The Fibonacci Quarterly 44(2006), no.2, PRIMALITY TESTS FOR NUMBERS OF THE FORM k 2 m ± 1. Zhi-Hong Sun

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1 The Fibonacci Quarterly 44006, no., PRIMALITY TESTS FOR NUMBERS OF THE FORM k m ± 1 Zhi-Hong Sun eartment of Mathematics, Huaiyin Teachers College, Huaian, Jiangsu 3001, P.R. China zhsun@hytc.edu.cn Homeage: htt:// Abstract. Let k, m Z, m, 0 < k < m and - k. In the aer we give a general rimality criterion for numbers of the form k m ± 1, which can be viewed as a generalization of the Lucas-Lehmer test for Mersenne rimes. In articular, for k = 3, 9 we obtain exlicit rimality tests, which are simler than current known results. We also give a new rimality test for Fermat numbers and criteria for 9 4n+3 ± 1, 3 0n+6 ± 1 or 3 36n+6 ± 1 to be twin rimes. 1. Introduction. For nonnegative integers n, the numbers F n = n +1 are called the Fermat numbers. In 1878 Pein showed that F n n 1 is rime if and only if 3 Fn/ mod F n. For rimes, let M = 1. The famous Lucas-Lehmer test states that M is a Mersenne rime if and only if M S, where {S n } is given by S 0 = 4 and S k+1 = Sk k = 0, 1,,.... In [1], [], [6] and [9], W. Borho, W. Bosma, H. Riesel and H.C. Williams extended the above two tests to numbers of the form k m ± 1, where 0 < k < m and k is odd. For examle, we have the following known results. Theorem 1.1 Let = k m + 1 with m, 0 < k < m, k and Z with the Jacobi symbol =. Then is rime if and only if / mod. In articular, if 3 k we may take = 3. Let {S n x} be given by S 0 x = x and S k+1 x = S k x k 0. Then we have Theorem 1. Let = k m 1 with m 3, 0 < k < m and k ±1 mod 6, and let x = + 3 k + 3 k. Then is rime if and only if S m x. Here we oint out that the x in Theorem 1. is also given by x = k/ k r=0 k r k r r r 4 k r. In this aer we rove the following main result. 1.1 For m let = k m ± 1 with 0 < k < m and k odd. If b is an integer such that +b = b =, then is rime if and only if S m k/ k k r r=0 k r r r b k r. As alications of 1.1 we have many new simle rimality criteria for numbers of the form k m ± 1k = 1, 3, 9. Here are some tyical results. 1. For n 1 the Fermat number F n is rime if and only if F n S n. 1

2 1.3 Let m 3 be a ositive integer. If m 0 mod or m, 11 mod 1, then 9 m 1 is comosite. If m 1, 3, 7, 9 mod 1, then 9 m 1 is rime if and only if 9 m 1 S m x, where 778 if m 1, 9 mod 1, x = if m 3 mod 1, if m 7 mod Let n be a nonnegative integer. Then 9 4n+3 1 and 9 4n are twin rimes if and only if 9 4n+3 1 S 4n Throughout this aer we use the following notations: Z the set of integers, N the set of ositive integers, d the Jacobi symbol, m, n the greatest common divisor of m and n, S n x the sequence defined by S 0 x = x and S k+1 x = S k x k 0.. Basic Lemmas. For P, Q Z the Lucas sequences {U n P, Q} and {V n P, Q} are defined by and U 0 P, Q = 0, U 1 P, Q = 1, U n+1 P, Q = P U n P, Q QU n P, Q n 1 V 0 P, Q =, V 1 P, Q = P, V n+1 P, Q = P V n P, Q QV n P, Q n 1. Let = P 4Q. It is well known that U n P, Q = 1 { P + n P n } 0.1 and P + n P n. V n P, Q = +. Set U n = U n P, Q and V n = V n P, Q. From the above one can easily check that From [] we also have V n = P U n QU n = U n+1 P U n..3 U n = U n V n, V n = V n Q n and V n U n = 4Q n..4 If is an odd rime not dividing Q, it is well known that [] U P, Q 0 mod and U P, Q mod.. Let be an odd rime such that Q = 1 and. following stronger congruence see [4] or [9,.8]:. H. Lehmer roved the U /P, Q 0 mod..6 efinition.1 Let P, Q Z, and be an odd rime such that Q. efine r P, Q to be the smallest ositive integer n such that U n P, Q. From [, IV.17] or [9,.87] we know that U m P, Q if and only if r P, Q m. This can also be deduced from [9, 4..9,.81]. Using. and.6 we have

3 Lemma.1. Let P and Q be integers, = P 4Q, and let be an odd rime such that Q. Then r P, Q. Moreover, if Q = 1 and, then r P, Q. From.4 and induction we have Lemma.. Let P, Q Z, Q 0 and n N. Then S n P Q = Q n V np, Q. Lemma.3. Let P, Q Z and n N. Let be an odd rime such that QP 4Q and S n P/ Q 0 mod. Then P 4Q mod n+3+ Q /. Proof. In view of Lemma. we have V np, Q and so U n+1p, Q by.4. From.4 we see that U np, Q. Thus, r P, Q = n+1. This together with Lemma.1 gives the result. Lemma.4. Let P, Q Z and n N, and let > 1 be an odd integer such that, QP 4Q = 1 and S n P/ Q 0 mod. Let α = n + or n + 1 according as Q is a square or not. If < α 1, then is rime. Proof. If is comosite, then has a rime divisor q such that q. Since q and S n P/ Q 0 mod we see that S n P/ Q 0 mod q. It follows from Lemma.3 that q P 4Q q mod n+3+ Q q / and so q n+3+ Q q / 1. Thus, q n+3+ Q q / 1. This contradicts the assumtion. So must be rime. Let [x] denote the greatest integer not exceeding x. Using induction one can easily rove Lemma. [9, 4..36]. Let P, Q Z and n N. Then V n P, Q = [n/] r=0 n n r n r r P n r Q r. 3. The general rimality test for numbers of the form k m ± 1. Lemma 3.1. Let P, Q Z and = P 4Q. Let be an odd rime such that Q. Suose Q = 1 and so c Q mod for some integer c. Then P + c i V P, Q P + c ii V + P, Q P Proof. For b, c Z it is clear that 1 c mod, c mod. b ± b 4bc b c ± b c 4c = b. 3

4 Thus, alying. we see that V n b, bc = b n V n b c, c. 3.1 Hence, if is an odd rime such that b 4bc and ε = b 4bc, by [9, 4.3.4] we obtain V ε b c, c = b ε V ε b, bc b ε 1 ε bc = b 1 ε c b c 1 ε mod. Now suose b = P + c and c Q mod. Then b 4bc = P 4c P 4Q mod and so ε =. From the above we see that V ε P, Q V ε b c, c P + c 1 ε c mod. This roves i. From.1 and. we see that V + /P, Q = 1 { P V Q 1 / Thus, by i and.6 we obtain V + /P, Q 1 P Q 1 / P + c /P, Q + 1 c P U /P, Q }. P + c c mod. This roves ii and hence the roof is comlete. Remark 3.1 Lemma 3.1 can also be easily deduced from [7, Lemma 3.4] or [8, Lemma 3.1]. Lemma 3.. Let P, Q Z and be an odd rime with QP 4Q. = 1 and so c Q mod for some integer c. Then Q Suose V 4 P, Q 0 mod if and only if Q + cp Q cp = =. Proof. From Lemma 3.1 we have V P, Q P 1 P +c c c P +c mod if 4Q P = 1, 1+ 4 mod if 4Q P =.

5 Thus, alying.4 we obtain V 4 P, Q = V c 1 c 1+ P, Q + Q 4 V { P +c { P P +c P, Q + c 1 c + c } mod if 4Q P = 1, + c c } mod if 4Q P =. Since P 4Q and c Q mod we see that P P +c c c mod. Hence, V 4 This roves the lemma. P, Q 4Q P P + c = 1 and Q + cp Q cp = =. Lemma 3.3. Suose P, Q, k, n Z with k, n 0. Then V kn P, Q = V n V k P, Q, Q k. Proof. Set V m = V m P, Q. From [9, 4..8] we know that c = V r+k = V k V r Q k V r k and so V km+1 = V k V km Q k V km. Now we rove the result by induction on n. Clearly the result is true for n = 0, 1. Suose the result holds for 1 n m. By the above and the inductive hyothesis we have V km+1 = V k V m V k, Q k Q k V m V k, Q k = V m+1 V k, Q k. So the result holds for n = m + 1. Hence, the lemma is roved by induction. Theorem 3.1. For m {, 3, 4,... } let = k m ± 1 with 0 < k < m and k odd. If b, c Z,, c = 1 and c+b = c b = c, then is rime if and only if S m x, where x = c k V k b, c = k/ r=0 r b/c k r. Proof. Set U n = U n b, c and V n = V n b, c. From Lemmas 3.3,. and. we have k k r k r r V /4 = V k m = V m V k, c k = c k m S m V k /c k = c k m S m x. If is rime, it follows from Lemma 3. that V /4. So S m x 0 mod. Now suose S m x = S m V k /c k 0 mod. From.4 we have Vn b 4c Un = 4c n. Thus, U n, V n 4c n. As, c = 1 and V k m we find, U k m = 1. It is well known that see [] and [9] U r U rn for any ositive integers

6 r and n. Thus, U k U k m and so, U k = 1. Hence,, V k 4ck = 1 by.4. Set P = V k, Q = c k and n = m. If 0 < k < m, then clearly = k m ±1 < m. By Lemma.4, is rime. If = m 1 m ± 1 is comosite, by Lemma.3 we know that any rime divisor q of satisfying q ±1 mod m. It is easy to check that m ± 1. Thus m 1 m + 1. This is imossible. So is rime. This comletes the roof. Taking b = 4 and c = 1 in Theorem 3.1 we obtain the Lucas-Lehmer test for Mersenne rimes and Theorem 1.. From Theorem 3.1 we also have the following criterion for Fermat rimes, which is similar to the Lucas-Lehmer test. Corollary 3.1. For n N the Fermat number F n is rime if and only if F n S n. Proof. Since F n mod 3 and F n 3, mod 7 we see that 3 = F n 7 = and = F n F n 3 F n 7 =. Thus utting = F n, k = 1, b = and c = 1 in Theorem 3.1 we obtain the result. Remark 3. In 1960 K. Inkeri[3] showed that the Fermat number F n n is rime if and only if F n S n The rimality criterion for numbers of the form 9 m ± 1. In the section we use Theorem 3.1 to obtain exlicit rimality criterion for numbers of the form 9 m ± 1. Theorem 4.1. Let m 3 be a ositive integer. If m 0 mod or m, 11 mod 1, then 9 m 1 is comosite. If m 1, 3, 7, 9 mod 1, then 9 m 1 is rime if and only if 9 m 1 S m x, where 778 if m 1, 9 mod 1, x = if m 3 mod 1, if m 7 mod 1. Proof. Clearly the result is true for m = 3. Now assume m 4. If m = n for some integer n, then 9 m 1 = 3 n + 13 n 1 and so 9 m 1 is comosite. If m, 11 mod 1, then 7 9 m 1 since 3 1 mod 7. If m 1, 3, 7, 9 mod 1, once setting 3 if m 1, 9 mod 1, b = if m 3 mod 1, 11 if m 7 mod 1 one can easily check that + b 9 m = 1 From Lemma. we know that b 9 m =. 1 V 9 b, 1 = b 9 9b 7 + 7b 30b 3 + 9b = b 3 3bb 3 3b 3 = x. Alying Theorem 3.1 in the case c = 1 we get the result. In a similar way, alying Theorem 3.1 we have 6

7 Theorem 4.. Let m 3 be a ositive integer. If m 0 mod 4, then 9 m + 1. If m 10 mod 1, then 13 9 m + 1. If m mod 8, then 17 9 m + 1. If m 0 mod 4, m 10 mod 1 and m mod 8, then 9 m + 1 is rime if and only if 9 m + 1 S m x, where x is given by Table 4.1. Table 4.1 m b x = V 9 b, 1 = b 3 3bb 3 3b 3 m 1, 9 mod m mod m 3, 6, 7 mod m 11 mod m 17, 6 mod m 41 mod Remark 4.1 For m 4 let = 9 m + 1 and if m 0,, 3 mod 4, 7 if m 1, 9, 13, 1 mod 4, = 17 if m mod 4, 41 if m 17 mod 4. In [] W. Bosma showed that is rime if and only if / mod. Theorem 4.3. Let n be a ositive integer. Then 9 4n+3 1 and 9 4n are twin rimes if and only if 9 4n+3 1 S 4n Proof. Let b = 3. Then + b = 17 and b = 3. Since 3 9 4n+3 ±1 = 1 and 4 mod 17 we find 9 4n+3 ±1 = + b n+3 ± 1 4 n ± 1 9 4n+3 = ± 1 9 4n+3 = = =, ± b 9 4n+3 ± 1 7 ± 1 9 4n+3 = ± 1 9 4n+3 = ± = ± =. ± 1 Thus, alying Theorem 3.1 we see that 9 4n+3 ±1 is rime if and only if 9 4n+3 ±1 S 4n+1 V 9 b, 1. To see the result, we note that 9 4n+3 + 1, 9 4n+3 1 = 1 and that V 9 b, 1 = b 9 9b 7 + 7b 30b 3 + 9b = b 3 3bb 3 3b 3 = Remark 4. If 9 m ± 1m > 1 are twin rimes, then m 3 mod 4. If m 11 mod 1, then 7 9 m and so 9 m ±1 cannot be twin rimes. If m 3 mod 1, by taking b = 1 and c = 1 in Theorem 3.1 we can rove that 9 m and 9 m +1 are twin rimes if and only if 9 m 1 S m It is known that 9 m 1 and 9 m + 1 are twin rimes when m = 1, 3, 7, 43, 63, 11. o there exist only finitely many such twin rimes? 7

8 . The rimality criterion for numbers of the form 3 m ± 1. Theorem.1. Let m 3 be a ositive integer such that m mod If m 1 mod 4, m 46 mod 7 or m 86 mod 1440, then 3 m 1 is comosite. If m 1 mod 4, m 46 mod 7 and m 86 mod 1440, then 3 m 1 is rime if and only if 3 m 1 S m x, where x is given by Table.1. Table.1 m b x = V 3 b, 1 = b 3 3b m 0, 3 mod m, 6 mod m 10 mod m mod m 70 mod m 14 mod m 86, 74 mod m 110 mod m 1438, 878, 4318, 7198 mod m 78 mod m 8638 mod Proof. If m 1 mod 4, then 3 m ; if m 46 mod 7, then 37 3 m ; if m 86 mod 1440, then 11 3 m. Now suose m 1 mod 4, m 46 mod 7 and m 86 mod Let b be given by Table.1. One can easily check that + b b 3 m = 1 3 m =. 1 Thus the result follows from Theorem 3.1 by taking c = 1 and = 3 m 1. Remark.1 If m N and m 0, mod 3, in 1993 W. Bosma[] showed that 3 m is rime if and only if 3 m 1 S m m. In a similar way, using Theorem 3.1 we can rove Theorem.. Let m 3 be a ositive integer such that 180 m. If m 1 mod 3, then 7 3 m + 1; if m 3 mod 4, then 3 m + 1; if m mod 1, then 13 3 m + 1; if m 144 mod 180, then 61 3 m + 1. If m 1 mod 3, m 3 mod 4, m mod 1 and m 144 mod 180, then 3 m + 1 is rime if and only if 3 m + 1 S m x, where x is given by Table.. Table. m b x = V 3 b, 1 = b 3 3b m mod m 6 mod m 8 mod m 9 mod m 1, 4 mod m 36 mod m 7 mod m 108 mod

9 Theorem.3. Let n be a nonnegative integer. Then 3 0n+6 1 and 3 0n are twin rimes if and only if 3 0n+6 1 S 0n Proof. Let b = 4. Then + b = 44 and b = 40. Since 3 0n+6 ±1 = 1 and mod 11 we find + b n+6 ± 1 6 ± 1 3 0n+6 = ± 1 3 0n+6 = ± = ± =, ± b 3 0n+6 ± 1 1 ± 1 3 0n+6 = ± ± 1 3 0n+6 = ± = ± =. ± 1 Thus, alying Theorem 3.1 in the case b = 4 and c = 1 we see that 3 0n+6 ± 1 is rime if and only if 3 0n+6 ± 1 S 0n+4 V 3 b, 1. To see the result, we note that 3 0n+6 + 1, 3 0n+6 1 = 1 and that V 3 b, 1 = b 3 3b = = In the same way, utting b = 17 and c = 1 in Theorem 3.1 we get Theorem.4. Let n be a nonnegative integer. Then 3 36n+6 1 and 3 36n are twin rimes if and only if 3 36n+6 1 S 36n References k r mod m 1. W. Borho, Grosse Primzahlen und befreundete Zahlen: über den Lucas-test und Thabit Regeln, Mitt. Math. Ges. Hamburg 1983, W. Bosma, Exlicit rimality criteria for h k ± 1, Math. Com , , MR94c: K. Inkeri, Tests for rimality, Ann. Acad. Sci. Fenn. Ser. A , 1-19, MR# H. Lehmer, An extended theory of Lucas functions, Ann. Math , P. Ribenboim, The Book of Prime Number Records, nd ed., Sringer, Berlin, 1989, H. Riesel, Lucasian criteria for the rimality P of N = h n 1, Math. Com , n 7. Z.-H. Sun, Combinatorial sum k and its alications in number theory III, J. Nanjing Univ. Math. Biquarterly 1 199, 90-10, MR96g: , Values of Lucas sequences modulo rimes, Rocky Moun. J. Math , H.C. Williams, Édouard Lucas and Primality Testing, Canadian Mathematical Society Series of Monograhs and Advanced Texts V ol., Wiley, New York, 1998, AMS Classification Numbers: 11Y11,11B39,11B0. 9