#A12 INTEGERS 13 (2013) THE DISTRIBUTION OF SOLUTIONS TO XY = N (MOD A) WITH AN APPLICATION TO FACTORING INTEGERS


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1 #A1 INTEGERS 13 (013) THE DISTRIBUTION OF SOLUTIONS TO XY = N (MOD A) WITH AN APPLICATION TO FACTORING INTEGERS Michel O. Rubinstein 1 Pure Mthemtics, University of Wterloo, Wterloo, Ontrio, Cnd Received: 3/6/09, Revised: 9/1/1, Accepted: 3/6/13, Published: 3/15/13 Abstrct We consider the uniform distribution of solutions (x, y) to xy = N mod, nd obtin bound on the second moment of the number of solutions in squres of length pproximtely 1/. We use this to study new fctoring lgorithm tht fctors N = UV provbly in O(N 1/3+ ) time, nd discuss the potentil for improving the runtime to subexponentil. 1. Introduction Let gcd(, N) = 1. A clssic ppliction of Kloostermn sums shows tht the points (x, y) mod stisfying xy = N mod become uniformly distributed in the squre of side length s. In this pper we investigte n ppliction of this fct to the problem of fctoring integers. We give new method to fctor the integer N which bets tril division, nd prove tht it runs in time O(N 1/3+ ). Unlike mny existing fctoring lgorithms tht ttempt to fctor N by considering vrious congruences modulo N, the method presented here revisits tril division nd exploits informtion vilble on filed tril division, nmely it considers the reminder of N when divided by n integer, nd lso by its neighbor, 1. While the complexity of our method is not exciting, considering the existence of severl probbilistic subexponentil fctoring lgorithms, the runtime here is provble. For comprison, the best known provble fctoring lgorithm, Pollrd Strssen, runs in time O(N 1/4+ ). Shnk s clss group method runs in time O(N 1/5+ ) ssuming the GRH. It hs comprble runtime to Lehmn s method [7], but the method is different. Our lgorithm is described in Section. Furthermore, proving this runtime requires understnding the finer distribution of solutions to xy = N mod, nd our results in this regrds re interesting in their own right. We discuss the problem on uniform distribution in Sections 4 nd 5. Finlly, ll existing subexponentil fctoring lgorithms hve grown out of much weker exponentil lgorithms, nd we hope tht the fctoring ides presented here 1 Support for work on this pper ws provided by n NSERC Discovery Grnt
2 INTEGERS: 13 (013) will be improved. In Section 3 we discuss some needed improvements to chieve better runtime. We hve not implemented the lgorithms described in this pper. The purpose of this pper is to present new pproch to fctoring integers nd nlyze its runtime.. Algorithm Hide nd Seek Let N be positive integer tht we wish to fctor. Sy N = UV where U nd V re positive integers, not necessrily prime, with 1 < U V. For simplicity, ssume V < U, so tht V < (N) 1/. The generl cse, without this restriction, will be hndled t the end of this section. The ide behind the lgorithm is to perform tril division of N by couple of integers, nd to use informtion bout the reminder to determine the fctors U nd V. Let be positive integer, 1 < < N. By the division lgorithm, write U = u 1 + u 0, with 0 u 0 < V = v 1 + v 0, with 0 v 0 <. (1) Assume tht u 0 is reltively prime to, nd likewise for v 0, since otherwise we esily extrct fctor of N by tking gcd(, N). If, for given, we cn determine u 0, u 1, v 0, v 1 then we hve found U nd V. Consider N = u 0 v 0 mod. One cnnot simply determine u 0 nd v 0 from the vlue of N mod, becuse φ() pirs of integers (x, y) mod stisfy xy = N mod (if x = mu 0 mod, then y = m 1 v 0 mod, where gcd(m, ) = 1). However, sy is lrge, (N) 1/3 > V /3, so tht v 1 nd u 1 re comprtively smll, u 1, v 1 V 1/3, i.e., both re < 1/. If we consider N mod δ N = UV = (u 1 δ + u 0 )(v 1 δ + v 0 ) mod δ, () for δ = 0, 1, we get, s solutions (x, y) to xy = N mod δ, two nerby points, (u 0, v 0 ) nd (u 0 + u 1, v 0 + v 1 ), whose coordintes re within 1/ of one nother. This pir of points is just one pir mongst the mny pirs of solutions to the bove equtions, for δ = 0, 1. However, the fct tht the solutions re nerby reduces the mount of checking tht we need to do in order to find the pir of points, (u 0, v 0 ) nd (u 0 + u 1, v 0 + v 1 ), tht we seek. Figures 1 nd illustrte this fct, for N = = , nd = 157. Thus, U = 1061, u 0 = 119, u 1 = 8, nd V = 1801, v 0 = 74, v 1 = 15. Rther thn just depict the solutions to xy = N mod δ, for δ = 0, 1, we lso plot the solutions for δ =, 3 (though our lgorithm below only mkes use of solutions for
3 INTEGERS: 13 (013) 3 δ = 0, 1). Plotting four sets of solutions, for δ = 0, 1,, 3 mkes it esier for the humn eye to tell the points (u 0, v 0 ) = (119, 74), (u 0 + u 1, v 0 + v 1 ) = (15, 85), (u 0 + u 1, v 0 + v 1 ) = (131, 96), nd (u 0 + 3u 1, v 0 + 3v 1 ) = (137, 107) from the rndom coincidences of nerby points s these ll lie eqully spced prt nd on one line. Solutions to xy = mod 157. Solutions to xy = mod y 80 y x x Solutions to xy = mod 155. Solutions to xy = mod y 80 y x x Figure 1: The solutions (x, y) to xy = mod 157 δ, for δ = 0, 1,, 3 So, we cn set, sy, = (N) 1/3, nd prtition the Crtesin plne into squres of side length 1/, ech squre being of the form {(x, y) R m 1/ x < (m + 1) 1/, n 1/ y < (n + 1) 1/ }, where m, n Z. We then list ll φ() pirs of integers (x, y), with 1 x, y, tht stisfy xy = N mod, throwing them into our squres of side lengths 1/. We cn
4 INTEGERS: 13 (013) 4 Solutions to xy = mod 157,156,155,154 superimposed y Figure : The solutions (x, y) to xy = mod 157 δ, for δ = 0, 1,, 3, superimposed. The four circled points re (u 0, v 0 ) = (119, 74), (u 0 + u 1, v 0 + v 1 ) = (15, 85), (u 0 + u 1, v 0 + v 1 ) = (131, 96), nd (u 0 + 3u 1, v 0 + 3v 1 ) = (137, 107). x ssume tht gcd(, N) = 1, becuse, otherwise we esily extrct fctor of N. We cn compute ll inverses mod, nd hence ll (x, y) = (x, x 1 N) mod in O() opertions mod. To compute ll inverses, strt with m =, multiply mod by m until we rrive t 1, or hit residue clss lredy encountered (in which cse m is not invertible). Then, tke the first residue not yet encountered nd repet the previous step until ll residue clsses re exhusted. Hving produced ll solutions for the modulus, we then repet the process for the modulus 1. For ech solution (x 1, y 1 ) to xy = N mod 1, we determine which 1/ 1/ squre it flls within, nd consider ll nerby (with ech coordinte within 1/, wrpping to the opposite side of the lrger squre if needed) solutions (x 0, y 0 ) to xy = N mod from our list of stored solutions. We set µ 0 = x 0, ν 0 = y 0, µ 1 = x 1 µ 0, ν 1 = y 1 ν 0, nd check whether (µ 1 +µ 0 )(ν 1 +ν 0 ) = N. If so, we hve determined nontrivil fctor of N nd quit.
5 INTEGERS: 13 (013) 5 How much work does compring pirs of points (x 0, y 0 ) nd (x 1, y 1 ) entil? There re φ( 1) solutions to xy = N mod 1, nd, typiclly, we expect there to be only hndful of solutions to xy = N mod whose coordintes re ech within 1/. Ech such pir of solutions gives us cndidte vlues µ 0, ν 0 nd µ 1, ν 1 for u 0, v 0 nd u 1, v 1, nd we check to see whether they produce N = (u 1 + u 0 )(v 1 + v 0 ). On verge, ech 1/ 1/ squre contins O(1) points, the overll time to check ll squres nd points is roughly predicted to be O(). In Section 4 we obtin runtime bound of O( 1+ ). This lgorithm termintes successfully when the true points (u 0, v 0 ) nd (u 0 + u 1, v 0 + v 1 ) re found. Since = O(N 1/3 ) this gives running time tht is provbly O(N 1/3+ ). The ide tht lies behind the lgorithm suggests the nme Hide nd Seek. The solutions tht we seek (u 0, v 0 ) nd (u 0 + u 1, v 0 + v 1 ) re hiding mongst mny solutions in the lrge squre, but, like children who hve hidden next to one nother while plying the gme Hide nd Seek, they hve become esier to spot. We summrize the bove in the following lgorithm. Algorithm.1. (Hide nd Seek) Let N = UV be positive integer, nd ssume tht 1 < U V < U, with U, V Z. Thus V < (N) 1/. For given positive integers N, r, define Step 1 Set = (N) 1/3. H N, = {(x, y) xy = N mod, 0 x, y < }. (3) Step Use the Eucliden lgorithm to compute gcd(n, δ) for δ = 0, 1. If either gcd is > 1 then we hve determined nontrivil fctor of N nd quit. Step 3 Compute nd store in n rry ll φ() points of H N,. This cn be done using O() rithmetic opertions mod s described bove. Step 4 For 0 m, n < 1/, initilize doubly indexed rry, Bin. Ech element, Bin[m,n], will contin list of points nd be used to prtition H N,. Ech is initilly set to empty. Step 5 Prtition the elements of H N, ccording to squres of side length 1/ by computing, for ech (x, y) H N,, the vlues m = x/ 1/ nd n = y/ 1/, nd ppending the point (x, y) to Bin[m,n]. Step 6 Compute the φ( 1) elements of H N, 1. For ech (x 1, y 1 ) H N, 1 : Step 6 Determine which bin it corresponds to by computing m = x 1 / 1/ nd n = y 1 / 1/. Step 6b Loop through the nerby points (x 0, y 0 ) of H N, whose coordintes lie, left nd downwrds, within 1/. Typiclly, this entils exmining the four bins Bin[m 1, n ], where 1, {0, 1}. However, slight cre
6 INTEGERS: 13 (013) 6 is needed when crossing over n edge of the squre one should wrp to the opposite side of the squre. Step 6c Set µ 0 = x 0, ν 0 = y 0, µ 1 = x 1 µ 0, ν 1 = y 1 ν 0, nd check whether (µ 1 + µ 0 )(ν 1 + ν 0 ) = N. If so, we hve determined nontrivil fctor of N nd quit. The storge requirement of O(N 1/3 ) cn be improved to O(N 1/6 ) by generting the solutions (x, y) to xy = N mod δ lying in one verticl strip of width O( 1/ ) t time (esy to do since we cn choose x s we plese, which then determines y). In generl, we re then no longer free to generte ll modulr inverses t once, nd must compute inverses in intervls of size 1/, one t time, t cost, using the Eucliden lgorithm, of O( ) per inverse. A slight improvement on the runtime of the lgorithm cn be chieved my modifying the choice of so tht φ() or φ( 1) re comprtively smll. This cn be chieved by selecting n or 1 with smller distinct prime fctors..1. Vrint: 1 < U V < N Without Restriction Sy U = N α, V = N 1 α, with 1/3 < α 1/. We my ssume tht α > 1/3, for, if not, we cn find U by performing O(N 1/3 ) tril divisions. Let = N 1/3 (we do, here, men N 1/3, rther thn (N) 1/3 of the previous section, s explined below). Insted of working with smll squres of side length 1/, prtition the squre into rectngles of width w nd height h, with wh = N 1/3. We would like to select w roughly equl to N α 1/3 nd hence h = N 1/3 /w roughly equl to N /3 α. These rough vlues of w nd h re needed to mke sure tht, using the sme nottion s before, (u 0, v 0 ) nd (u 0 + u 1, v 0 + v 1 ) re in the sme, or neighboring, rectngles. More precisely, sy N α 1/3 < w N α 1/3. Then h = N 1/3 /w N /3 α /. Then, in (1), u 1 = U/ N α /N 1/3 N α 1/3 / < w, nd v 1 = V/ N 1 α /N 1/3 N /3 α / h. Thus the xcoordintes of (u 0, v 0 ) nd (u 0 + u 1, v 0 + v 1 ) re < w prt nd the ycoordintes re h prt. Since we do not, priori know α, we cnnot simply set w nd h. Insted, we use n exponentilly incresing set of w s, for exmple strting with w =, nd, repetedly pplying the bove procedure, ech time doubling the size of w, until w > N α 1/3 nd one successfully fctors N. The re of ech rectngle is N 1/3, nd of the squre is pproximtely N /3, so there re O(N 1/3 ) rectngles (t the top nd right edges these will typiclly be truncted), nd, on verge, ech contins O(1) solutions to xy = N mod δ. Running through ech rectngle nd its immedite neighbors, checking ll pirs of points in these rectngles suggests O(N 1/3 ) opertions re needed for prticulr choice of w nd h. Since we might hve to repet this few times, doubling the size of w, the overll running time gets multiplied by O(log N) which is O(N ).
7 INTEGERS: 13 (013) 7 In Section 5, running time equl to O(N 1/3+ ) is proven. The steps described in this section re summrized below. Algorithm.. Let N = UV be positive integer, nd ssume tht 1 < U V < N, with U, V Z. Step 1 Crry out tril division on N up to N 1/3. found, quit. Step Set = N 1/3. If nontrivil fctor of N is Step 3 Use the Eucliden lgorithm to compute gcd(n, δ) for δ = 0, 1. If either gcd is > 1 then we hve determined nontrivil fctor of N nd quit. Step 4 Compute nd store, in two rrys, ll the points of H N, nd H N, 1. Step 5 Set j = 0. While we hve not succeeded in finding nontrivil fctor of N: Step 5 Increment j by 1 nd set w = j nd h = N 1/3 /w. Step 5b For 0 m < /w nd 0 n < /h, initilize doubly indexed rry, Bin, whose elements, Bin[m,n], will contin lists of points nd be used to prtition H N,. Ech bin is initilly set to empty. Step 5c Prtition the elements of H N, ccording to rectngles of width w nd height h by computing, for ech (x, y) H N,, the vlues m = x/w nd n = y/h, nd ppending the point (x, y) to Bin[m,n]. Step 5d For ech (x 1, y 1 ) H N, 1 : Step 5d1 Determine which bin it corresponds to by computing m = x 1 /w nd n = y 1 /h. Step 5d Loop through the nerby points (x 0, y 0 ) of H N, whose coordintes lie, left nd downwrds, within w nd h respectively. Typiclly, this entils exmining the four bins Bin[m 1, n ], where 1, {0, 1}. However, slight cre is needed when crossing over n edge of the squre one should wrp to the opposite side of the squre. Step 5d3 Set µ 0 = x 0, ν 0 = y 0, µ 1 = x 1 µ 0, ν 1 = y 1 ν 0, nd check whether (µ 1 + µ 0 )(ν 1 + ν 0 ) = N. If so, we hve determined nontrivil fctor of N nd quit. Step 5e Free up the memory used by Bin. 3. Towrds Subexponentil Bound The bove lgorithm exploits the fct tht when is lrge, nd δ is smll, the points with coordintes (U, V ) mod δ re close to one nother. In fct they lie eqully spced on line with common horizontl difference u 1, nd verticl difference v 1.
8 INTEGERS: 13 (013) 8 An obvious thing to try is to reduce the size of. However, s decreses, u 1 nd v 1 increse so tht not only do the points (u 0, v 0 ) nd (u 0 + u 1, v 0 + v 1 ) move fr prt, the ltter point soon flls fr outside the squre of side length. To fix this, one cn view (1) s the bse expnsion of U nd V. When is smller, one could insted use polynomil expnsion U = u d1 d u 1 + u 0, 0 u i < V = v d d v 1 + v 0, 0 v i <, (4) with u d1 = 0 nd v d1 = 0. For simplicity in wht follows, ssume tht the degrees of both polynomils re equl, d 1 = d = d, so tht both U nd V stisfy d U, V < d+1. A polynomil of degree d is determined uniquely by d + 1 vlues. Imitting the pproch in Section 1, we evlute N mod δ for d + 1 vlues of δ. A nturl choice might be δ = 0, ±1, ±,..., but, to keep our polynomil vlues positive, we consider nonnegtive vlues of δ, nd, for good mesure, tke extr vlues, δ = 0, 1,,... d (by extr, we men δ d rther thn d leqd). Now, N = UV = (u d δ d u 1 δ + u 0 )(v d δ d v 1 δ + v 0 ) mod δ. (5) Since 0 u j <, we hve where u d δ d u 1 δ + u 0 < λ(d, δ) (6) λ(d, δ) = δ d + δ d = (δ d+1 1)/(δ 1) δ d, s δ. (7) nd similrly for the v j s. For ech δ one lists ll solutions (x, y) to xy = N mod δ (8) 0 < x, y < λ(d, δ). (9) The number of points (x, y) for given δ is φ( δ) per squre, nd hence, overll, equls φ( δ)λ(d, δ) = O((d) d ). (10) We re gin ssuming tht gcd( δ, N) = 1, otherwise one esily pulls out fctor of N. We need method to recognize the solutions tht we seek (u d δ d +...+u 0, v d δ d v 0 ) hiding mongst ll the (x, y) s. This leds to the question: Let X > 0 nd let S 0, S 1,..., S d be d + 1 sets of points Z ll of whose coordintes re positive nd X. Assume tht mongst these points there exists
9 INTEGERS: 13 (013) 9 d + 1 points, one from ech S δ, whose coordintes re described by polynomils u(δ), v(δ) Z[δ] of degree d. More precisely, for ech 0 δ d there exists point (x δ, y δ ) S δ such tht x δ = u(δ) = u d δ d u 0 y δ = v(δ) = v d δ d v 0. (11) Cn one find these d+1 points much more efficiently thn by exhustively serching through ll possible d + 1 tuples of points? For exmple, cn one find these points in time O(X α d βd ) for some α, β > 0? In our ppliction, X = O((d) d ). Since N = UV nd d U < V < d+1, we hve < N 1/(d). Assuming tht there is n O(X α d βd ) time lgorithm for finding points with polynomil coordintes, on tking d proportionte to 1/ log N (1) log log N one gets fctoring lgorithm requiring exp γ(log N log log N) 1/ (13) time nd storge, for some γ > 0. One cn cut bck bit on the serch spce, by noting, for exmple, tht the coefficients of u(δ) nd v(δ) re integers (this imposes divisibility restriction on finite differences between points lying on the polynomil), nd, in our prticulr ppliction, tht the coefficients re nonnegtive nd bounded, nd this restricts the rte of growth of the polynomils. However, to get down to running time polynomil in X, one needs to do much better. 4. Uniform Distribution Let gcd(, N) = 1. A clssic ppliction of Kloostermn sums shows tht the points (x, y) mod stisfying xy = N mod become uniformly distributed in the squre of side length s. While the tools used in this section re firly stndrd, they will lso be pplied in the next section to estimte the running time of the Hide nd Seek lgorithm. Similr theorems cn be found in the literture ([1], [], [3], [4], [8], [9], [11]), often with restrictions to prime vlues of or to N = 1. Consider the following identity which detects pirs of integers (x, y) such tht xy = N mod : 1 1 k 1 if xy = N mod e (y xn) = (14) 0 otherwise k=0
10 INTEGERS: 13 (013) 10 where e(z) = exp(πiz), nd where x stnds for ny integer congruent to x 1 mod, if the inverse exists. Recll tht we hve ssumed gcd(, N) = 1 so tht ny solution to xy = N mod must hve gcd(x, ) = 1. Thus, for such solutions, x 1 mod exists. Let R be the rectngle bounded horizontlly by x 1, x Z nd verticlly by y 1, y Z, where 0 x 1 < x nd 0 y 1 < y : R = R(x 1, x, y 1, y ) = {(x, y) Z x 1 x < x, y 1 y < y }. (15) Let c R (N, ) denote the number of pirs of integers (x, y) tht lie in the rectngle R, nd stisfy xy = N mod : c R (N, ) = 1. (16) The identity bove gives c R (N, ) = 1 1 k=0 (x,y) R xy=n mod (x,y) R gcd(x,)=1 k e (y xn). (17) Notice tht we only need to restrict x to gcd(x, ) = 1 nd tht y runs over ll residues in y 1 y < y. This will llow us to del with the sum over y s geometric series. The k = 0 term provides the min contribution while the other terms cn be estimted using bounds for Kloostermn sums. We require two lemms. The first considers the min contribution, nd the second bounds the remining terms. Lemm 4.1. The k = 0 term in (17) equls for ny > 0. re(r) φ() + O( ) (18) Proof. The k = 0 term is 1 (x,y) R gcd(x,)=1 Using the Mobius function we hve 1 = x 1 x<x gcd(x,)=1 = d x 1 x<x 1 = (y y 1 ) d gcd(x,) µ(d) = d µ(d) x 1 x<x gcd(x,)=1 x 1/d x<x /d 1. (19) µ(d)((x x 1 )/d + O(1)) = (x x 1 ) 1/p) + O(τ()), (0) p (1 1
11 INTEGERS: 13 (013) 11 where τ() equls the number of divisors of nd is O( ) for ny > 0. This implies tht the k = 0 contribution to c R (N, ) equls which gives the lemm. re(r) φ() + O((y y 1 ) 1+ ) (1) The next lemm bounds the contribution of the k 1 terms in (17). Lemm 4.. For ny > 0 we hve 1 1 k=1 (x,y) R gcd(x,)=1 k e (y xn) = O( 1/+ ). () Proof. One cn seprte the sum over y nd evlute it s geometric series obtining for the lefthnd side bove 1 1 e k y e k y 1 e k k=1 1 x 1 x<x gcd(x,)=1 Tking bsolute vlues we get n upper bound of 1 1 sin πk (y y 1 sin πk k=1 x 1 x<x gcd(x,)=1 k e xn. (3) k e xn. (4) Next, notice tht the terms k nd k give the sme contribution, so we my restrict our ttention to just the terms 1 k ( 1)/. If 1 is odd, the middle term is left out t cost of O(1), nd the bound becomes 1 k ( 1)/ x 1 x<x gcd(x,)=1 sin πk (y y 1 sin πk x 1 x<x gcd(x,)=1 k e xn + O(1). (5) The second sum bove over x cn be expressed in terms of Kloostermn sums, nd using estimtes for Kloostermn sums one hs k e xn = O( 1/+ gcd(k, ) 1/ ). (6) For proof, see Lemm 4 on pge 36 of Hooley s book [5] where proof is given (his r corresponds to our, nd his l is kn. Also recll tht we re ssuming gcd(n, ) = 1 so tht N does not pper in the gcd of the O term).
12 INTEGERS: 13 (013) 1 Furthermore, using the Tylor expnsion of sin(x) one obtins the two inequlities sin(x) min(x, 1), x 0, 1/ sin(x) < /x, 0 < x < π/. (7) For the second inequlity, use 0 < x/ < x x 3 /3! < sin(x) in the stted intervl. Applying (7) nd (6) gives n upper bound for (5) of O 1/+ min( πk (y y 1 ), 1) πk gcd(k, )1/ + 1. (8) 1 k ( 1)/ Breking up the sum into 1 k /(π(y y 1 )) nd /(π(y y 1 )) < k ( 1)/, the sum over k in the O term equls (y y 1 ) gcd(k, ) 1/ + gcd(k, ) 1/ /k. π 1 k /(π(y y 1)) 1 k X d k /(π(y y 1))<k ( 1)/ (9) Both kinds of sums cn be esily hndled (the first cn lso be found in Hooley). Let X > 0. Then, gcd(k, ) 1/ d 1/ 1 X d 1/ = O(X ). (30) 1 k X d d Next, let 0 < X 1 < X. Then gcd(k, ) 1/ /k X 1<k X d d 1/ X 1 <k X d k = O log(x X 1 + ) d d 1/ 1/k (31) which equls Applying (30) nd (3) to (9), we hve tht (8) is completing the proof. O(log(X X 1 + ) ). (3) These two lemms together give the following theorem. O( 1/+ ), (33) Theorem 4.3. Let gcd(n, ) = 1 nd R s described in (15). Then, c R (N, ), the number of solutions (x, y) to xy = N mod with (x, y) lying in the rectngle R, is equl to re(r) φ() + O( 1/+ ) (34) for ny > 0.
13 INTEGERS: 13 (013) 13 This theorem shows tht the points (x, y) stisfying xy = N mod re uniformly dense in the sense tht the rectngle R contins its fir shre of solutions, so long s the re of R is of lrger size thn 3/+. For exmple, if R is squre, it needs to hve side length t lest 3/4+ to contin its fir shre of points. This is considerbly lrger thn the side length of 1/ tht is used in the lgorithm of Section 1. The ppers of Shprlinski ([8], [9]) contin mny references to the problem of uniform distribution nd discusses improved results on verge over N. 5. Second Moment nd Running Time We now exmine the ssertion mde in Section 1 tht O( 1+ ) time is needed to scn cross ll squres of side length 1/ nd their immedite neighbors, compring ll pirs of points contined in sid squres. The running time of the lgorithm in Section.1, i.e., in the cse U V < U depends on how the solutions to xy = N mod nd x y = N mod ( 1) re distributed mongst the smll squres of side length 1/. In Section 5.1 we will consider the running time of the vrint in Section.1 which is used for the generl sitution 1 < U V < N. Let S denote one such squre, i.e., of side length 1/. Then the running time needed to exmine just the squre S, looking t ll pirs of points (x, y), (x, y ) in S is O(c S (N, )c S (N, 1)), which, by the rithmetic geometric inequlity is O(c S (N, ) + c S (N, 1) ). The lgorithm lso requires us to compre points in neighboring squres, sy S 1 nd S, which, similrly, tkes O(c S1 (N, ) +c S (N, 1) ) time. Hence, the overll running time to compre pirs of points is O c S (N, ) + c S (N, 1), (35) S the sum being over the roughly squres of side length 1/ tht prtition the squre {(x, y) Z 0 x, y < } (t the top nd right edges we get rectngles, unless 1/ is n integer). Consider now the contribution from the points mod : c S (N, ). (36) S For convenience, rther thn del with squres S of side length 1/, we will estimte (36) by mking smll djustment nd prtitioning the squre into squres of side length b = 1/. (37)
14 INTEGERS: 13 (013) 14 We lso ssume tht gcd(b, ) = 1. If not, replce b with b + 1 until this condition holds. By eqution (0), this will not tke long to occur, so tht, for ny > 0, b = 1/ + O( ). Thus, consider the squres B ij = {(x, y) Z ib x < (i + 1)b, jb y < (j + 1)b} (38) with 0 i, j < /b 1. Since b, these will not entirely cover the squre, but the number of points (x, y) Z stisfying xy = N mod tht re neglected t the right most nd top portions of the squre is, by (0), O(φ()b/), nd these therefore contribute O(φ() b / ) = O(φ()) to (36). The points (x, y) Z belonging to n 1/ 1/ squre S re contined entirely in t most four squres, sy B i1j 1, B ij, B i1j 3, B i4j 4, of side length b. Therefore, c S (N, ) (c Bi1 j 1 (N, ) + c Bi j (N, ) + c Bi3 j 3 (N, ) + c Bi4 j 4 (N, )) (39) which, by the CuchySchwrtz inequlity is 4(c Bi1 j 1 (N, ) + c Bi j (N, ) + c Bi3 j 3 (N, ) + c Bi4 j 4 (N, ) ). (40) Since ech B squre overlps with O(1) S squres, we thus hve tht c S (N, ) = O φ() + c B (N, ), (41) S B the φ() ccounting for the contribution from the neglected portion t the right most nd top portions of the squre. A similr considertion for the points stisfying xy = N mod ( 1), prtitioning the lrger squre into squres D of side length d, where d is the smllest integer greter thn ( 1) 1/ which is coprime to, gives the sme kind of sum c S (N, 1) = O φ( 1) + c D (N, 1). (4) S D Therefore, we need to estimte the second moment c B (N, ) (43) B where B rnges over ll /b squres of the form (38). To prove tht the running time of the hide nd seek lgorithm of Section 1 is O(N 1/3+ ) we need to prove tht (43) is O( 1+ ).
15 INTEGERS: 13 (013) 15 Theorem 5.1. Let Then P = {B ij } 0 i,j</b 1 (44) c B (N, ) = O( 1+ ). (45) B P Proof. Rther thn look t just the squre, it is helpful to consider the b b squre {(x, y) Z 0 x, y < b}. The dvntge of looking t the lrger squre will become pprent when we turn to the discrete Fourier trnsform, nd will be summing over ll the th roots of unity. This lrger squre cn be prtitioned into b squres of side length. Becuse the solutions to xy = N mod repet mod, we cn count ech c B (N, ) once per squre, by summing c B (N, ) = c B (N, ) over ll b trnsltes B = B + (r 1, r ) of B, with 0 r 1, r < b. On the other hnd, we cn lso prtition the b b squre into squres of side length b: P = {B ij } 0 i,j 1 (46) with B ij given by (38). Ech trnslte of B squre, B = B + (r 1, r ), is covered by t most four B ij P, nd ech B ij P overlps t most four such trnsltes of B. Hence, pplying the CuchySchwrtz inequlity s before, b c B (N, ) = O c B (N, ). (47) B P B P To study c B (N, ) we multiply eqution (17) by its conjugte, giving c B (N, ) = 1 k1 e (y 1 x 1 N) k (y x N). 0 k 1,k 1 (x 1,y 1 ) B gcd(x 1,)=1 (x,y ) B gcd(x,)=1 (48) Next, sum over ll B ij P, nd brek up ech sum over (x, y) B ij into double sum ib x < (i + 1)b, jb y < (j + 1)b, B P c B (N, ) = 1 0 k 1,k 1 1 i=0 ib x 1,x <(i+1)b gcd(x 1,)=gcd(x,)=1 1 j=0 jb y 1,y <(j+1)b ) e N (k 1 x 1 k x k1 y 1 k y e. (49)
16 INTEGERS: 13 (013) 16 Now, the inner most sum, jb y 1,y <(j+1)b is product of two geometric series nd equls We understnd e(kb/) 1 e(k/) 1 1 gives k1 y 1 k y e, (50) e((k 1 k )jb/) e(k 1b/) 1 e( k b/) 1 e(k 1 /) 1 e( k /) 1. (51) to equl b if k = 0 mod. Summing (51) over 0 j e(k1b/) 1 e(k 1/) 1 if k 1 = k mod 0 otherwise (recll tht we hve chosen b so tht gcd(b, ) = 1). Therefore, only the terms with k 1 = k contribute to (49) nd it equls e N (k( x e(kb/) 1 1 x )) e(k/) 1. (5) k=0 i=0 ib x 1,x <(i+1)b gcd(x 1,)=gcd(x,)=1 The k = 0 term gives, on seprting the sum over x 1 nd x, b 1 1 i=0 ib x<(i+1)b gcd(x,)=1 (53) which, by (0) nd using b 1/, equls b (φ()b/ + O( )) = O(φ() ). (54) Next, we del with the terms 1 k 1. The sum over i in (5) equls where 1 A x1,x (t) = i=0 ib x 1,x <(i+1)b A x1,x ( Nk), (55) 0 if gcd(x 1 x, ) > 1 e(t( x 1 x )/) otherwise. To nlyze this sum, we use the two dimensionl discrete Fourier trnsform Â m1,m (t) = A x1,x (t)e m 1x 1 + m x, (56) 0 x 1,x 1
17 INTEGERS: 13 (013) 17 so tht A x1,x (t) = 1 0 m 1,m 1 nd (55) equls, on chnging order of summtion, 1 Â m1,m ( Nk) 0 m 1,m 1 m1 x 1 + m x Â m1,m (t)e, (57) 0 i 1 ib x 1,x <(i+1)b m1 x 1 + m x e. (58) The brcketed sum over i is similr to the sum over j worked out bove nd equls Therefore, (58) equls e(m1b/) 1 e(m 1/) 1 if m = m 1 mod 0 otherwise. 1 1 Â m, m ( Nk) m=0 So, (5), nd hence (49), equls k=0 m=0 Â m, m ( Nk) e(mb/) 1 e(m/) 1 e(mb/) 1 e(m/) 1. (59) e(kb/) 1 e(k/) 1. (60) But, Â m, m ( Nk) = = 0 x 1,x 1 gcd(x 1 x,)=1 0 x 1 gcd(x,)=1 e Nk ( x 1 x ) e mx 1 mx e Nk x + mx. (61) However, the sum on the righthnd side is Kloostermn sum, Nk x + mx e = S( m, Nk, ), (6) 0 x 1 gcd(x,)=1 nd re known [10] [6] to stisfy the bound S( m, Nk, ) τ() gcd(m, k, ) 1/ 1/ = O( 1/+ gcd(k, ) 1/ ) (63)
18 INTEGERS: 13 (013) 18 (recll we re ssuming tht gcd(n, ) = 1 so tht N does not pper on the righthnd side of this inequlity). Applying this bound to Âm, m( Nk), shows tht (60) is O 1 1 k=1 m=0 gcd(k, ) e(mb/) 1 e(m/) 1 e(kb/) 1 e(k/) 1 + φ(). (64) The φ() terms comes from the k = 0 contribution, (54). We must isolte this term, otherwise the estimte below will be too lrge. Seprting sums gives O 1 gcd(k, ) e(kb/) 1 1 e(mb/) 1 e(k/) 1 e(m/) 1 k=1 m=0 + φ(). (65) Both sums cn be bounded using the sme pproch s for (4) in the previous section, nmely: combining terms k nd k (similrly for the m sum, but tking the m = 0 term lone), breking up the sum into the terms with k /(πb) 1/ /π (respectively, m), pplying inequlities (7), estimting the resulting sums, using b 1/, we find, for ny > 0, tht (65) equls O(b 1+ ). (66) We hve thus estimted the sum tht ppers on the righthnd side of (47). The sum tht we wish to bound ppers on the lefthnd side of (47) but with n extr fctor of b. Hence, dividing the bove by b gives O( 1+ ) for the sum in theorem. Remrk: In certin cses, such s when = p, with p prime, one cn improve the bove estimte for the second moment to O() by tking b = p nd, for x = jp + l, with gcd(l, p) = 1, using x = l (l jp) Running Time of the Vrint for 1 < U V < N Insted of prtitioning the squre into smller squres of side length b 1/, we prtition it into rectngles R of width w < nd height h <, where w, h Z nd, for convenience, gcd(w, ) = gcd(h, ) = 1. We prtition the squre nd lso the lrger w h rectngle into smller rectngles R: R = R ij = {(x, y) Z iw x < (i + 1)w, jh y < (j + 1)h} Q = {R ij } 0 i</w 1 0 j</h 1 Q = {R ij } 0 i,j 1.
19 INTEGERS: 13 (013) 19 As in Section 4.1, we hve wh c R (N, ) = O c R (N, ). (67) R Q R Q with wh ppering on the lefthnd side since the lrge w h rectngle hs tht mny copies of the squre. Using the discrete Fourier trnsform, s before, 1 1 S( m, Nk, ) e(mw/) 1 e(m/) 1 e(kh/) 1 e(k/) 1. c R (N, ) = 1 R Q k=0 m=0 (68) This useful identity expresses the second moment for the lrger w h rectngle s sum involving Kloostermn sums. The k = 0 term cn be estimted s in (54) nd symptoticlly equls h w φ(). (69) For the k 1 terms, we use bound (63) to estimte the Kloostermn sums nd seprte the double sum bove to get contribution of 1 O gcd(k, ) e(kh/) 1 1 e(mw/) 1 e(k/) 1. (70) e(m/) 1 k=1 The first sum is estimted to equl O(τ()h) while the second sum is O(w), giving, for k 1, contribution of m=0 O( 1+ wh) (71) for ny > 0. Putting (69) nd (71) together, then dividing the lefthnd side of (67) by wh gives the following estimte for the second moment: Theorem 5.. Let 1 < w, h <, with gcd(w, ) = gcd(h, ) = 1. Then, using the nottion bove, we hve n estimte for the second moment tht depends on the re wh of the rectngles R: c R (N, ) O( 1+ ) if wh = O( 1+ ), = O(whφ() / ) if wh λ (7) for some λ > 1. R Q Remrk: if gcd(w, ) = gcd(h, ) = 1 does not hold, one cn bound the lefthnd side of (7) by compring with the sme kind of sum, but where w nd h re incremented, s before, by t most O( ) until this gcd condition holds. So long s w, h to begin with, the estimtes in the bove theorem re unffected.
20 INTEGERS: 13 (013) 0 In Section 1., our choice of w nd h hs wh = O(), nd the estimte for the second moment is thus O( 1+ ), s in the previous section. The second estimte of the theorem (not relevnt for our prticulr ppliction), O(whφ() / ), cn probbly be turned into n symptotic formul nd centrl limit theorem proven. This will remin n inquiry for the future. Acknowledgements I wish to thnk Andrew Grnville, Jeffrey Lgris, Crl Pomernce, nd Mtthew Young for helpful feedbck, nd the referee for mking number of suggested improvements to the presenttion nd description of the lgorithms. References [1] J. Beck, nd M. R. Khn, On the uniform distribution of inverses modulo n, Periodic Mthemtic Hungric 44 (00), [] F. P. Boc, C. Cobeli nd A. Zhrescu, Distribution of Lttice Points Visible from the Origin, Commun. Mth. Phys. 13 (000), [3] A. Fujii, nd Y.Kitok, On plin lttice points whose coordintes re reciprocls modulo prime, Ngoy Mth. J. 147 (1997), [4] D.R. HethBrown, Arithmetic pplictions of Kloostermn sums, Nieuw Arch. Wiskd. (5) 1 (000), no. 4, [5] C. Hooley, Applictions of sieve methods to the theory of numbers, Cmbridge University Press (1976). [6] H. Iwniec, E. Kowlski, Anlytic Number Theory, Colloquium Publictions 53, Americn Mth. Soc., Providence, RI, 004, Chpter 11. [7] R.S. Lehmn, Fctoring lrge integers, Mth. Comp. 8 (1974), [8] I. Shprlinski, Distribution of points on modulr hyperbols, Siling on the Se of Number Theory: Proc. 4th ChinJpn Seminr on Number Theory, Weihi, 006, World Scientific, 007, [9] I. Shprlinski, Modulr hyperbols, http : //rxiv.org/bs/ [10] A. Weil, On some exponentil sums, Proc. Nt. Acd. Sci. USA 34 (1948), [11] W. Zhng, On the distribution of inverses modulo n, Journl of Number theory 61 (1996),
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