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1 Booker, A. R., & Pomerance, C. (07). Squarefree smooth numbers and Euclidean rime generators. Proceedings of the American Mathematical Society, 45(), htts://doi.org/0.090/roc/3576 Peer reviewed version Link to ublished version (if available): 0.090/roc/3576 Link to ublication record in Exlore Bristol Research PDF-document This is the author acceted manuscrit (AAM). The final ublished version (version of record) is available online via AMS at htt:// Please refer to any alicable terms of use of the ublisher. University of Bristol - Exlore Bristol Research General rights This document is made available in accordance with ublisher olicies. Please cite only the ublished version using the reference above. Full terms of use are available: htt://

2 SQUAREFREE SMOOTH NUMBERS AND EUCLIDEAN PRIME GENERATORS ANDREW R. BOOKER AND CARL POMERANCE Abstract. We show that for each rime > 7, every residue mod can be reresented by a squarefree number with largest rime factor at most. We give two alications to recursive rime generators akin to the one Euclid used to rove the infinitude of rimes.. Introduction In [4], Mullin considered the sequence { k } k= defined so that, for every k 0, k+ is the smallest rime factor of + k. From the argument emloyed by Euclid to rove the infinitude of rime numbers, it follows that the k are airwise distinct, and Mullin s sequence can thus be viewed as an exlicit, constructive form of the roof. A natural question, which Mullin osed, is whether every rime eventually occurs in the sequence. Desite clear heuristic and emirical evidence that the answer must be yes, it aears to be very difficult to rove anything substantial to that end. With this setting in mind, in [5, Section..3] and [4], we (indeendently) described two variations of Euclid s argument that allow greater flexibility and lead to sequences that rovably contain every rime number. We recall these constructions in detail in Section 5. The main focus of this article is the following question, which arises naturally as an ingredient in both constructions, but is ossibly of indeendent interest: For rimes, are all residue classes mod reresented by the ositive integers that are both squarefree and -smooth? (Recall that an integer n is called y-smooth if every rime divisor of n is y.) Since there are π() squarefree, -smooth, ositive integers and only residue classes mod, one heuristically exects the answer to be yes, at least for large. (However, note that y = is best ossible, since the zero residue class mod is not attained by a y-smooth number for any y <.) We will show that, with two excetions, this is indeed the case: Theorem. Let be a rime different from 5 and 7, and a Z. Then there is a squarefree, -smooth, ositive integer n such that n a (mod ). The roof of Theorem consists of three largely indeendent stes that we carry out in Sections 4. Our key tools include a numerically exlicit form of the Pólya Vinogradov inequality, see Frolenkov and Soundararajan [7], and a combinatorial result of Lev [] on h-fold sums of dense sets. In Section 5 we aly Theorem to variants of Euclid s argument, and thus show how one can generate all of the rimes out of nothing. In the final section we mention a few related unsolved roblems. Date: November 4, Mathematics Subject Classification. Primary A4, Secondary A5, B5, L40. At least one of the authors thinks that Mullin s question is likely undecidable.

3 Related results on rime generators of Euclid tye may be found in [3], [5], [9], [0], and the references of those aers.. Large : character sums For a rime and a ositive integer d, let denote the subgrou of (Z/Z) of index d. H d, = {h (Z/Z) : h ( )/d (mod )} Proosition. Let > be a rime and suose d with d < log +. For each nonzero residue m (mod ) there is some squarefree number j < with j mh d,. Proof. We may assume that d >, since otherwise we can take j =. Let χ be a character mod of order d. Since χ i for i =,..., d runs over all of the characters mod of order dividing d, we have that () d d µ (j)χ i (j)χ i (m) i= is the number of squarefree numbers j < with χ(j) = χ(m), and so is the number of squarefree numbers smaller than in the coset mh d,. The rincile character (the term when i = d) contributes µ (j) d to the sum. This exression is 6 /d as. One can get an exlicit lower bound π valid for all via the Schnirelmann density of the squarefree numbers, see [7]. Thus, () d µ (j) 53 ( ). 88d Our task is then to show that the other terms in () are small in comarison. By recognizing a squarefree number by an inclusion-exclusion over square divisors, we have for i d, µ (j)χ i (j) = µ(v)χ i (v ) χ i (j). v /v We may discard the term v = since it is 0. For v > /4 +, we may use the trivial estimate χ i (j) /v < v < 3/4. /v v> /4 + v> /4 + v> /4 + For v/4 + we use an exlicit form of the Pólya Vinogradov inequality, see [7], where we may divide the estimate for even characters by since our character sum is over an initial interval. This gives χ i (j) v> /v ( ) π / log + / 4π 3/4 log + 3/4. v> v /4 + v /4 +

4 Thus, Hence i= µ (j)χ i (j) 3/4 d µ (j)χ i (j)χ i (m) d ( 4π log + 5 ). ( ) ( 3/4 d 4π log + 5 ), and we would like this exression to be smaller than the one in (). That is, we would like the inequality ( 53( ) > ) ( 3/4 88d d 4π log + 5 ), to hold true, or equivalently, ( 53( ) > (d ) 88 3/4 4π log + 5 ). Using d < log +, we see that this inequality holds for all > Remark 3. Instead of [7] for our estimate of the character sum, we might have used [6] or we might have used the smoothed version in [3]. The former would require raising the lower limit slightly, while the latter would likely lead to a reduction in the lower limit, but at the exense of a more comlicated roof. For a rime > and an integer d with d < log +, let C d, denote a set of squarefree coset reresentatives for H d, smaller than as guaranteed to exist by Proosition. Also, let S d, denote the set of rimes that divide some member of C d, and let S be the union of all of the sets S d, for d, d < log +. Let ω(n) denote the number of distinct rime divisors of n. It is known that ω(n) (+o()) log n/ log log n as n. We have the weaker, but exlicit inequality: ω(n) < log n for n > 6. To see this, note that it is true for ω(n), since it holds for n = 7, and for n 8 we have log n >. If ω(n) = k 3, then n 6 5 k, so that k (log n + log(5/6))/ log 5, which is smaller than log n for n. But k 3 imlies n 30. As a corollary, we conclude that under the hyotheses of Proosition, we have each #S d, < d log and #S < (log + )3. 3. Proof of Theorem for large Assume the rime exceeds Let S denote the set of rimes identified at the end of the last section, let N =, and let K = π(n) #S > π(n) (log + )3 denote the number of remaining rimes smaller than. For an integer m with 0 < m <, let f(m) denote the number of unordered airs of distinct rimes q, r with q, r <, qr m (mod ) and q, r S. Set A = {m (0, ) : f(m) > K/ } {}, 3 A = #A.

5 Lemma 4. For > 3 0 8, we have Proof. Evidently, (3) m= A > f(m) = N log N +. ( ) K = K(K ). Further, if two airs q, r and q, r are counted by f(m), then either they have no rime in common or they are the same air. Thus, for each m, (4) f(m) K. Since we have by (3) that Thus, from (4), A K/ f(m) (N A)K/, m A f(m) K(K ) (N A)K/. m A f(m) K (N A)/ > K. m A Since K > π(n) (log + )3, by using inequality (3.) in [8] and > 3 0 8, we have the inequality in the lemma. For a ositive integer k, let A k denote the set of k-fold roducts of members of A. Lemma 5. There are ositive integers d < log +, k < log + 3 such that A k contains a subgrou H d, of (Z/Z). Proof. Let g be a rimitive root modulo and let A denote the set of discrete logarithms of members of A to the base g. That is, j A with 0 j < N if and only if g j (mod ) A. We now aly a theorem of Lev [, Theorem ] to the set A. With κ := (N )/(A ), this result imlies that there are ositive integers d κ and k κ+ such that ka contains N consecutive multiles of d. Here, ka denotes the set of integers that can be written as the sum of k members of A. Thus, reducing mod N, the set ka contains a subgrou of Z/NZ of index d := (d, N). Hence, A k contains the subgrou H d, of (Z/Z). Further, from Lemma 4, we have d d κ < log +, which comletes the roof. Lemma 6. For the subgrou H d, of (Z/Z) roduced in Lemma 5, each member of H d, has a reresentation modulo as a squarefree number involving rimes smaller than and not in S (and so not in S d, ). Proof. Suose that m A k, so that (5) m = m m... m k (q r )(q r )... (q k r k ) (mod ), where each m i A and m i q i r i (mod ). This last roduct over rimes is -smooth, but is not necessarily squarefree. However, each m i A has many reresentations as q i r i, in fact 4

6 at least K/ reresentations, with each reresentation involving two new rimes. So, if k is small enough, there will be a reresentation of each m i so that the roduct of rimes in (5) is indeed squarefree. Now k < log + 3, so having at least (k ) + < 4 log + 5 reresentations for each member of A is sufficient. The number of reresentations exceeds K/ and by the same calculation that gave us the last ste in Lemma 4, we have this exression exceeding / log. This easily exceeds 4 log + 5 for > It is now immediate that every residue class mod contains a squarefree number with rime factors at most. Indeed this is true for 0 (mod ) take as the reresentative. For a nonzero class j (mod ), find that member m of C d, with j mh d,, and write j = mh with h H d,. We have seen that each member of H d, has a squarefree reresentative using rimes smaller than and not in S d,. Since m is squarefree and uses only rimes in S d, it follows that mh is also squarefree using only rimes smaller than. This comletes the roof of Theorem for rimes > Verification of Theorem for small It remains only to verify the theorem for < For > 0 4 we use the following simle strategy: Comute a rimitive root g (mod ), and find airwise corime, squarefree, -smooth numbers m i such that m i g i (mod ) for each nonnegative integer i log ( ). If this is ossible then, given any nonzero residue n (mod ), we have n g k (mod ) for some integer k [0, ]. Exressing k in binary, viz. k = 0 i log ( ) b i i for b i {0, }, we have n 0 i log ( ) mb i i (mod ). Since the m i are airwise corime, the residue class of n is thus reresented by a squarefree -smooth number, as desired. It is convenient to choose m i of the form q i r i for rimes q i, r i. To find these efficiently, for each i = 0,,,... we search through small rimes q, comute the smallest ositive r q g i (mod ), test whether r is rime, and ensure that qr is corime to m j for j < i. The only essential ingredient needed to carry this out is a fast rimality test; we used a strong Fermat test to base couled with the classification [6] of small strong seudorimes, which would allow us, in rincile, to handle any < 64. Heuristically, one can exect this method to succeed using O(log 3 ) arithmetic oerations on numbers of size, and we found it to be very fast in ractice; it takes just minutes to verify the theorem for all (0 4, ) on a modern multicore rocessor. For < 0 4 we fall back on a brute-force algorithm: For each integer a [, ], consider each of the numbers a, a +, a +,... until encountering one that divides q rime q. This q< takes only seconds to check for all < 0 4 other than 5 and 7. (For {5, 7} one can see directly that 4 + Z is not reresented, but all other residue classes are.) 5. Generating all of the rimes from nothing We give two alications of Theorem to Euclidean rime generators. described without roof in [5, Section..3]; we suly the short roof here. The first was Corollary 7. Starting from the emtyset, recursively define a sequence of rimes, where if n is the roduct of the rimes generated so far, take = min{q rime : q n, q d + for some d n} as the next rime. This sequence begins with, 3, 7, 5, and then roduces the rimes in order. 5

7 Proof. Let be the least rime not yet roduced after the 4th ste and let n be the roduct of the rimes smaller than. By Theorem, there exists some d n with d (mod ). Thus, we generate at the next ste. The second variant was described in [4]. With Theorem in hand, we can give a shorter roof. (As will be clear from the roof, there is an obstruction reventing the terms from aearing in strict numerical order in this case, so the conclusion is weaker than that of Corollary 7.) Corollary 8. Starting from the emty set, recursively generate a sequence of rimes where if n is the roduct of the rimes generated so far, take as the next rime some rime factor of some d + n/d, where d n. Such a sequence can be chosen to contain every rime. Proof. One such sequence begins with, 3, 5, 3, 7. Suose > 7 is a rime number and that the sequence constructed so far contains every rime smaller than and erhas some rimes larger ( than), but it does not contain. Let n be the roduct of the rimes generated so far. If = then there exists a (Z/Z) such that a + n/a = 0 (mod ). n By Theorem there exists d n belonging to( the) class of a, so we can choose as the next n rime. On the other hand, since > 5, if ) =, then [4, Lemma 3(i)] guarantees the existence of a (Z/Z) such that =. By Theorem there exists d n ( a+n/a belonging to( the ) class of a, and by multilicativity it follows that d + n/d ( has ) a rime factor q qn q satisfying =. Choosing q as the next rime, we thus have =, so by the above argument we can now take. This comletes the roof. 6. Comments and roblems The rime generator of Corollary 7 has its roots in a construction in the rimality test of []. There, a finite initial set of rimes is given with roduct I, and then one takes the roduct E of all rimes I with d + for some d I. The rimality test can be used for numbers n < E and runs in time I O(). Further, it is shown that there are choices of I, E with I = (log E) O(log log log E), so the test runs in almost olynomial time. The same I, E construction (with I no longer required to be squarefree) is used in the finite fields rimality test of Lenstra []. In Theorem we insist that the squarefree -smooth integers used be ositive. If negatives are allowed, then the rimes 5 and 7 are no longer excetional cases. Further, if d is allowed to be negative in the context of the rime generator in Corollary 7, the rimes are generated in order. (For this to be nontrivial, d should not be chosen as.) Suose we use the generator of Corollary 8 by always returning the least rime ossible, and say this sequence of rimes is q, q,.... Does {q k } contain every rime? Is there way of choosing the sequence in Corollary 8 such that every rime is generated and the kth rime generated is asymtotically equal to the kth rime? Is it true that any sequence containing all rimes as in Corollary 8 cannot contain the rimes in order starting from some oint? These questions might all be asked if we allow rime factors of d±n/d in Corollary 8 instead of just d + n/d. Presumably in Theorem, when is large, residues a (mod ) have many reresentations as squarefree -smooth integers. Say we try to minimize the largest squarefree -smooth used. 6

8 For > 7, let M() be the smallest number such that every residue mod can be reresented by a squarefree -smooth number at most M(). Our roof shows that M() O(log ). We conjecture that M() O(). We mentioned in the Introduction that the condition -smooth in Theorem cannot be relaxed to y-smooth for any y <, since otherwise the residue class 0 (mod ) will not be reresented. However, we may ask for the smallest number y = y() such that every nonzero residue class mod can be reresented by a y-smooth squarefree number. Via the Burgess inequality, it is likely that one can show that y() /(4 e)+o() as. Assuming the Generalized Riemann Hyothesis for Kummerian fields (as Hooley [9] did in his GRHconditional roof of Artin s conjecture), it is likely that one can rove that y() = O((log ) ). We note that if one dros the squarefree condition then these statements follow from work of Harman [8] unconditionally and Ankeny [] under GRH; see also the recent aer [0] for a strong, numerically exlicit version of the latter. Acknowledgments We thank Enrique Treviño and Kannan Soundararajan for some helful comments. The first-named author was artially suorted by EPSRC Grant EP/K034383/. References [] L. M. Adleman, C. Pomerance, and R. S. Rumely, On distinguishing rime numbers from comosite numbers, Ann. of Math. 7 (983), [] N. C. Ankeny, The least quadratic non residue, Ann. of Math. 55 (95), [3] A. R. Booker, On Mullin s second sequence of rimes, Integers (0), [4] A. R. Booker, A variant of the Euclid-Mullin sequence containing every rime, J. Integer Seq., Volume 9 (06), Article [5] R. E. Crandall and C. Pomerance, Prime numbers: A comutational ersective, second ed., Sringer, New York, 005. [6] J. Feitsma, Pseudorimes, htt:// [7] D. A. Frolenkov and K. Soundararajan, A generalization of the Pólya Vinogradov inequality, Ramanujan J. 3 (03), [8] G. Harman, Integers without large rime factors in short intervals and arithmetic rogressions, Acta Arith. 9 (999), [9] C. Hooley, On Artin s conjecture, J. Reine Angew. Math. 5 (967), [0] Y. Lamzouri, X. Li, K. Soundararajan, Conditional bounds for the least quadratic non-residue and related roblems, Math. Com. 84 (05), [] H. W. Lenstra, Jr., Galois theory and rimality testing, in Orders and their alications (Oberwolfach, 984), Lecture Notes in Math. vol. 4 (985), Sringer-Verlag, Berlin, [] V. F. Lev, Otimal reresentations by sumsets and subset sums, J. Number Theory 6 (997), [3] M. Levin, C. Pomerance, and K. Soundararajan, Fixed oints for discrete logarithms, ANTS IX Proceedings, Sringer LNCS 697 (00), 6 5. [4] A. A. Mullin, Recursive function theory, Bull. Amer. Math. Soc. 69 (963), 737. [5] P. Pollack and E. Treviño, The rimes that Euclid forgot, Amer. Math. Monthly (04), [6] C. Pomerance, Remarks on the Pólya Vinogradov inequality, Proc. Integers Conf. Oct. 009, Integers A (0), Article 9,. [7] K. Rogers, The Schnirelmann density of the squarefree integers, Proc. Amer. Math. Soc. 5 (964), [8] J. B. Rosser and L. Schoenfeld, Aroximate formulas for some functions of rime numbers, Illinois J. Math. 6 (96),

9 [9] S. S. Wagstaff, Jr., Comuting Euclid s rimes, Bull. Inst. Combin. Al. 8 (993), 3 3. [0] T. D. Wooley, A suerowered Euclidean rime generator, 06, rerint. School of Mathematics, Bristol University, University Walk, Bristol, BS8 TW, UK address: andrew.booker@bristol.ac.uk Mathematics Deartment, Dartmouth College, Hanover, NH 03755, USA address: carl.omerance@dartmouth.edu 8