x3 − 3x, (x2 − 1)y).

Lemmermeyer [4] considered the arithmetic of Pell conics indicating many inter- esting similarities with elliptic curves. The following theorem appears in [4] in a Δ more general form. We use p to mean the Legendre symbol.

Theorem 1.1. Let C : X2 − ΔY 2 =4, Δ be a fundamental discriminant, p be prime and N be an integer. Then: •C(Z) is an abelian group with identity O =(2, 0),andpointT =(−2, 0) of order 2. No other points have y =0or x = ±2. The inverse of (x, y) is (x, −y). •C(Z/N ) is an abelian group with identity O. •CF − Δ ( p) is a cyclic group of order p p .

Received by the editors June 11, 2009 and, in revised form, November 9, 2010 and March 15, 2011. 2010 Mathematics Subject Classiﬁcation. Primary 11Y11; Secondary 11G30. Key words and phrases. Pell conics, Lucas-Lehmer, primality.

c 2011 American Mathematical Society Reverts to public domain 28 years from publication 2653

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For multiplication by m ≥ 1, including points of Equation (2), we deﬁne polynomials

(3) f0 =2;f1 = x; fi+1 = xfi − fi−1,

(4) g0 =0;g1 =1;gi+1 = xgi − gi−1,

(5) F1 =1;F3 = x +1;F2i+3 = xF2i+1 − F2i−1. Lemma 1.2. Let C be a Pell conic and P =(x, y) ∈C(Z). Then for m ≥ 1, mP = fm(x),y· gm(x) ,andforoddm ≥ 1, 2 (6) fm(x)=(x − 2)Fm(x) +2. Proof. mP = fm(x),y· gm(x) may be proved by induction using Equations (1), (3), and (4). To prove Equation (6), induction shows that 2 − · 2 (7) F2i−1 x F2i−1F2i+1 + F2i+1 = x +2, and using Equation (7) in another induction proves Equation (6).

Corollary 1.3 is a direct result of Lemma 1.2. C(Fp)[m] is used to denote the set of points P of C(Fp) for which mP = O.

Corollary 1.3. Let C be a Pell conic, P =(x, y) ∈C(Fp),andp prime. Then for odd m ≥ 1, P ∈C(Fp)[m]\O if and only if Fm(x) ≡ 0(modp). Lemmermeyer [4] discussed primality proving using Pell conics, giving Theo- rem 1.5 as an analogue of Lucas’ theorem. It is assumed that the integers N are greater than 1 and coprime to 6.

− N−1 Theorem 1.4 (Lucas). If aN 1 ≡ 1(modN) but a q ≡ 1(modN) for every prime factor q of N − 1,thenN is prime. Theorem 1.5. Let N ≥ 5 and C : X2 − ΔY 2 =4be a Pell conic deﬁned over Z Δ − P ∈C Z /N with N = 1.ThenN is prime if and only if for some point ( /N ), P O N+1 P O (N +1) = ,but q = for every prime factor q of N +1. Lemmermeyer [4] remarked that there are Proth versions in which only part of N ± 1 needs to be factored and commented that in the same way Gross [2] gave an elliptic curve ‘Lucas-Lehmer’ test, the Lucas-Lehmer test itself may be proved with the Pell conic C : X2 − 12Y 2 = 4 and the point (4, 1). This method is extended here to more general primality proving.

2. Twin primes of the form 2nh ± 1 The theory of Pell conics is applied to a test similar to Riesel’s [6] generalization of the Lucas-Lehmer test to N =2nh − 1. Our test, however, also includes Proth’s Theorem. One advantage in their combination is a single primality certiﬁcate for a pair of twin primes. The uppercase letter Q will henceforth be used exclusively for a ﬁxed generator of C(Fp). Lemma 2.1. Let p be an odd prime and let C : X2 − ΔY 2 =4be a Pell conic such ≡ Δ that p p (mod 4). There is an exact sequence of group homomorphisms

(8) 0 −→ 2C(F ) −→ C (F ) −→θ { ± 1}× −→ 0, p p → x+2 where θ :(x, y) p .

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use GENERALIZED LUCAS-LEHMER TESTS USING PELL CONICS 2655 Proof. Let P1, P2 ∈C(Fp). We must prove that θ(P1) · θ(P2)=θ P1 + P2 .We will use the fact that θ(P) = 1 if and only if P ∈ 2C(Fp), but this also establishes that ker θ =2C(Fp). By Theorem 1.1, C(Fp) is cyclic, generated by Q,sothere exist integers 1 and 2 such that P1 = 1Q and P2 = 2Q.Ifθ(P1)=θ(P2)then ≡ P P Q ∈ C F 1 2 (mod 2) so 1 + 2 is even and 1 + 2 =(1 + 2) 2 ( p)sothat P P P P ≡ θ 1 + 2 =1.Ifθ( 1) = θ( 2)then 1 2 (mod 2) so 1 + 2 is odd and P P Q ∈ C F P P − 1 + 2 =(1 + 2) 2 ( p)sothatθ 1 + 2 = 1. It follows that θ is a group Q − C F − Δ ≡ Q homomorphism. θ( )= 1 since # ( p)=p p 0(mod4)andθ(2 )=1, so θ is surjective.

It follows that if p =2nh ± 1 is prime, whereh is odd and n ≥ 2, and the Pell C 2 − 2 ≡ Δ conic : X ΔY = 4 is chosen such that p p (mod 4), then: P • If x( )+2 =1,thenP ∈ 2C(F )and2n−1hP = O. p p • x(P)+2 − P ∈ C F n−1 P T If p = 1, then 2 ( p)and2 h = . n n−1 Observe that the order of C(Fp)isequalto2h,so2 hP = O or T.Now − − − P = Q,so2n 1hP =2n 1hQ = O if and only if 2nh | 2n 1h, because Q is the x(P)+2 − n−1 P T generator, if and only if is even. If p = 1, then 2 h = . Lemma 2.2. Let N =2nh ± 1,whereh is odd. Let C : X2 − ΔY 2 =4be a Pell − conic with Δ N.Ifthereexistsa P∈C(Z/N ) such that 2n 1hP = T, then every ≡ Δ n prime factor q of N satisﬁes q q (mod 2 ).

n−1 Proof. Suppose 2 hP = T in C(Z/N ). Let q be a prime divisor of N.Letoq(P) 2 2 denote the order of the point P on C : X − ΔY =4overFq. Clearly we also have n−1 P T C F P | n P n−1 2 h = in ( q). Therefore oq( ) 2 h, but oq( ) 2 h.Since h is odd, 2n | o (P). Now C(F )iscyclicoforderq − Δ ,soo (P) | q − Δ .Thatis, q q q q q n | − Δ ≡ Δ n 2 q q or simply q q (mod 2 ). · For the following, · will mean the Jacobi symbol. Theorem 2.3. Let N =2nh ± 1 where 0

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We will assume from here on that f, g,andF always refer to the polynomials defined in Equations (3), (4), and (5) respectively. We give an application of Theorem 2.3 to pairs of twin primes. Algorithm 2.4. To certify the primality of a pair of twin primes of the form 2nh ± 1: n n (1) Choose an integer n ≥ 2 and a positive odd integer h< 2 ,andset r =2 h. Δ − Δ (2) Find a fundamental discriminant Δ such that r−1 = 1 and r+1 =1. x+2 x+2 x−2 − x−2 − (3) Find an integer x such that r−1 = r+1 = r+1 = r−1 = 1. (4) Compute fr/2(x) modulo r −1 and r +1.Ifbothfr/2(x) ≡−2(modr ±1), then the points (x, ·) and Pell conic C : X2 − ΔY 2 =4certify that r ± 1 is apairoftwinprimes. ± ≡− ± Proof. If r 1 are composite, then by Theorem 2.3, fr/2(x) 2(mod r 1), so ± x+2 x+2 x−2 − x−2 − we must prove that if r 1 are prime and r−1 = r+1 = r+1 = r−1 = 1 P ∈C F P P ∈ C F then there exist points ( r±1) such that x = x( )and 2 ( r±1). Clearly, x2− Δ there exist points P ∈C(F ± ) such that x = x(P) if and only if 4 = r 1 r±1 r±1 x−2 ∓ if and only if r±1 = 1. Part (4) then follows from Theorem 2.3. In Step (4) of Algorithm 2.4, it is unnecessary to evaluate, at the integer x chosen in Step (3), each of the polynomials fi preceding fr/2, as illustrated in the following example. Example 2.5. The Pell conic C : X2 − 28Y 2 = 4 and points (17,y) ∈C(Z/r ± 1) certify the twin primes 191 and 193: (1) r =26 · 3. 28 − 28 C 2 − 2 (2) 191 = 1and 193 =1,so :X 28Y = 4 is suitable. 19 19 15 − 15 − ∈CZ (3) 191 = 193 = 193 = 191 = 1, so the points (17, 50) ( /191) and (17, 47) ∈C(Z/193) have the required properties. (4) To compute f25·3(17) modulo 191 and 193, write h = 3 in binary as b = 11. The k-th term tk of a sequence B is obtained by taking the first k digits of b from left to right, B = {1, 11}. To compute fh(17) (mod 191) and fh(17) (mod 193), for each tk ∈ B, we wish to compute a sequence of pairs (f (17),f (17)) evaluated modulo 191 and 193, where tk 1+tk 2 (f − 2,ft · f1+t − x)iftk+1 is even, (9) (f ,f )= tk k k tk+1 1+tk+1 (f · f − x, f 2 − 2) if t is odd tk 1+tk 1+tk k+1 2 and f1(x)=x and f2(x)=x − 2. We have f3(17) ≡ 87 (mod 191) and f3(17) ≡ 37 (mod 193). Using f2n−1h = f2n−1 (fh), we iterate terms of the sequences determined by repeated doubling, n − 1 = 5 times, modulo 191 and 193: {87, 118, 170, 57, 0, 189} (mod 191) and {37, 16, 61, 52, 0, 191} (mod 193). Equation (9) has been used by Williams [7] in the same way to efficiently evaluate Lucas sequences.

3. Easy-to-find primality certificates for mnh ± 1 The main result of this section diﬀers from the previous since we use solved Pell conics over integers to certify primes of the form mnh ± 1. The smallest non-trivial

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point of C(Z) with x, y > 0, the fundamental solution, is usually a generator of C(Fp). We begin with the main result which builds on a result of Williams [7]. The lemmas supporting Theorem 3.4 are included below. Remark 3.1 allows a comparison with Lemma 2.1.

Remark 3.1. Let p be an odd prime, m be an odd integer and C : X2 − ΔY 2 =4be ≡ Δ a Pell conic such that p p (mod m). Let μm denote the multiplicative group of m-th roots of unity, generated by ω. There is an exact sequence φ 0 −→ mC(Fp) −→ C (Fp) −→ μm −→ 0, where φ : P = Q → ω.

Proof. Let P1, P2 ∈C(Fp). Writing P1 = 1Q and P2 = 2Q, φ(P1) · φ(P2)= 1 2 1+2 ω ω = ω = φ(P1 + P2). Again letting P = Q, φ(P) = 1 if and only if ω = 1, if and only if m | , if and only if P ∈ mC(Fp). So ker(φ)=mC(Fp).

Lemma 3.2. Let p = mnh ± 1 be prime, where m is odd, h is even, not divisible ≥ C 2 − 2 Δ ≡ by m and n 2.Let : X ΔY =4be a Pell conic where p p (mod m). If P =(x, y) ∈C(Fp) but P ∈ mC(Fp),thens ≡ fmn−1h(x)(modp) satisﬁes Fm(s) ≡ 0(modp).IfP ∈ mC(Fp),thens ≡ 2(modp). n−1 n Proof. If P ∈ mC(Fp), then m hP = m hQ = O so that s ≡ 2(modp). Points n−1 P ∈C(Fp)satisfym hP ∈C(Fp)[m]. Assuming P ∈C(Fp) \ mC(Fp)wemust show that mn−1hP = O.NowP = Q for some positive integer ,somn−1hP = mn−1hQ = O if and only if mnh | mn−1h, because Q is a generator, if and only n−1 if m | .SinceP ∈ mC(Fp), m ,som hP ∈C(Fp)[m] \ O, and it follows that s ≡ fmn−1h(x)(modp) satisﬁes Fm(s) ≡ 0(modp) by Corollary 1.3. The proof of Lemma 3.3 is similar to the proof of Lemma 2.2.

Lemma 3.3. Let N = mnh ± 1,wherem is odd, h is even, not divisible by m and ≥ C 2 − 2 Δ ≡ n 2.Let : X ΔY =4be a Pell conic satisfying N N (mod m).If P ∈C Z ≡ − P ≡ there exists a ( /N ) such that s fmn 1h(x( )) (mod p) satisfies Fm(s) 0 ≡ Δ n (mod N), then every prime factor q of N satisfies q q (mod m ). For the purpose of certifying primes of the form mnh ± 1 in the case where it is not checked whether the point P belongs to mC(Z/N ), it should be assumed that the Pell conic satisfies Δ > 0andP is the fundamental solution of C(Z), reduced modulo N to an element of C(Z/N ), in order to increase the chance that P ∈ mC(Z/N ).

Theorem 3.4. Let N = mnh ± 1,wherem is odd, h is even, not divisible by m, n ≥ C Δ ≡ 0

(1) If Fm(s) ≡ 0(modN),thenN is prime. (2) If s ≡ 2(modN) and Fm(s) ≡ 0(modN),thenN is composite. (3) If s ≡ 2(modN),thenN is either prime or a Lucas pseudoprime. Another Pell conic is required. (4) If it is known that P ∈ mC(Z/N ),thenN is prime if and only if Fm(s) ≡ 0 (mod N).

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Proof. If N is prime, then by Lemma 3.2, Fm(s) ≡ 0(modN)ors ≡ 2(modN). Conversely, suppose Fm(s) ≡ 0(modN) while N is composite. By Lemma 3.3 every prime factor q of N, and hence every factor, satisfies q ≡±1(modmn). If n n n N ≡ 1(modm), then N may factor as N =(m h1 +1)(m h2 +1) or N =(m h1 − n n 1)(m h2 − 1). These correspond to h = ±(h1 + h2)+h1h2m = n n n n h1(m ± 1) + h2(m ± 1) − m +(h1 − 1)(h2 − 1)m .Nowh1 and h2 must be even since h is even so that h>mn, a contradiction, and N must be prime. If N ≡−1 n n (mod m), then N may factor as N =(m h1 +1)(m h2 − 1). This corresponds n n to h = h2 − h1 + h1h2m .Ifh1 = h2 =2,thenh =4· m so that m | h,a n n contradiction. Writing h = h2 + h1(m − 1) + h1(h2 − 1)m and noting that h2 must be even, h>mn, a contradiction, so N must be prime. This completes the proof of case (1). If the conditions of case (2) are satisfied, then mnhP = O,so N is composite by Theorem 1.1. Case (3) follows from Lemma 3.2, noting that N may be a Lucas psuedoprime since the fi are terms of a Lucas sequence. If N is prime and P ∈ mC(FN ), then by Lemma 3.2, Fm(s) ≡ 0(modN). This, together with case (1), completes the proof of case (4). C Δ ≡ Given that N is prime and the Pell conic ,withΔ> 0and N N (mod m), is randomly chosen while always using the fundamental solution of C(Z) reduced modulo N to a point of C(Z/N ) as the point P of Theorem 3.4, this result leads to − 1 a primality certificate with a probability of 1 m . This attests to the occurrence of item (3) of Theorem 3.4 being unlikely. The fm(x) are in fact the Dickson polynomials of the first kind Dm(x, a), with a = 1. The following comes from [5]. ∗ Theorem 3.5 (Dickson). Let a ∈ Fq ,whereq is the power of a prime. The 2 Dickson polynomial Dm(x, a) permutes the field Fq if and only if gcd(m, q −1) = 1. If p = mnh ± 1isprimeandm is odd while h is even, then gcd(m, p2 − 1) = m and fm(x)=Dm(x, 1) cannot permute Fp. That is, there exists a P ∈C(Fp)such P ∈ C F C 2 − 2 that m ( p), where : X ΔY = 4 is a Pell conic with the required property Δ ≡ p p (mod m). Corollary 3.6. Let p = mnh± 1 be prime where m is odd. Let C : X2 − ΔY 2 =4 Δ ≡ 1 be a Pell conic satisfying p p (mod m). Then the proportion m of the points P ∈C(Fp) satisfy P ∈ mC(Fp).

Proof. Let Q be the generator of the cyclic group C(Fp). Suppose Q = mP.Then all of the elements of C(Fp) are divisible by m, contradicting Theorem 3.5. Thus Q C F Z n 1 is not divisible by m. Under the isomorphism ( p) /m h, the proportion m of the points of C(Fp) are divisible by m. Remark 3.7. Let h be an integer. Neglecting the computation of the binary sequence B of Example 2.5 and modular additions, the evaluation of fh modulo an integer N requires at most 2 log2(h) modular multiplications [5]. Induction may be used with Equations (5) and (7) to establish the identities · − 2 F4j−1 = F2j+1 F2j−1 F2j−1 +1, 2 − · − (10) F4j+1 = F2j+1 F2j+1 F2j−1 1, − 2 − F4j+3 =(x 1)F2j+1 F2j+1F2j−1 +1.

We give an example showing how the Fm may be computed using Equation (10).

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Example 3.8. We show how F795(x) may be evaluated. The binary representation of m = 795 is m2 = 1100011011. The k-th term uk of a sequence U = {uk} is obtained by taking the ﬁrst k + 1 digits of a from left to right from k =1upto, − but not including m2,tok = log2(m) 1: U = {11, 110, 1100, 11000, 110001, 1100011, 11000110, 110001101}.

Deﬁne a new sequence V = {vk} by vk =2 uk/2 : V = {10, 110, 1100, 11000, 110000, 1100010, 11000110, 110001100} = {2, 6, 12, 24, 48, 98, 198, 398} in base 10.

The pairs (F−1+vk ,F1+vk ) may be evaluated recursively by using 2 F F− − F +1 ifv ≡ 0(mod4), 1+vk 1+vk −1+vk k+1 (11) F−1+vk+1 = 2 F − F1+v F−1+v − 1ifvk+1 ≡ 2(mod4), 1+vk k k 2 F − F F− − 1ifv ≡ 0(mod4), 1+vk 1+vk 1+vk k+1 (12) F1+vk+1 = 2 (x − 1)F − F F− +1 ifv ≡ 2(mod4), 1+vk 1+vk 1+vk k+1

which follow from Equations (10), until the value of (F397,F399) is known. Finally · − 2 F795 = F399 F397 F397 +1. 5 · − 13 Example 3.9. We certify the primality of N = 795 1588 1. Now N = −1 ≡ N (mod 795), so the Pell conic C : X2 − 13Y 2 = 4 is suitable for applying Theorem 3.4. The point (11, 3) is the fundamental solution of C(Z). We must 4 evaluate s ≡ f7954·1588(11) (mod N). The binary representation of 795 · 1588 is 10010000001110110010111101110001101010001011110100. As in Example 2.5 we recursively evaluate modulo N, the sequence of pairs (f1(11),f2(11)), (f2(11),f3(11)), (f4(11),f5(11)), (f9(11),f10(11)), ...,(f7954·794(11),f7954·794+1(11)) (mod N) using Equation (9) ≡ (119, 1298), (14159, 154451), (2186871698, 23855111399), ...,(363731798219995454, 244122496081218988) (mod N),

s ≡ 3637317982199954542 − 2 ≡ 388011723344360862 (mod N).

To evaluate F795(s)(modN), we proceed as in Example 3.8 using Equations (11) and (12):

{(F1(s),F3(s)), (F5(s),F7(s)), (F11(s),F13(s)),...,(F397(s),F399(s))} (mod N) ≡{(1, 388011723344360863), (329280233123969461, 62155946453030219), ...,(139048125143085063, 364686195778250318)} (mod N). · − 2 ≡ Now F795(s)=F399(s) F397(s) F397(s)+1 0(modN), so N is prime.

4. The primality test for 3nh ± 1 using X2 +3Y 2 =4 The recursions of both the classical Lucas-Lehmer test and that of the primality test [1] for 3nh ± 1 coincide with repeated duplication and repeated multiplication by 3 respectively. We have a geometric variation of the test of Berrizbeitia and Berry [1]. Theorem 4.2 relies on the cubic reciprocity law in the unique factorization domain Z[ω]whereω is a primitive cube root of unity. See [3] for the following and for the · other various identities of the cubic residue symbol · 3.

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Theorem 4.1 (Eisenstein) . If α and π are primes of Z[ω] which are congruent to ± π α 1(mod3),then α 3 = π 3. Below we simply associate the primality test [1] for 3nh ± 1 with the curve X2 +3Y 2 = 4. This avoids case (3) of Theorem 3.4 altogether, since one may not wish to accept 2 chances in 3 for obtaining a primality certiﬁcate. Theorem 4.2. Let N =3nh + ,where = ±1, h is even, not divisible by 3, 0

3 − The ﬁrst term of the recursion Tk+1 = Tk 3Tk in the statement of Theorem 4.2 h P according to [1] is the trace Tr(β )whichis fh(x( β)) so that the computation of s ≡ f3n−1h(x(Pβ)) ≡ f3n−1 fh(x(Pβ)) (mod N) requires repeated multiplication by 3 on X3 +3Y 2 =4overZ/N .

Acknowledgments The author would like to thank Victor Scharaschkin for supervision of this project, which has been supported by the University of Queensland. Franz Lem- mermeyer has given instructive advice about Pell conics and related topics and has suggested many improvements to the presentation of this material, for which the author is grateful.

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6. H. Riesel, Lucasian criteria for the primality of N = h · 2n − 1, Math. Comp. 23 (1969), 869–875. MR0262163 (41:6773) 7. H. Williams, Eﬀective primality tests for some integers of the form A5n − 1 and A7n − 1, Math. Comp. 48 (1987), no. 177, 385–403. MR866123 (88b:11089)

School of Mathematics and Physics, University of Queensland, St. Lucia, Queens- land, Australia 4072 E-mail address: [email protected]

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