Biquadratic reciprocity and a Lucasian primality test

Let {sk, k ≥ 0} be the sequence defined from a given initial value, the seed, s0, by the recurrence sk+1 = s 2 k − 2, k ≥ 0. Then, for a suitable seed s0, the number Mh,n = h · 2n − 1 (where h < 2n is odd) is prime iff sn−2 ≡ 0 mod Mh,n. In general s0 depends both on h and on n. We describe a slight modification of this test which determines primality of numbers h·2n±1 with a seed which depends only on h, provided h 6≡ 0 mod 5. In particular, when h = 4m − 1, m odd, we have a test with a single seed depending only on h, in contrast with the unmodified test, which, as proved by W. Bosma in Explicit primality criteria for h · 2k ± 1, Math. Comp. 61 (1993), 97–109, needs infinitely many seeds. The proof of validity uses biquadratic reciprocity. The Lucasian sequence with seed s0 is the sequence {sk} defined from the given initial value s0 by the recurrence sk+1 = sk − 2, k ≥ 0. A Lucasian primality test is a primality test involving a Lucasian sequence. The terminology comes from the Lucas-Lehmer test for Mersenne primes (see [4] for historical details): Theorem 1 (Lucas-Lehmer). Let p be an odd prime, and let Mp = 2 − 1 be the corresponding Mersenne number. Let {sk} be the Lucasian sequence with seed 4. Then Mp is prime iff sp−2 ≡ 0 mod Mp. Let n, h ∈ N with h odd, h < 2, and let Mn,h = h · 2 − 1. The Lucas-Lehmer test generalizes to a Lucasian primality test for M = Mh,n as follows: Theorem 2. Suppose n ≥ 2. Let d ∈ Z satisfy ( d M ) = −1, where ( . M ) is the Jacobi symbol. Let K = Q( √ d), and let OK be the ring of integers of K. Let α ∈ OK satisfy (αα M ) = −1, where α denotes the conjugate of α in K. Then the following are equivalent: (1) M is prime. (2) (α/α) ≡ −1 mod M . (3) sn−2 ≡ 0 mod M , where sk is the Lucasian sequence with seed s0 =(α/α)+ (α/α) = TrK/Q(α/α) . For a proof see [3]. The Lucas-Lehmer test is the special case d = 3, α = −1 + √ 3 of this theorem. For then TrQ(3)/Q(α) = −4 whence (the sign being clearly irrelevant) s0 = 4. The generalization differs from the original Lucas-Lehmer test in two respects. First, in the generalized test the seed is a rational number, which may not be an integer; this is not a serious difficulty, since, by inverting the denominator mod M , one can replace the rational seed by an integer seed. Second, Received by the editor May 3, 2002 and, in revised form, January 10, 2003. 2000 Mathematics Subject Classification. Primary 11A51, 11Y11. c ©2003 American Mathematical Society 1559 License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 1560 PEDRO BERRIZBEITIA AND T. G. BERRY in the generalized test, the seed s0 depends on n as well as h, while in the LucasLehmer test the seed 4 works for all odd p. It is usually easy to find a seed by trial and error. However, following the analogy with Mersenne numbers, our philosophy is to fix h and search for primes in the family Mh,n with n increasing. Thus it is certainly desirable to have a seed independent of n, if possible. We will have a seed independent of n if we can solve the following problem: For given h, find d ∈ Z, α ∈ OK where K = Q( √ d), such that ∀n , ( d Mh,n ) 6= 1; ( αα Mh,n ) 6= 1. All the above considerations apply also, mutatis mutandis, to numbers of the form M h,n = h ·2+1, when the primality test in question is Proth’s generalization of Pépin’s test for Fermat numbers: let M = M h,n, and let d be an integer such that ( d M+ ) = −1. Then M is prime iff d(M+−1)/2 ≡ −1 mod M. The problem in this case is to find d, depending only on h, such that ∀n, ( d M h,n ) 6= 1. If n ≥ 3, h 6≡ 0 mod 3, then it is easy to see that, for the generalized Mersenne numbers Mh,n, as for Mersenne numbers, d = 3, α = −1 + √ 3 solves the problem; for the M h,n the corresponding problem is solved by d = 3. The case h ≡ 0 mod 3 is studied in [1] and [3]. In [3] tables of seeds are given for Mh,n. In [1] for each h ≡ 0 mod 3, h < 10, but h not of the form 4 − 1, Bosma exhibits a finite set of pairs (dk, αk), dk ∈ Z, αk ∈ Q( √ dk), such that, for any n, one of the pairs solves the problem for Mh,n. On the other hand, for h of the form 4− 1 he proves there is no such finite set of pairs. Similar results are obtained for the M h,n. We shall show that, despite Bosma’s results, a small modification of the algorithm of Theorem 2 allows us, for fixed h 6≡ 0 mod 5, to test primality of Mh,n and M h,n by means of a Lucasian sequence with a seed independent of n. In particular when h = 4 − 1, m odd, we have a single seed. For any odd integer k we set k∗ = (−1 k ) k. This notation allows us to treat the cases h ·2±1 simultaneously. Note that, if M = h ·2±1, then M∗ = (±h)2+1. We shall prove: Theorem 3. Let M = Mh,n = h · 2± 1, where h < 2n−2− 1 is odd, h 6≡ 0 mod 5, and n ≥ 3. Let α = −1 + 2i ∈ Z[i] and let {sk} be the Lucasian sequence with seed s0 = (α/α) + (α/α). Then M is prime iff • either M∗ ≡ ±2 mod 5 and sn−2 ≡ 0 mod M • or M∗ ≡ −1 mod 5 and sn−3 ≡ 0 mod M . The proof of the theorem uses the biquadratic power residue symbol, whose properties we summarize in the following section. Details can be found in [2], Chapter 9. Biquadratic reciprocity. Let K = Q[i], and let R = Z[i] be the ring of integers. Recall that a rational prime q splits in R iff q ≡ 1 mod 4. Let p be a prime ideal of R lying over an odd rational prime, and let β ∈ R. The biquadratic residue symbol ( β p )