LEGENDRE TYPE FORMULA FOR PRIMES GENERATED BY QUADRATIC POLYNOMIALS

Transcription

1 Ann. Sci. Math. Québec 33 (2009), no 2, LEGENDRE TYPE FORMULA FOR PRIMES GENERATED BY QUADRATIC POLYNOMIALS TAKASHI AGOH Deicate to Paulo Ribenboim on the occasion of his 80th birthay. RÉSUMÉ. On en connaît très peu sur l ensemble es entiers m tels que le polynôme quaratique f(x) = ax 2 + bx + c, avec a, b et c copremiers, est premier lorsqu évalué en X = m. Dans cet article, basé sur le crible Ératosthène écalé, nous présentons une formule e type Legenre pour trouver explicitement la carinalité es valeurs u polynôme f(x) qui sont es nombres premiers ne épassant pas t pour t > 0. ABSTRACT. Very little is known about the set of integers m such that the quaratic polynomial f(x) = ax 2 + bx + c with relatively prime integer coefficients has a prime value at X = m. In this paper, base on the shifte sieve of Eratosthenes, we introuce a Legenre type formula for counting explicitly the number of prime values taken by f(x) that are less than or equal to t for any t > Introuction Consier an irreucible quaratic polynomial f with integer coefficients: (1) f(x) := ax 2 + bx + c, a > 0, (a, b, c) = 1. There are only few corroborative results on the set of integers m such that the polynomial f(x) of (1) has a prime value for X = m; moreover, no quaratic polynomial proucing infinitely many primes is known. However, Iwaniec [2] prove in 1978 that there are infinitely many integers n such that n has at most 2 prime factors. For other interesting results an heuristic arguments incluing ample numerical computations see, e.g., [5, 6]. Concerning the istribution of prime values of the polynomial f(x), there is a conjecture pose by ( Hary-Littlewoo [1], which is firmly supporte by ample numerical evience. Let p) be the Legenre symbol. If D := b 2 4ac is not a square an a + b, c are not both even, then the number of prime values less than t taken by the Reçu le 30 avril 2009 et, sous forme éfinitive, le 28 novembre 2009.

2 116 LEGENDRE TYPE FORMULA FOR PRIMES polynomial f(x) is, asymptotically, π f (t) ε C a t log t p>2 p (a,b) p p 1, where 1 if 2 a + b, ε := 2 if 2 a + b an C := p>2 p a ( 1 ( ) ) D 1 p p 1 We now introuce the Legenre formula to count the number of primes less than t for any real t > 0. Let p i be the i-th prime (i.e., p 1 = 2, p 2 = 3, p 3 = 5, an so on), an P := {p i i = 1, 2, 3,...} (the set of all primes), P m := {p 1, p 2,..., p m }, Q m := p 1 p 2 p m, π(t) := #{p P p t}. Base on the sieve of Eratosthenes, Legenre, in 1808, showe the following formula for π(x). Denoting P m := {p P p x} for m = π( x) an Φ(x; P m ) := #{n Z 1 n x, (n, Q m ) = 1}, it follows that an hence Φ(x; P m ) = [ x ] Q m (2) π(x) = m 1 + Φ(x; P m ), where µ is the Möbius function an [x/] is the greatest integer less than or equal to x/. Using the fact that Φ(x; P m ) = Φ(x; P m 1 ) Φ(x/p m ; P m 1 ), Meissel [4] obtaine a more efficient formula to compute π(x). Letting m = π( x) an l = π( 3 x), it follows that π(x) = Φ(x; P l ) + l(m l + 1) + (m l)(m l 1) 2 1 m i=l+1 ( ) x π p i By means of this formula, Meissel himself obtaine π(10 8 ) = , an subsequently Lehmer [3] compute π(10 10 ) = with an improve approach. It is the main purpose of this paper to euce a Legenre type formula analogous to (2) for explicitly counting the number of prime values taken by the polynomial f(x) of (1). We shoul note beforehan that our metho is very elementary an we will apply only the shifte sieve of Eratosthenes for fining primes.

3 T. Agoh Prime values of quaratic polynomials For a real t > 0, we enote π f (t) := #{n 1 f(n) P, f(n) t}. In what follows, we assume that a + b an c are not simultaneously even an that D = b 2 4ac is not a square. To simplify our iscussion, we also assume that 0 < f(n) < f(n + 1) for each integer n 0, which is equivalent to a > 0, a + b > 0 an c > 0. If a, b an c are integers that satisfy any one of the conitions (i) a, b are even an c is o, (3) (ii) all of a, b an c are o, then f(x) 0 (mo 2) has no solution. Moreover, if p ( 3 an ) p a, then we easily see that f(x) 0 (mo p) has no solution if an only if D p = 1. Base on these observations, we have the following proposition. Proposition 2.1. Let f be the quaratic polynomial inicate above. ( ) Let a, b an c be such that they satisfy any one of the conitions in (3) an D p = 1 for p = p 2,..., p m. Then f has prime values at X = 0, 1,..., n, where n is the largest integer satisfying f(n) < p m+1. Proof. Inee, if 0 k n, then f(k) 0 (mo p i ) for i = 1,..., m an this implies f(k) P as esire. Example 2.2. Consier ( ) the polynomial f(x) = X 2 + 3X In this case we have D = 67 an D p = 1 for each p = 3, 5, 7, 11, 13, but ( D 17) = 1. Since the largest integer n satisfying f(n) < 17 is n = 14, f(x) must have prime values at X = 0, 1,..., 14. Inee, these values are 19, 23, 29, 37, 47, 59, 73, 89, 103, 127, 149, 173, 193, 227 an 257 which are all primes, but f(15) = 17 2 is not a prime. For a prime q P, consier the congruence (4) f(x) = ax 2 + bx + c 0 (mo q), 0 X q 1, where a, b an c satisfy the conitions state above. Here an in what follows, we shall use the following notations: (i) ρ(q) is the number of istinct solutions of (4), hence ρ(q) = 0, 1 or 2, (ii) S := {q P ρ(q) 0} is the infinite subset of P corresponing to f, (iii) S m := {q 1, q 2,..., q m } is the set of the first m primes in S, hence ρ(q i ) = 1 or 2, (iv) Q m := Q m (S m ) = q 1 q 2 q m is the prouct of all primes in S m.

4 118 LEGENDRE TYPE FORMULA FOR PRIMES Given n 1, if q m is the largest prime in S such that q m f(n), then it is obvious that f(n) is prime if an only if f(n) 0 (mo q) for all q S m. For a prime q S m, suppose that the solutions of (4) are u 1 (q) (mo q) when ρ(q) = 1, (5) X u 1 (q), u 2 (q) (mo q) when ρ(q) = 2. Let 2 be a ivisor of Q m. Since is square-free an there are ρ(q) istinct solutions of (4) for each q S m, we know from the Chinese Remainer Theorem that the number of non-negative common solutions less than or equal to 1 moulo is equal to ρ() := q ρ(q). Without loss of generality, we orer all istinct solutions of the congruence (6) ax 2 + bx + c 0 (mo ), 0 X 1, as follows: 0 u 1 () < u 2 () < < u ρ() () 1. Before iscussing π f (t), we calculate, as a preliminary, the value of ρ() (7) ε m := δ i (), Q m i=1 where, for i = 1, 2,..., ρ(), (8) δ i () := { 0 if ui () = 0, 1 if u i () 0. Using the smallest solution u 1 (q) in (5), we efine k := #{q S m ρ(q) = 1} an s := #{q S m u 1 (q) = 0}. Then the above ε m can be calculate as follows. Lemma 2.3. We have 0 if k 0 an s 0, 1 if k 0 an s = 0, ε m = ( 1) m if k = 0 an s 0, ( 1) m 1 if k = 0 an s = 0. Proof. Since ρ(q i ) {1, 2}, we have m 0 if k 1, ρ() = (1 ρ(q i )) = Q m i=1 ( 1) m if k = 0.

5 T. Agoh 119 Now, let ρ () {0, 1} enote the number of u i () s equal to 0. Then m 0 if s 1, ρ () = (1 ρ (q i )) = Q m i=1 1 if s = 0. Since ε m = Q m (ρ() ρ ()), the result follows. For a real number x 1 an an integer m 1, put Φ(x; S m ) := #{n 1 n x, f(n) 0 (mo q) for all q S m }. Base on the shifte sieve of Eratosthenes, we can easily euce an explicit formula for Φ(x; S m ). Proposition 2.4. We have Φ(x; S m ) = ρ() [ ] x ui () + ε m. Q m i=1 Proof. For each ivisor 1 of Q m, set η(x; u()) := #{n 1 n x, n u() Then for i = 1, 2,..., ρ(), we have [ ] x ui () η(x; u i ()) = + δ i (), (mo )}. where δ i () = 0 or 1 as efine in (8). Therefore, it follows from the principle of inclusion-exclusion that which completes the proof. ρ() Φ(x; S m ) = η(x; u i ()) Q m i=1 = ρ() ([ x ui () Q m i=1 = ρ() [ x ui () Q m i=1 ] ) + δ i () ] + ε m, Given n 1, we assume that S m is the set of all primes in S up to f(n). Then Φ(n; S m () expresses the number of prime values taken by f(x) = ax 2 + bx + c in the f(n), ] interval f(n). Using this function, we are able to euce an explicit formula for π f (t). For any t f(1) = a + b + c, we first look for the largest prime q m1 S such that q m1 t 1/2. Next, if q m1 q 1, then we look for the largest prime q m2 S such that q m2 t 1/22. We repeat this proceure until we arrive at q mk = q 1 to obtain the sequence (9) q 1 = q mk t 1/2k < < q m2 t 1/22 < q m1 t 1/2 < t

6 120 LEGENDRE TYPE FORMULA FOR PRIMES an the escening chain of subsets {q 1 } = S mk S mk 1 S m2 S m1 S. With the above notations, we can state the following theorem. k ) Theorem 2.5. We have π f (t) = Φ ( t 1/2i 1 1; S mi. i=1 ( ) Proof. Since Φ t 1/2 i 1 1; S mi [ 1, ] t 1/2i 1 1 is the number of integers n in the interval with f(n) 0 (mo q) for all q S mi, this expresses the num- ( t 1/2i, t 1/2i 1]. Connecting these ber of prime values taken by f(x) in the interval intervals for i = 1, 2,..., k, we obtain immeiately the above formula. Proposition 2.6. For any fixe m 1, it follows that lim x Φ(x; S m) =. Proof. Since ε m 2 by Lemma 2.3 an ρ is multiplicative for relatively prime ivisors of Q m, we obtain, for a fixe m, Φ (x; S m ) = ρ() [ ] x ui () + ε m Q m i=1 = ρ() x u i () + ε m + O ρ() Q m i=1 Q m = x ρ() ρ() u i () + ε m + O (4 m ) Q m Q m i=1 = x ( 1 ρ(q) ) + O(1). q q S m The right-han sie iverges as x tens to infinity which, however, is absur. We give here an example of how to compute π f (x) using the shifte sieve metho of Eratosthenes. Example 2.7. For the special polynomial f(x) = X 2 + 1, we see that there ( ) exists a solution of X (mo q), 0 X q 1, if an only if q = 2 an 1 q = 1 for an o q P. Hence we have S = {2} {q P q 1 (mo 4)} an all the solutions of this congruence are given, for each q S, as u 1 (q) = 1 for q = q 1 = 2, X = u 1 (q), u 2 (q) for q 1 (mo 4). Here note that if q 2, then u 1 (q) u 2 (q) an u 1 (q) + u 2 (q) = q.

7 T. Agoh 121 As a practical application, we will now compute π X 2 +1(t) for t = First fin the sequence of all primes q mi S, i = 1, 2,..., k, given in (9). Then we have q m1 = q 12 = 97, q m2 = q 2 = 5, q m3 = q 1 = 2, an so k = 3. Therefore, consier the following chain of subsets of S: S 1 = {2} S 2 = {2, 5} S 12 = {2, 5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97}. We enumerate all solutions u i (q) of X (mo q) for each q S 12 in the following table: q u 1 (q) u 2 (q) [ 10 Write all positive integers less than or equal to 99 = 1] an cross out integers n such that n u 1 (q), u 2 (q) (mo q) for all q S 12. Then we get the set of integers that survive: M 1 := {10, 14, 16, 20, 24, 26, 36, 40, 54, 56, 66, 74, 84, 90, 94}. Hence the number of primes in (10 2, 10 4 ] of the form X is ( ) Φ ; S 12 = #M 1 = 15. Similarly, we have M 2 := {4, 6} an M 3 := {2}, which are the sets of integers in the intervals [1, 99] an [1, 3], respectively, not satisfying (4). Hence ( ) Φ 99; S2 = #M 2 = 2, Φ (3; S 1 ) = #M 3 = 1. Noticing that = 2 is a prime, we finally have π X 2 +1(10 4 ) = # (M 1 M 2 M 3 ) + 1 = 19. Inee, the set M of primes in the interval [1, 10 4 ] of the form f(x) = X is precisely given by M = { n n {2} M 1 M 2 M 3 }. The next proposition seems to be trivial, but its proof gives us a certain piece of information about the value of each term in the function Φ(x; S m ). Proposition 2.8. Let n 2, m 1 an u(q) be a solution of (4). If n u(q) (mo q) for all q S m, then Φ(n; S m ) = Φ(n 1; S m ) + 1. Proof. By the ivision algorithm we have [ ] (n 1) u(q) (10) (n 1) u(q) = q + r q (n 1), 0 r q (n 1) < q. q If we choose especially an integer n 2 satisfying n u(q) (mo q) for all q S m, then r q (n 1) q 1, i.e., 0 r q (n 1) q 2. Inee, if r q (n 1) = q 1, then n u(q) (mo q), which is contrary to the assumption. From (10) we obtain [ ] (n 1) u(q) n u(q) = q + r q (n 1) + 1, q

8 122 LEGENDRE TYPE FORMULA FOR PRIMES where 1 r q (n 1) + 1 q 1. This implies r q (n) = r q (n 1) + 1 an hence [(n u(q))/q] = [((n 1) u(q))/q], for all q S m. Since the conition n u(q) (mo q) for all q S m yiels n u() (mo ) for all ivisors > 0 of Q m, we also have [(n u())/] = [((n 1) u())/]. Consequently, it follows from Proposition 2.4 that Φ(n; S m ) = n + ρ() [ n ui () which completes the proof. Q m 1 i=1 ] + ε m = (n 1) + ρ() [ (n 1) ui () Q m 1 = Φ(n 1; S m ) + 1, i=1 ] + ε m + 1 To conclue this paper, we woul like to mention that all above iscussions can be extene to a polynomial f(x) with arbitrary egree 2. Inee, let S an S m be the sets of primes corresponing to f, similarly to the quaratic case. Then it will be possible to compute Φ(x; S m ) an π f (t) without ifficulty by the same metho as above, except for a treatment of the number ρ(q) (where q S) of istinct solutions of the congruence (11) f(x) 0 (mo q), 0 X q 1. It is easy to show that the value of ε m in this case is formally given by ε m = g(1) (with 0 0 = 1) for the polynomial g(x) := (X ρ(q)) X m s (X 1) s, q Sm where s := #{q S m u 1 (q) = 0} an u 1 (q) is the smallest solution of (11). Acknowlegements. The author woul like to express his gratitue to the anonymous referee for the useful suggestions which le to various improvements of this paper. This research was supporte in part by the Grant-in-Ai for Scientific Research (C), No , Japan Society for the Promotion of Science. REFERENCES [1] G.H. Hary an J.E. Littlewoo, Some problems of Partitio Numerorum, III: On the expression of a number as a sum of primes, Acta Math. 44, 1923, Collecte papers of G.H. Hary, I, Clarenon Press, Oxfor, 1966,

9 T. Agoh 123 [2] H. Iwaniec, Almost-primes represente by quaratic polynomials, Invent. Math. 47, 1978, [3] D.H. Lehmer, On the exact number of primes less than a given limit, Illinois J. Math. 3, 1959, [4] E. Meissel, Über ie Bestimmung er Primzahlmenge innerhalb gegebener Grenzen, Math. Ann. 2, 1870, [5] P. Ribenboim, The new book of prime number recors (3r e.), Springer-Verlag, New York, [6] P. Ribenboim, The little book of bigger primes (2n e.), Springer-Verlag, New York, T. AGOH, DEPT. OF MATH., TOKYO U. OF SCIENCE, NODA, CHIBA , JAPAN agoh_takashi@ma.noa.tus.ac.jp