On the equation x 2 + dy 2 = F n

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1 ACTA ARITHMETICA 27.2 (2007 On the equation 2 + dy 2 = F n by Christian Ballot (Caen and Florian Luca (Morelia. Introduction. Let (F n n 0 be the Fibonacci sequence given by F 0 = 0, F = and F n+2 = F n+ + F n for all n 0. Let d be any fied rational integer. Using standard sieve methods it is easy to establish that, for d not an integer, most positive integers m are not representable as m = 2 + dy 2 with and y integers. In this paper, we look at those positive integers m which are both members of the Fibonacci sequence and are representable as 2 +dy 2 for some integers and y. That is, we investigate the set ( N d = {n > 0 : F n = 2 + dy 2 for some integers and y}. Clearly, N 0 consists of the positive integers n such that F n is a perfect square and Cohn [] showed that N 0 = {, 2, 2}. When d =, using the formula (2 F 2n+ = F 2 n + F 2 n+, we see that N contains all odd positive integers. Furthermore, since F n and F n+ are coprime, every odd prime factor of F 2n+ is congruent to modulo 4. In [2], it was shown that for most even positive integers n, F n admits a prime factor q 3 (mod4. Here, we go one step further. In order to settle the case of N, we first prove the following result. Proposition. For all even positive integers n ecept a set of asymptotic density zero, there eists a prime q 3 (mod4 such that q F n and the eact power of q that divides F n is odd. Since for q 3 (mod4, is a quadratic nonresidue (mod q, Proposition immediately implies that the asymptotic density of N is precisely /2. Note also that if d is a perfect square, then N d has positive lower asymptotic density. Indeed, if we write (d for the rank of appearance of d in (F n n 0, i.e., (d is the minimal positive integer k such that d F k, then formula (2 implies that if d is a perfect square, then the set N d contains 2000 Mathematics Subject Classification: B39, N37. [45] c Instytut Matematyczny PAN, 2007

2 46 Ch. Ballot and F. Luca the set {2n + : n 0, (mod (d}, which is of positive asymptotic density. But N d also has positive lower asymptotic density if d is the opposite of a perfect square. Indeed, N d then contains {n : (4d n}. If we put d = t 2, then F n /t 2 is an integer multiple of 4 for n divisible by (4d. As such, F n /t 2 can be written as ( y( + y. Hence F n = (t 2 (ty 2 = (t 2 + dy 2. Therefore we have shown the following result. Theorem 2. For any d which is plus or minus a perfect square, the set N d has positive lower asymptotic density. The asymptotic density of N is /2. We put D = {d Z : N d has positive lower asymptotic density}. Theorem 2 implies that D is an infinite set. However, in this paper, we show that most integers do not belong to D. For a positive real number we write D( for the set of d D with d. Theorem 3. There eists a positive constant C such that if > is any real number then #D( C (log 3. By a standard procedure of partial summation, Theorem 3 implies that d < d D (note that 0 D. We would like to make the following conjecture. Conjecture 4. D contains only finitely many integers not a square or the negative of a square. For integers a and b with b > 0 odd, we write ( a b for the Jacobi symbol of a with respect to b. We state another related conjecture. Conjecture 5. For all but finitely many of the integers d not a square or the negative of a square, there is a prime q 5 such that ( d =. F q The argument used in the proof of Lemma 9 below shows that Conjecture 5 implies Conjecture 4. If true, Conjecture 4 would imply a stronger bound on the cardinality of D( than the one provided by Theorem 3. We

3 On the equation 2 + dy 2 = F n 47 would like to leave these conjectures as problems to the reader. In fact, it may be that Conjecture 5 is true without eceptions. Throughout this paper, we assume familiarity with basic properties of Fibonacci and Lucas numbers. The nth Lucas number is denoted by L n. We recall here that for a prime p, the rank of appearance (p of p in the Fibonacci sequence divides p e p, where e p is the Legendre symbol of 5 with respect to p. Also, we use the Vinogradov symbols and and the Landau symbols O and o with their regular meanings. The constants implied in them are absolute. For a positive real number, we use log for the maimum between the natural logarithm of and. We write π( for the number of primes p, and for coprime integers a b we write π(; a, b for the number of primes p congruent to a modulo b. We use p, q and r to denote prime numbers. For a set A of positive integers we put A( = A [, ]. Acknowledgements. This paper was written in part during a very enjoyable visit by the second author to the Laboratoire Nicolas Oresme of the University of Caen; he wishes to epress his thanks to that institution for the hospitality and support. During the preparation of this paper, F. L. was also partly supported by grants SEP-CONACYT 46755, PAPIIT IN04505 and a Guggenheim Fellowship. 2. The proofs. For any positive integer n we let P(n denote the largest prime factor of n, and for real numbers y we put Ψ(, y = {n : P(n y}. The numbers belonging to Ψ(, y are usually referred to as smooth numbers. The following estimate for the number of smooth numbers (see Section III.5.4 of Tenenbaum s book [3] will play a crucial rôle in our proofs. Lemma 6. Let ε > 0 be fied. Uniformly for we have ep((log log 5/3+ε y #Ψ(, y = ep( ( + o(ulog u, where u = log log y. Let a b be fied coprime integers. For a positive real number we put A(; a, b = {n : if p n and p > log, then p a (modb}, that is, n is in A(; a, b if no prime factor of n larger than log is congruent to a (modb. We will need the following estimate.

4 48 Ch. Ballot and F. Luca Lemma 7. If a b are coprime, then there eists a,b such that #A(; a, b (log log 2 (log /φ(b for > a,b. Proof. Let be a large real number and let y= /log log, u=log /log y = log log. We put A ( = A(; a, b Ψ(, y. Then, by Lemma 6, (3 #A ( #Ψ(, y = ep( u( + o( log u < log for large. We now put A 2 ( = A(; a, b \ A (. To bound #A 2 (, let n A 2 ( and write n = Pm, where P = P(n > y. Then m < /y. Thus, fiing m, we see that the number of choices for P is π(/m m log(/m m log y = log log m log. Note that m is an integer which is free of primes p a (modb larger than log. Write M( for the set of such positive integers m. Then, summing up over all possible choices of m M(, we get (4 #A 2 ( = = = log log log log log log log log log log log log log log log (log log 2 (log /φ(b, m m M( ( p p a(mod b p p a(mod b ( ep α 0 ( p p p a (mod b ( φ(b ep φ(b p α p + p log p a(mod b p log p a(mod b p log p a(mod b ( α 0 p α ( p p + O( log log + log log log + O( φ(b where we used the fact that for any fied A > 0, the estimate ( log log y log b (5 = + O p φ(b b p y p c(mod b holds uniformly in the range y 3 and c b < (log y A with c and b

5 On the equation 2 + dy 2 = F n 49 coprime. In particular, the above estimate also holds when b is fied. This completes the proof of the lemma. We will also need the following lemma. Lemma 8. Let B( be the set of integers n divisible by a product of primes pq, where p > log and q ± (modp. Then #B( log log. log Proof. Let be large and fi two primes p and q such that p > log, q ± (modp and pq <. The number of positive integers n such that pq n is /pq. Summing up over all possible choices of p and q, we get (6 #B( pq = (S + S 2, log <p q q ±(mod p where S is the contribution to the double sum from primes p < (log 3, and S 2 is the contribution from primes p (log 3. Let T p = pq. q q ±(mod p Using estimate (5 with A = 3 when p < (log 3, we get log log T p p 2. We use the trivial estimate T p ( p pk + + pk p 2 k log p 2, k /p k when p (log 3. Thus (7 S + S 2 T p + log <p<(log 3 log log p 2 + log <p log log log, where we used the trivial bound p 2 t n t p n 2 t (log 3 p (log 3 p ds s 2 = t T p log p 2

6 50 Ch. Ballot and F. Luca with t = log and with t = (log 3. Estimates (6 and (7 now lead to the desired conclusion. Proof of Proposition. Let be a large real number and let C( = {n : if q 3 (mod4 and q α F 2n, then α is even}. Lemmas 7 and 8 yield #A(; 5, 6 + #B( = o(. We now show that C( A(; 5, 6 B(, which, together with the previous estimate, will prove the proposition. Let n C( and assume n A(; 5, 6. Then there eists a prime p > log with p 5 (mod6 such that p n. But 2p 2n, and 2p 4 (mod6. Since the Fibonacci sequence is periodic modulo 4 with period 6, and F 4 = 3, we find that F 2p 3 (mod4. Thus, there eists a prime q 3 (mod4 such that q a F 2p, where a is odd. Since 2p 2n, we infer that q a F 2n. Now since n C(, we must have q a+ F 2n. Now q F 2n /F 2p with q F 2p implies, by the well-known law of appearance of powers of primes in Lucas sequences, that q n/p. However, since q F 2p, the rank (q is either p or 2p, which in both cases implies that q ± (modp. Hence, pq n, q ± (modp, and p > log. Therefore, n B(. This completes our proof. The following lemma will be useful for the proof of Theorem 3. Lemma 9. Let d be a nonzero integer. Suppose that p is a prime number not dividing 2 (d such that ( d =. Then N d is of asymptotic density zero. Proof. Note that p 3, so that is odd and the Jacobi symbol of d with respect to is well-defined. Let q = 2 (dk+p for some nonnegative integer k. By the addition formula 2F m+n = F m L n + L m F n, we have 2F q = F 2 (dk L p + L 2 (dk. Clearly, 6 F 2 F 2 (dk and d F (d F 2 (dk. Furthermore, since L 2n = 5F 2 n + 2( n, L 2 (dk = 5F 2 6 (dk + 2 is congruent to 2 both modulo 6 and modulo d. The above arguments show that 2F q 2 (modlcm[6, d], therefore F q (modlcm[8, d].

7 On the equation 2 + dy 2 = F n 5 These congruences imply the Jacobi symbols identity ( ( d d =. F q We now show that N d ( A(; p, 2 (d B(, which will prove that #N d ( = o(. Let n N d ( and assume that n A(; p, 2 (d, so that there eists a prime q > log with q n and q = 2 (dk + p for some k 0. Assume also that n B(, so that we now seek a contradiction. Write F q = δ q λ 2 q, where δ q and λ q are positive integers with δ q squarefree. Note that δ q is odd and > because F q is odd and not a square. Any prime r dividing δ q satisfies r α r F q for some odd eponent α r. If r α r+ F n, then r F n /F q, and hence r n/q, so that qr n and r ± (modq (because (r = q and, assuming log 5, we cannot have r = q = p = 5. Thus, n B(, a contradiction. Therefore r α r F n. So, there eist m, y, z N such that (8 y 2 + dz 2 = mλ 2 qδ q = F n, where gcd(m, δ q =. If g = gcd(δ q, yz, then, having in mind that δ q is square-free and gcd(δ q, d = (since ( d F q = 0, we get g gcd(y, z, λq. Hence, dividing out relation (8 by g 2 yields (9 y 2 + dz 2 = mµ 2 qδ q, for some integers y, z, µ q with gcd(δ q, y z =. But equation (9 implies that ( d δ q =. Because Fq is odd and F q = F(q /2 2 + F (q+/2 2, we have F q (mod4. Therefore ( ( ( ( d d d d = = = = =, which is a contradiction, and our proof is complete. F q Remark. For d {±2, ±3, ±5, ±6, ±7, ±8, ±0}, N d is of asymptotic density 0 since ( ( ( ( ( ( = = = = = =. F 5 F 5 In what follows, we put F 5 F 7 F q F 7 δ q D = {d D : d is square-free}. We approach the proof of Theorem 3 by first proving the following somewhat weaker statement. F 9

8 52 Ch. Ballot and F. Luca Theorem 0. The estimate #D ( (log 3 holds for all sufficiently large values of. Proof. Let be large and y = /log log, u = log /log y = log log. Let D 2 ( = {d D ( : d Ψ(, y}. By Lemma 6, (0 #D 2 ( 2#Ψ(, y = 2ep( ( + o(ulog u < (log 3, when is large. For a positive integer k, we write ω(k for the number of distinct prime factors of k. Let v = 25(log log 2 and put D 3 ( = {d D ( : ω( (d v}. We now bound D 3 (. Let d D 3 (. Because d F n if and only if (d n, we deduce that (d p d (p. Therefore (d p d(p e p, where e p = ( 5 p. Since ω( (d v, it follows that either d has at least w = 5log log distinct prime factors, or there eists p d such that p e p has at least w distinct prime factors. In the first case, the number of such numbers d does not eceed 2 m < 2 m ω(m w ω(m w m 2 k w m< ω(m=k In the second case, let p < be a prime such that p e p has at least w prime factors. The number of numbers d with d which are multiples of p does not eceed 2 p 4 p e p. Summing up over all such primes and noting that for every m the equation p e p = m can have at most two solutions p, we find that in this case the number of acceptable d s is Hence, if we write 8 k w m ω(m=k m. S(; k = m, m ω(m=k m.

9 then On the equation 2 + dy 2 = F n 53 ( #D 3 ( k w S(; k. Using the multinomial formula, we get a bound for S(; k: S(; k ( k k! p α = ( (2 k! p + O( Furthermore, p α = k! (log log + O(k. p (log log + O( k /k! (log log + O( k+ /(k +! = k + (log log + O( > 2 if k w and is large, therefore by estimates ( and (2, and Stirling s formula, we get (3 #D 3 ( S(; k (log log + O( w w! k w ( elog log + O( w ( e 5log log < w 5 (log 3 for large because 5log(5/e = > 3. Let D 4 ( = D ( \ (D 2 ( D 3 (. Let d D 4 ( and write it as d = εpm, where P = P(d > y, m is a positive integer < /y, and ε {±}. We fi the number m and let D4 m( be the subset of D 4( that contains the d s for which d = ±mp(d. Assume D4 m ( is not empty. Let z = 300(log log 2 log log log and let P = {p : p z}. For large, the cardinality of P satisfies π(z = ( + o( z log z > 25(log log 2 = 5v. k = 50( + o((log log 2 Let Q be a fied subset of P having precisely 5 v primes in it. Because D4 m ( is not empty, there is a subset T of Q of cardinality 4 v such that every prime number in T is coprime to 2 (m. Indeed, since there is a d in D 4 ( such that m d, we know that (m divides (d, so that any p coprime to 2 (d is coprime to 2 (m. Thus, let Q m be the set of subsets of Q of cardinality 4 v whose (prime elements are all prime to 2 (m. Choose a T in Q m and put D m,t 4 ( = {d D4 m ( : gcd(p, 2 (d =, p T }. We will bound D 4 ( by using the crude estimate #D 4 ( 4 (. m T Q m #D m,t

10 54 Ch. Ballot and F. Luca By Lemma 9, we have, for d D m,t 4 (, ( d = for all p T. The above condition means that ( ( εm P =. But again because p is not 3, is odd. And since is the sum of two squares we have (mod4, so that ( ( P m =. In the above relation, m is fied, and p is a prime in the fied set T. Let again = δ p λ 2 p. The above relation puts P into half of all possible φ(δ p arithmetic progressions with common differences δ p, which are all odd and >. Using the fact that the s are mutually coprime as p varies in T, we conclude that P lies in /2 #T of all admissible progressions of the form A T (modb T, where (4 B T = p T δ p p T ep(#t z = ep(30000(log log 4 log log log. Here, we used the fact that F n < e n for all positive integers n. By the Brun Titchmarsh theorem, the number of such primes P /m does not eceed 2/m 4log log 2 #T log(/mb T 2 4 v m log, where we used estimate (4 to conclude that /m > y > (B T 2 for large, therefore that /mb T > y /2. The number of subsets T Q m is less than ( 5 v 4 v so that the number of acceptable primes P when m is fied is 2 4 v ( 5 v 4 v 4log log m log, and summing up over all possible values of m we get ( 4log log 5 v #D 4 ( log 2 4 v 4 v m m log log 2 4 v By Stirling s formula, the above inequality leads, for large, to ( 5 5 v (5 #D 4 ( log log < (log 3, ( 5 v. 4 v

11 where we used the fact that On the equation 2 + dy 2 = F n 55 25log(5 5 /( = < 3. The conclusion of the theorem now follows from estimates (0, (3 and (5. Proof of Theorem 3. Let d D, and write it as d = d d 2 0, where d is square-free. It is clear that N d N d, therefore d D as well. Thus, if is large, then #D( d 0 #D (/d 2 0. By Theorem 0, #D(/d 2 0 d 2 0 (log(/d When d 0 < /3, we have /d 2 0 > /3, therefore #D(/d 2 0 (log 3. Otherwise, we use the trivial inequality #D (/d 2 0 2/d2 0 to get #D( + 2 (log 3 d 2 d 0 /3 0 (log 3 d 2 d 0 0 d which completes the proof of the theorem. d 2 /3 d 0 0 /3 dt t 2 (log 3, References [] J. H. E. Cohn, Square Fibonacci numbers, etc., Fibonacci Quart. 2 (964, [2] F. Luca, Prime factors of Fibonacci numbers, solution to Advanced Problem H546, ibid. 42 (2004. [3] G. Tenenbaum, Introduction to Analytic and Probabilistic Number Theory, Cambridge Univ. Press, 995. Laboratoire Nicolas Oresme Université de Caen F-4032 Caen Cede, France Christian.Ballot@math.unicaen.fr Instituto de Matemáticas Universidad Nacional Autónoma de Méico C.P , Morelia, Michoacán, Méico fluca@matmor.unam.m Received on and in revised form on (52