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Using Fibonacci Factors to Create Fibonacci Pseudoprimes 3

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Using Fibonacci Factors to Create Fibonacci Pseudoprimes 3 USING FIBONACCI FACTORS TO CREATE FIBONACCI PSEUDOPRIMES JUNHYUN LIM, SHAUNAK MASHALKAR, AND EDWARD F. SCHAEFER Abstract. Carmichael showed for sufficiently large L, that FL has at least one prime divisor that is ±1(mod L). For a given FL, we will show that a product of distinct odd prime divisors with that congruence condition is a Fibonacci pseudoprime. Such pseudoprimes can be used in an attempt, here unsuccessful, to find an example of a Baillie-PSW pseudoprime, i.e. an odd Fibonacci pseudoprime that is congruent to ±2(mod5) and is also a base-2 pseudoprime. 1. Introduction 5 For all odd prime numbers p we have p|F − 5 , where ( ) is the Legendre symbol. This p ( p ) p is well-known and can be proved using the lemmas in Section 2. An odd composite integer 5 n is said to be a Fibonacci pseudoprime if n|F − 5 where ( ) is the Jacobi symbol, which n ( n ) n generalizes the Legendre symbol. Let us list the six smallest Fibonacci pseudoprimes, their prime factorizations, and the smallest positive Fibonacci number that each divides. We have 323 = 17 · 19|F18, 377 = 13 · 29|F14, 1891 = 31 · 61|F30, 3827 = 43 · 89|F44, 4181 = 37 · 113|F19, and 5777 = 53 · 109|F27. For each of the six smallest Fibonacci pseudoprimes n, if FL is the smallest positive Fibonacci number for which n is a divisor (i.e. L = ordf (n)), then the prime divisors of n are each ±1(mod L). This is not always the case. In fact the seventh smallest Fibonacci pseudoprime is 6601 = 7 · 23 · 41|F120 and none of its prime divisors are ±1(mod 120). Nevertheless, we can use the above observation to create many Fibonacci pseudoprimes. We construct them using the theorem below. Theorem 1.1. Let L be a positive integer. Let p1,...,pk, for some k ≥ 2, be distinct odd primes dividing FL with the property that for each i we have pi ≡ ±1(mod L), assuming at k least two such primes exist. Then P := Qi=1 pi is a Fibonacci pseudoprime. It does seem quite common for prime divisors of FL to be ±1(mod L). For example, four of the prime divisors of F100 are 101, 401, 3001, and 570601 and F7560 has at least 29 prime arXiv:2105.13513v1 [math.NT] 28 May 2021 divisors that are ±1(mod7560) (note F7560 has a composite factor with 711 decimal digits that is still unfactored). All Fibonacci numbers up to F1408 have been factored completely and complete or partial factorizations of FL for 1409 ≤ L ≤ 9999 have been given (see [FF]). For 1 ≤ L ≤ 1408, the average number of odd prime divisors of FL that are ±1(mod L) is 7279 1408 ≈ 5.17. Applying Theorem 1.1 to the fully and partially factored Fibonacci numbers FL for 1 ≤ L ≤ 9999, we can create approximately 231 Fibonacci pseudoprimes. However they will 2010 Mathematics Subject Classification. Primary 11B39; Secondary 11A51. The authors are grateful to Carl Pomerance for challenging us to the $620 problem and useful conversations. 1 2 JUNHYUNLIM,SHAUNAKMASHALKAR,ANDEDWARDF.SCHAEFER not all be distinct. For example, F19 = 4181 = 37 · 113. Not only are both prime divisors ±1(mod19), they are both ±1(mod 38). So the Fibonacci pseudoprime 4181 will appear for both L = 19 and L = 38. One reason for the appearance of such prime divisors of Fibonacci numbers comes from [Ca, Thm. XXVI]. Carmichael proved that for every L =16 , 2, 5, 6, 12 there is a prime divisor p ≡ ±1(mod L) of FL which divides no FK for K < L. We leave it to the analytic number theorists to study the expected number of prime divisors of FL that are ±1(mod L) as a function of L. In Section 2 we prove Theorem 1.1. It provides a new way of creating Fibonacci pseu- doprimes and thus can be added to the list of methods for constructing them found in [CG, DFM, Le, Pa, Ro, SW]. There is much interest in finding an integer that is an odd Fibonacci pseudoprime, congruent to ±2(mod 5), and simultaneously a base-2 pseudoprime. These are sometimes referred to as Baillie-PSW pseudoprimes. There is a heuristic argu- ment that an example exists (see [CG], which refers to ideas in [Po1]). However, no example is known and there is a $620 prize (payable by Carl Pomerance, Sam Wagstaff, and the Number Theory Foundation, see [Po2]) for those who either find an example, or prove that none exists. In Section 3 we discuss how we used our theorem in our unsuccessful attempt at finding an example. 2. Proof of the Theorem We will use the following well-known lemmas. Lemmas 2.1 and 2.2 are proven in [Lu, p. 297] Lemma 2.1. Let p ≡ ±1(mod5) be prime. Then p|Fp−1. Lemma 2.2. Let p ≡ ±2(mod5) be prime. Then p|Fp+1. Lemma 2.3 is proven in [Ha, p. 35]. Lemma 2.3. Let m, n be positive integers. If m|n then Fm|Fn. Lemma 2.4 is proven in [Wa, Thm. 3]. Lemma 2.4. Let n be a positive integer and let L be an integer such that n|FL. Then ordf (n)|L. Lemma 2.5 follows from Gauss’ law of quadratic reciprocity. Lemma 2.5. 5 Let p be an odd prime. If p ≡ ±1(mod 5) then ( p )=1. If p ≡ ±2(mod 5) 5 then ( p )= −1. The following lemma is new. Lemma 2.6. Let p be an odd prime and L be a positive integer with p|FL. Assume p ≡ 5 ±1(mod L). Then p ≡ ( p )(mod L). Proof. Let p = 5. Then from Lemmas 2.3 and 2.4, p|FL if and only if 5|L. So the result is vacuously true as p 6≡ ±1(mod L). Assume p =6 5 is an odd prime, that p|FL, and p ≡ ±1(mod L). Let p ≡ ±1(mod5). From Lemma 2.1 we have p|Fp−1. Assume p ≡ −1(mod L). Then L|p+1. From Lemma 2.3 we have USING FIBONACCI FACTORS TO CREATE FIBONACCI PSEUDOPRIMES 3 FL|Fp+1. Since p|FL we have p|Fp+1. As p|Fp+1 and p|Fp−1 we have p|(Fp+1 − Fp−1) = Fp. Since p divides two consecutive Fibonacci numbers, p divides all Fibonacci numbers - a contradiction. So p ≡ 1(mod L). The result follows from Lemma 2.5. 5 Similarly, assuming p ≡ ±2(mod 5) gives p ≡ −1=( p )(mod L), though we use Lemma 2.2 instead of Lemma 2.1. In other words, consider the odd prime divisors of FL that are ±1(mod L). Those that are 1(mod L) are those that are ±1(mod5). Those that are −1(mod L) are those that are ±2(mod 5). We are now ready to prove Theorem 1.1. Proof. For each i we have pi|FL and pi ≡ ±1(mod L). So from Lemma 2.6, for each i we have p ≡ ( 5 )(mod L). Taking the product of both sides over all i gives P ≡ ( 5 )(mod L). i pi P 5 Thus L|(P − ( )). From Lemma 2.3 we have FL|FP −(5/P ). Since P |FL we get P |F − 5 . P P ( P ) Note, that the construction described in Theorem 1.1 is related to, though not the same as, the construction in [CG] of Fibonacci pseudoprimes. 3. The search for a Baillie-PSW pseudoprime There is a $620 prize for a Baillie-PSW pseudoprime or a proof that none exists. That is an odd Fibonacci pseudoprime that is ±2(mod 5) and also a base-2 pseudoprime. This problem was originally posed in [PSW]. Jan Feitsma and William Galway (see [FG]) have computed all base 2 pseudoprimes up to 264. Sam Wagstaff has checked all of those to determine if there were any $620 winners and there were none (see [Po2]). The search is also described in [BW, CG, CP, MK, Po1, PSW, SW]. Fix a Fibonacci number FL. For i = −1, 1 we let Si be the set of odd prime divisors of FL that are congruent to i modulo L. Recall that the primes in S−1 are ±2(mod5) and those in S1 are ±1(mod 5). When possible, we created products of at least two distinct primes from S−1 ∪ S1 such that the product contains an odd number of primes from S−1. That way the product is ±2(mod 5). For example, for F258 we have |S−1| = 5 and |S1| = 4. Thus we can 5 4 5 4 5 4 create 1·(2 −1) +3·2 +5·2 = 251 different products, all of which are odd Fibonacci pseudoprimes that are ±2(mod 5). For our search, we used the complete and partial factorizations of Fibonacci numbers FL into prime divisors for 1 ≤ L ≤ 9999 found in [FF]. Using the construction described in the previous paragraph and these factorizations we created approximately 230 odd Fibonacci pseudoprimes that are ±2(mod 5). We then checked each to see if the Fibonacci pseudoprime P is a base-2 pseudoprime, i.e. if 2P ≡ 2(mod P ). Alas, none were base-2 pseudoprimes. Some of the Fibonacci pseudoprimes we created are huge. For example F9967 is squarefree and all of its prime divisors are ±1(mod9967). Therefore F9967 is a Fibonacci pseudoprime 6918 and it is −2(mod5). Note F9967 ≈ 2 . k If P = Qi=1 pi is a huge Fibonacci pseudoprime, then using the relatively fast repeated squares algorithm to reduce 2P (mod P ) can still be quite slow.
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