infinitely any values of. This is alost certainly true, but will be hopelessly difficult to prove. Let g(n) be the sallest integer t? (if it exists) f

Transcription

1 ON THE PRIME FACTORS OF \k/ AND OF CONSECUTIVE INTEGERS P. Erdős and A. Sárközy. First we introduce soe notations. c l, c 2,... denote positive absolute constants. The nuber of eleents of the finite set S is denoted by ISI. We write ex = exp x. We denote the i th prie nuber by p,. v(n) denotes i the nuber of distinct prie factors of n. k is the sallest integer for which v(k» > k. P(n) is the greatest, p(n) the sallest prie factor of n. nk the sallest integer n for which is P(n+i) > k for all < i <_ k. In L3, Erdös, Gupta, and Khare proved that () Mk > c I k 2 log k. We prove THEOREM. Por k > k 0 we have (2) k > c 2 k 2 (log k) 4/3 (log log k)-4/3(logloglog k) -/3 It sees certain that k > k2+c. It follows fro results of Selfridge and the first author [4 that, by an averaging process, k < ke+e The exact deterination of, or even good inequalities for, k sees very difficult. Denote by k the least integer such that for every? k k )I? k Szeerédi and the first author proved in L5 that k < (e+c ) k for every c > 0 if k > k0. Very likely k < k c but we could not even prove k < (e-f) k. Schinzel conjectured that v((k)) = k holds for UTILITAS MATHEMATICA Vol. 6 (979), pp

2 infinitely any values of. This is alost certainly true, but will be hopelessly difficult to prove. Let g(n) be the sallest integer t? (if it exists) for which n-t has ore than one prie factor greater than t (or the sae definition for n+t) ; surely g(n) exists for both definitions but it is not known (see [4). In a previous paper C2, p.273, P. Erdős outlined the proof of (3) nk < k log k / log log k No reasonable lower bound for nk was known. We prove THEOREM 2. If k > k 0, then nk > k 5/2 /6. It sees certain that for k > k 0 (t), n k > k t and, in fact, perhaps the upper bound (3) is close to the "truth". Soe other related probles and results can be found in []. 2. Proof of Theore. We need two leas. LEMMA. If x? 0, y > 2 and d is a positive integer, then the nuber of sozutions of p-q = d, x < p << x+y, where p, q are prie nubers, is Zess than c 3 T[ I + y 2 < c 4 log log (d+2) y 2 p/d p log y log y This lea can be proved by Brun's sieve (see e.g. [6 or 7). LEMMA 2. Let N? 2 be a positive integer and s l < s 2 <...< sn e any integers. Put s N s l sozutions of = S and, for fixed d, denote the nuber of s - s = d x y by F(d). Then there exists a positive integer D such that D <_ NS and F(D) > 48S 2

3 Proof of Lea 2. For j =, 2,..., [N/4] +, let I j = [sl+(j-)y,sl+jy]. Then s l +(I s i+'y > s l + y y-s l+s=s N, and thus [S/y]+i U I j D Ls l, SN ; j= hence (4) lsj<_[n/4]+ J <-j5[n/4]+l J <j5[n/4]+l N.>-2 J N.= J =N- ([N/4]+)?N- 2= 2 N, since [ N ]± < N + N = N for N? 4 and holds alsó for N = 2, 3. Let N, J? 2 for soe N. Then for all the ( 2 J pairs, j and define z by s z i I J., sz+ E I j < x < y < N., we have. <_ s -sz+x < s z +N' J - sz+ < 4S/N z+ Y. Thus, with respect to (4) and using Cauchy's inequality, we obtain that lsd<_4s/n F(d) > 5j5[N/4]+ N,?2 J N `) 2j N,- N,/2 <-j<[n/4]+l J 2 lsjs[n/4]+ 2 N,>2 N.?2

4 4 lsj<_ln/4]+ > 4(2) N,?2 2 <j<ln/4]+l 2 N, E l< - j< - [N/4]+l J <j< - [N/4]+ N.?2 N.?2 J J > N2 N + > N 2 3N - N 6 4 ) hence F(d) <d<4s/n > N/2 = N 2 ax F(d) >_ < - d<-4s/n E 4S/N 48S ' < - d<-4s/n which copletes the proof of the lea. In order to prove Theore, it suffices to show that if e is sufficiently sall then the assuptions (5) v «M» > k, (6) : ck 2 (log k) 4/3 (log log k) -4/3 (log log log k) -/3 lead to a contradiction for all k? k 0 (c). By () we ay assue that (7) > c l k 2 log k also holds. For all 0 < i < k-, we write -i = a,b,, where P(a i ) < k and p(b i ) > k (if, for exaple, all the prie factors of -i are less than or equal to k then we write b, = ). i Furtherore, let S denote the set consisting of the integers 0 <_ i < k- such that P(-i) k log k log k and let S 2 = {0,,...,k-}-SI (i.e. i e S 2 if and only if 0 < i <- k- and P(-i) > k log k (log log k) -l ).

5 )-4/3 thus First we give a lower estiate for II a í (k) is an integer ; ics 2 k' k- k- (-)...(-k+) = II a, II b,, i=0 i=0 k- and, obviously, ~k!, R b, i=0 =. Thus we have ki and hence by Stirling's forula, (8) II i=0 a i > k. > (k, k 3 l (for sufficiently large k). Fro (5), we have (9) k < v ((k )' < (v((-)... (-k+l) ) + - p<-k k<p<_k log k (log log k) p I (-)...(-k+l) pim(-)...(-k+l) + ~ - k log k(log log k) <p p (-)...(-k+l) Here the first ter is less than or equal to 7(k). Furtherore, if k log k (log log k) - < p and p I (-)...(-k+l) then p-i for soe i e S 2, and by (6), we have -i p k log k (log log k) - sk 2 (log k)4/3 (log log (log log log k) -l/.3 k log k (log log k) - = ck(log k) /3 (log log k) -/3 (log log log k) -/3 < k log k (log log k) -.

6 Thus for all í c S 2, -i has only one prie factor greater than k log k (log log k) -. than or equal to IS2. This iplies that the third ter in (9) is less Finally, we write We obtain fro (9) that hence E - =R. k<p<k log k (log log k) pi(-)... (-k+l) k < Tí(k)+R+IS 2 I ; (0) R? (k-is2i)- TT(k) = S - Tr(k) Obviously, we have S () = - < ies l S b i II - p k<p<k log k (log log k) pl(-)...(-k+l) <_ i=tt(k)+r p p í = Ti (k)+ By the prie nuber theore, we have p i - i log i. Furtherore, by the prie nuber theore and the definition of R, R < Tr(k log k (log log k) ) < k (for large k). Thus by (0) and using the prie nuber theore and Stirling's forula, we obtain, for sufficiently large k and S I I > c 8, that 7T(k)+R Ti(k)+R (2) II p, > II í=7i (k)+ L í=tt(k)+ ' - i log i Tr (k) +R R í=tr (k)+ ' - )i log (TT(k)+)

7 R(log ~ (k)+r - i log k = 9 k)r (Tr(k)+R)' 0) (0) ( (k)). i=7r(k)+ (2_)k IS I I-7(k) IS I I : (log k) IO r(k) 7r k) ( - Ti > ( 9~) k (log k) ISl (7(k) log k) (k) k 9 (ISII (IO ) log S l i IS l i SI k) Is e -Ti(k) log (Tr(k) log k) 3 -k 3 9 k ( 0) (ISI log (9 0 ) k(isi log k) I S I e -2k 3 -k Is -k -k k) 93= 30(ISIIlog -k k) S l () and (2) yield that, for IS I I > c 8, (3) R ai < icsl Is ISI I k IS 30k IS IS I I k log k 30 -k (IS I Ilog k). By (0) and the prie nuber theore, we have (4) IS s R+TT(k) _ E + 7T(k) k<psk log k(log log k - pim(-)...(-k+l) 5 E (k) k<p`k log k(log log k) _ (Ti(k log k (log log k) - ) - TT(k)) + 7T (k) = Tr(k log k (log log k) -I)< 2k(log log k) -I - If a > 0 then an easy discussion shows that the function f(x) _ (a/x) x is increasing in the interval 0 < x < a/e. If we put a = /(k log k) and x = ÍS then fro (7) and (4) we have a c k 2 log k _ > 7 = rl k > 2k(log log k) > e e k log k e k log k e IS I

8 for sufficiently large k. Thus, for large k and ISI > c 8, (3) and (4) yield that (5) IS k a i < 30 ie S IS Ik log k / ISI < k IS 2k(log log k) - 30k( 2k(log log k) - k log k For large k, this holds also for 0 < IS 5 c 8 since in this case, fro (6) and (7) we have while IS I 2k(log log k) - < k 30k k 2 log k (log log k) - 2k(log log k) - kisi30k (k 2 log k (log log k) - c l k 2 log k > 30k (k 2 log k (log log k) - 2k(log log k) - k > 30, F a, < n = ISI < (k3) ics ics S I < k 3c 8 < 30k for sufficiently large k. (8) and (5) yield that k- (6) n a. _ ;I a a i(s2 i=0 iesi - - k 2k(log log k) kikisi30k ~3/ (k 2 log k (log jog k) - kk- IS -2k(log log k) k k 2 log k (log log k) - = k S k ~ 2k(log log k) - (k 2 log k(log log k) -

9 Let S,, denote the set consisting of the integers 0 <- i < k-, i E S 2, and ai < 0-6 k. Then we have i satisfying (7) R a, = R a. R ies2 ES 2,i~S 3 ics k b igs 2,i~S 3 ics 3 R 0 6k R k kp(-i) ics 2,i~S 3 ies 3 5 k Is 2I R 0-6 ' ics 2, U S 3 ics 3 k 2 log k (log log k) - Is2i (0-6) k ` 2 I-T T (k 2 log k (log log k) - kis2i(0-6)k-t ( _ k` log k (,log log k) - T where (.6) and (7) yield that - 2k(log log k) k is2i 9p k (k 2 log k (log log k) - ) < k S2I(lp-6)k-T( k 2 log k (log log k) - T hence (8) (0 6 ) k-t 90 -k < ( k 2 log k(log log k) - T+2k(log log k)- We are going to show that this inequality iplies that (9) T > k(log log k) -. If T? k/2 then this inequality holds trivially for large k ; thus we ay assue that T < k/2. Then we obtain fro (8) that

10 k 2 log k(lo g log k) - T+2k(log log k) - (0 6 ) k-k/2 90 -k > 03k 0-2k = 0 k. Thus, fro (6), we have T+2k(log log k)- > _ klog0 log k 2 log k(log log k) - klog0 log Ek 2 (logk) 4/3 (loglogk) -4/3 (loglog logk) -/3 k 2 log k (log log k) - k log 0 log (e(log k) /3 (log log k) -/3 (log log log k) -/3 ) > klog0-3klog0 > 3k(log log k) - log log k tog (log k) /3 for sufficiently large k, which yields (9). By the definition of a., we have a.l-í. If then > 0-6 k also holds. But, if a > 0-6 k, then a ay have at ost k+ < k + < < al a ultiples aongst the nubers, -,., -k+l. Thus, for fixed a, a i a, E S 3, has less than solutions. Fro (9), this iplies that the nuber of the distinct a í 's with í c S 3 is greater than F, ics3 T k(2.0 6 log log k) > By the definitions of the sets S 2 and S 3, i ( S 3 (á S 2 ) iplies that k log k (log log k) - < P(-i) < b i ;

11 hence, fro (6), (20) (0-6k <)a. -i = i bi k log k (log log k) I < ek 2 (log k) 4/3 (log log Q - 4/3 (log Iog log k) /3 - k log k (log log k)- ek (log k) I/3 (log log k) -I/3 (log log log k) -/3 Let us write = k (log k) -5/3 (log log k) 2/3 (log log log k) 2/3 and for all (ek(log k)/3(log log k) -/3 (log log log k)-/3 + l, t L let us for the interval I J, _ ((j-)t,jt7. By (20), these intervals cover all the a i 's (with i E S 3 ), and the nuber of these intervals is ~ek(log k)/3(log log k) -/3 (log log logk) -/3, + I t l J ~ek(log k) /3(log log k) -/3 (log log log k)-/3 + k(log k) -5/3 (log log k) 2/3 (log log log k) 2/3 c(log k) 2 (log log k) - (log log log k) -I + 2e(log k) 2 (log log k) - (log log log k) -I (for large k). Thus the atchbox principle yields that there exists an interval I, which contains ore than j k(2.0 6 log log k) -I 2e(log k) 2 (log log k) -I (log log log k) - _ ( ) - c -I k (log k) -2 log log log k distinct a I 's. In other words, there exist indices i l, i 2, satisfying i Q e S 3 for i <_ Q <_ N, N

12 (2) (j-)t < a. < a, <...< a. S jt 2 N and (22) N > ( ) - c - k (log k) -2 log log log k. Let us apply Lea 2 with the set a., a,,..., a, in place 2 N of the set s l, s 2,...,s N. (By (2), N? 2 holds trivially.) Then fro (2), we have S = s N -s l = a -a i l IV < t = k(log k) -5/3 (log log k) 2/3 (log log log k) 2/3 Thus, fro (22), we obtain that there exists an integer (23) D < 4S < 4k(log) k) -5/3 (log log k) 2/3 (log log log k) 2/3 N (4.0 6 ) - e - k (log k) -2 log log log k c(log k) /3 (log log k) 2/3 (log log log k) -/3 a. - a, = D x y has at least (25) F(D)? N2 > ( )-2e-2k2(log k) -4 (log log log k) 2 48S 48k(log k) -5/3 (log log k) 2/3 (log log log k) 2/3 solutions. that for large k > 0-5 c -2 k(log k) -7/3 (log log k) -2/3 (log log log k) 4 / 3 By the definition of the set S 3, i Q E S 3 iplies, fro (6),

13 -2 R a Q 0-6 k < 0 6 e k(log k) 4/3 (log log k) -4/3 (log log log k) -/3 < k(log k) 4/3 and, on the other hand, (26) b í? P(-í R ) > k log k (log log k)-. Q By the definition of b,, we have p(b, ) > k ; thus these inequalities Q Q iply that b i = q. ust be a prie nuber. Then q l, g 2' *** ' qn are 2 distinct since by (26), q Q > k log k (log log k) -I (for all R). Thus q. ay have at ost one ultiple aongst the nubers, -,...,-k+l. Furtherore, we have -iz a, < ('-)t (it can be shown easily that j > ) and qq -iq R > _ ai a i al jt k 0-6k jt Q Q Q Thus all the pries q R belong to the interval (27) 6 I= r j t - 0 (j - ) t Fro a i > 0-6 k, (5), and the definition of t, the length of this interval is

14 Q-) a i a a (28) ~~ _ 6 6 t - jel- - lo = 0-) jt + lo < t < t < t (it-t)jt (aiz-t)a (0-6 k-t)0-6 k < tk-2 < ek 2 (log k) 4/3 (log log k) -4/3 (log log log k) -/3 k (log k) -5/3 (log log k) 2/3 (log log log k) 2/3 k ek(log k) -/3 (log log k) -2/3 (log log log k) /3 Furtherore, if a, and a, satisfy (24) then we have x y a. -ai -i -i i i i Y x x= Y + x Y qy-qx, a, a, a,, i i i i i y x x y x y D +x - a i (ai +D) a, a, Y Y l x l y Hence (29) l) (4 y - q x ) - a l (a +D) Y Y < + < 2 k6 = a a y i i 0 k x y On the other hand, by (6) and (23), we have (30) D _ D a i (a i +D) jt(jt+d) Y Y jt-a, jt+a. +D = D l Y l Y a i (a i +D)jt(jt+D) Y Y L D (jt) 2- (a i ) 2) +D(jt-a i )I Y Y a. (a. +D)jt(Jt+D) l y l Y D jt-a, I(jt+a, ) y x a i a i jt(jt+d) y x < D t 2 jt a, a, it a y x x 2Dt = Dtk -3 (0-6 k) 3 < ek 2 (log k) 4/3 (log log k) -4/3 (log log log k) -/ e(log k) /3 (log log k) 2/3 (log log log k) -/3 k (log k) -5/3 (log log k) 2/3 (log log log k) 2/3 k e 2 = 32.0

15 For sall E, (29) and (30) yield that (q q ) - D Y - x it(jt+d) (q -q ) - D y x a i (a i +D) Y Y D D a i (a i +D) jt(jt+d) Y Y < E2 < Thus for all the F(D) solutions x, y of (24), q y -qx is in the interval < D+D q -q < D jt(~ ) y x it(jt+d) The length of this interval is ; thus, by the atchbox principle, there exists an integer d such that (fro (6) and (23)) (3) d < jt(jd+d) < -6 k) (0 < 0 2 ck 2 (log k) 4/3 (log log k) -4/3 (log log log k) -/3 6 /3 2/ E(log k) (log log k) (log log log k) -/3.k E 2 (log k) 5/3 log log k) -2/3 (log log log k) -2/3 and, denoting the nuber of solutions of (32) qx q y = d, q x E, q y E by G(d), we have fro (25) (33) G(d) > F(D) > 0-5 c -2 k(log k) -7/3 (log log k) -2/3 (log log log k) 4 / > 0 22 e -2 k(log k) - /3 (log log k) -2/3 (log log log k) 4/3 On the other hand, by (28), Lea yields that the nuber of solutions of (32) is

16 (34) G(d) < c log log (d+2) III 4 log 2 < c4 log log(6.08e2(log k)5/3(log log k)-2/3(log log log k)-2/3+2) 2.0-2ek(log k)-/3(log log k)-2/3(log log log k)/3 log 2(2.0-2Ek(log k)-/3(log log k)-2/3(log log log k)/3) < 2c tog tog tog k 2'0 2 ek(log k)- /3 (log log k)- 2/3 (log log log k) / log2k = s4 k(log k)-7/3(log log k)-2/3(log log log k) 4/3 hence (33) and (34) yield that 0-22e-2k(log k)-7/3(log log k)-2/3(log log log k) 4/3 < 8.0-2E c4 k(log k)-7/3(log log k)-2/3(log log log k) 4/ c- < c3 But for sufficiently sall c, this inequality cannot hold, and this contradiction proves Theore. (In fact, as this inequality shows, Theore holds,for exaple,with c2 = 0-4 c4/á ) 3. Proof of Theore 2. We have to prove that if k > k0 and (35) 0 <_ n <_ k5/2/6, then there exists an integer u such that (36) n < u < n+k and (37) P(u) <- k. We are going to show that there exists an integer u satisfying (36) and of the for

17 where the integers d i satisfy (38) 0 < di <_ k for i =, 2, 3. Obviously, an integer u of this for also satisfies (37). Let us define the positive integers x and z by z = Ck /2 /3] and x = C(n/z) /2 ] +. Then x 2 > ((n/z)/2)2 = n/z ; hence (39) x 2 - n > 0 z Let us define the non-negative integer y by (40) y < Finally, let (x2 - n)/2 < y+. z and d l = z, d 2 = x - y, d 3 = x+y. Then d l > 0 and d 3 > 0 hold trivially while d 2 > 0 will follow fro d l > 0, d 3 > 0 and u = d d 2 d 3 > n. Furtherore, d l = z = [k l/2 /3] < k holds trivially. Finally, in view of (35) and (39), we have

18 d 2 =,/2 x-y < x+y = d < (x + x2 - n zz 3 (n/z)/2+ + «(n/z)/2+,2 - \/2 n z (n/z)/2++(2(n/z)/2+)/2 k 5/2 /2 /2 5/2 /2 / k /2 \ 6[k /3J 6Ck /37 ) _ (+0())3 /2 4 - k++0(k í/2 ) _ (+0())3 /2 4 - k for large k -> + -, Here 3 /2 4 - < ; thus also d 2 <- d 3 < k holds for k which copletes the proof of (38) (provided that u > n). By (40), we have (4) u = dld2d3 = z(x-y)(x+y) = z(x 2 -y 2 ) > z/x2- ( x 2- n)l/2 2 = z,z = n. Finally, by (40), hence x 2 - z ~ (Y+) 2, z(x2-(y+)2) ~ n Thus, fro (35), we have (42) u = d l d 2 d 3 = z(x-y)(x+y) = z(x 2- y 2 ) n + (z(x2-y2)-n) -< n+(z(x2-y2)-z(x2-(y+)2) n+(2y+)z < n+(2 ( x2~ n)l/2 + z l I z < n+(2 ((n/z)/2+)2 - ntl/2 + z n+(2(2(n/z)/2+)/2+,z n /2 + )/2 + k l/2 n+ 2 (2( ( \ `Ck l/2 /37! 3 ( k 5/2 /6\ /2 /2 k /2 < n \kí/2/4 3 / n+'2 (k+)/2+) k 3/2 < n+3k/2 - k 3/2 n+k for sufficiently large k./ (4) and (42) yield (36) and this copletes the proof of Theore 2.

19 REFERENCES [7 P. Erdős, Uber die AnzahZ der Prifaktoren von (k), Archiu der Math. 24 (973), [2] P. Erdős, Probles and results on consecutive integers, Publ. Math. Debrecen. 23 (976), [37 P. Erdős, H. Gupta and S. P. Khare, On the nuber of distinct prie factors of (k), Utilitas Math. 0 (976), [4] P. Erdős and J. Selfridge, Soe probles on the prie factors of consecutive integers, Illinois J. Math. IZ (967), [5] P. Erdős and E. Szeerédi, Proble 92, Mat. Lapok 25 (974), [6 H. Halbersta and H. E. Richert, Sieve Methods, Acadeic Press, London - New York - San Francisco, 974. [77 H. Halbersta and K. F. Roth, Sequences, VoZ. I, Oxford, Clarendon Press, 966. Matheatical Institute Hungarian Acadey of Sciences Budapest Received February, 979.