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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. Instead we note that by the P P Chinese Remainder Theorem, 2 ≡ 2(mod P ) if and only if 2 ≡ 2(mod pi) for each i. We P order the prime divisors of P as p1 < p2 <...

Email address: [email protected] Department of Mathematics and Computer Science, Santa Clara University, Santa Clara, CA 95053, USA.

Email address: [email protected] USING FIBONACCI FACTORS TO CREATE FIBONACCI PSEUDOPRIMES 5

Department of Mathematics and Computer Science, Santa Clara University, Santa Clara, CA 95053, USA.