On the Congruence ax + by 1 Modulo xy

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1 On the Congruence ax + by Modulo xy J. Brzezińsi, W. Holsztyńsi, and P. Kurlberg CONTENTS. Introduction. Congruences. An Upper Bound 4. An Algorithm 5. Reduced Solutions 6. On the Rate of Growth of f (n) Acnowledgments References We give bounds on the number of solutions to the Diophantine equation ( +/x)(y +/y) =n as n tends to infinity. These bounds are related to the number of solutions to congruences of the form ax + by modulo xy.. INTRODUCTION Eri Ljungstrand ased the first author about estimates of the number of solutions to the equation ( n = + )( Y + ), ( ) x y where n,, x, Y, y are positive integers satisfying n>, x >, and y >. His computations suggested that the number of such solutions, when symmetric solutions obtained by transposing (, x) and (Y,y) are identified, is always less than n. It is easy to see that y divides x +andx divides yy +. Denoting the corresponding quotients by b and a, we get the following system: where ab = n. Thus, ax = yy +, by = x +, ax (mod y) and by (mod x). ( ) It is clear that the integers x, y satisfying these congruences are relatively prime, and the system is equivalent to ax + by (mod xy). ( ) 000 AMS Subject Classification: Primary D45; Secondary A5, D7 Keywords: Diophantine equation, linear congruence, divisor function It is also clear from the equations above that x y, so when counting the solutions, we may assume x<y. It is not difficult to see that the problem of finding all solutions to Equation ( ) with <x<yis equivalent to the problem of finding all solutions to the systems of linear congruences ( ) for all a, b such that ab = n with x, y satisfying the same conditions (see Section ). One of the aims of the present paper is to prove E. Ljungstrand s observation concerning the number f(n) c A K Peters, Ltd /005 $ 0.50 per page Experimental Mathematics 4:4, page 9

2 9 Experimental Mathematics, Vol. 4 (005), No. 4 of solutions to Equation ( ). The proof is a combination of an estimate of f(n) (see Theorem.), proving the result for relatively large values of n, and a portion of numerical computations, which together prove the inequality f(n) <nfor all n. The systems of linear congruences ( ) or the congruence ( ) (for fixed a, b) seem to be interesting in their own right. In this paper, we study the sets of solutions to these congruences and give some estimates for their size from both above and below. We also give a reasonably effective algorithm for finding all solutions of ( ) in positive integers and attach some numerical results. In the last part of the paper, we further study the rate of growth of f(n) and present some numerical data.. CONGRUENCES Our objective is to estimate the number of solutions with x, y > to the congruence ax + by (mod xy) when ab = n is fixed. Theorem.. Let a, b be fixed positive integers and ab = n>. Let ρ(a, b) denote the number of pairs (x, y) of integers x, y such that xy ax + by, <x<y.then, for every n and for every real number n, ρ(a, b) < n log(n)+ (+ 0.6 ) n + (n ) n. Before we prove Theorem., we need two preparatory results. Let τ(n) denote the number of divisors to n. Lemma.. Let n 484 be a natural number and n a real number. Then, n τ(n ) < = n log(n)+ (+ 0.6 ) n. Proof: We have (see, e.g., [Hardy and Wright 79, page 47]) n τ(n ) = = n = d n n d= n = ( ) n + d d n d n n(log n +0.6) + n = 0.6 n log n + ( + ) n, where the last inequality follows by noting that ( z ) log z is decreasing and less than 0.6 when z. Lemma.. Let a, b, x, y be positive integers such that ab = n, ax (mod y), by (mod x), andx, y >. Let ax =yy, by =x, andax + by =xy. Then, (a) = n Y, (b) x = b+y and y = a+, (c) max(x, y) n, (d) n+. Proof: We have xyy =(ax )(by ) = abxy ax by+ = abxy xy. Dividing by xy, we get (a). Now ax yy = by x gives x(a + ) =y(b + Y ), so a+ y = b+y x.but ( ) ax xy = ax + by = + b y =(Y + b)y y shows that both fractions are equal to, which proves (b). We have y = a + = a + ab Y ab + ab b + Y ab + ab (ab ) =, where the last inequality follows from b + Y = x, and the first is equivalent to ab a = a(b ) ab ab Y b + Y = ab b Y b + Y = b x, that is, a(x ), when b. This is equivalent to ax a + = y, which immediately follows from ax = yy + >y. By symmetry, we get the corresponding inequality with y replaced by x, which proves (c). Since x, y and, of course, x y, we have max(x, y). Thus, (c) implies (d).

3 Brzezińsi et al.: On the Congruence ax + by Modulo xy 9 Proof of Theorem.: Let <x<ybe integers such that xy ax+by. Notice that given y there is only one x satisfying the necessary condition ax (mod y) and therefore at most one pair (x, y) such that xy ax+by. Using notation from Lemma., we have Y = ab = n < n. Observe that and Y are positive, since x>andy>. We consider contributions to the numbers of solutions in two cases. First of all, let n, where n. Then according to Lemma.(c), we get y n n (n ) = n n. Since every y gives at most one x, we have less than possibilities for (x, y) in this case. (n ) n Assume now that < n is fixed. Then, since n, we get at most τ(n ) possibilities for the choice of (x, y). But and uniquely define y and, consequently, x. Therefore, the number of possibilities for (x, y) in this case is at most Lemma. is less than n = n log(n)+ (+ 0.6 τ(n ), which according to ) n. Thus, the total number of possible (x, y) isatmost n log(n)+ (+ 0.6 ) n (n ) + n. Notice that if we fix < n and choose as a divisor of n, then x and y are uniquely determined regardless of whether x<yor x>y. In fact, and uniquely determine y, Y (from Y = n ), and, consequently, x from Lemma.(b). Thus, if we are interested in the total number of solutions to ( ) without the assumption x<y, then we have to count twice the number of solutions corresponding to n (they may correspond to x<yor x>y) plus the number of solutions corresponding to < n. Thus, we have: Theorem.. Let a, b be fixed positive integers and ab = n. Let ρ (a, b) denote the number of pairs (x, y) of integers x, y such that xy ax + by, x, y >. Then, for every integer n and every real n, ρ (a, b) < n log(n)+ (+ 0.6 ) n + (n ) n. For completeness of our discussion of the congruence ( ), we note: Proposition.4. The congruence ax + by (mod xy) has infinitely many solutions in positive integers x, y if and only if a =or b =. Proof: As we already now, there are only finitely many solutions with x, y >. Therefore, if we have infinitely many solutions, then in infinitely many of them x = or y =. If for example, x = then infinitely many y divide a, so a =. The converse is trivial.. AN UPPER BOUND In this section, we discuss the number of solutions to Equation ( ), give an estimate of it, and prove that for large values of n it is always less than n. Theorem.. (a) The solutions (, x, Y, y) to Equation ( ) with <x<yare in a one-to-one correspondence with the quadruples (x, y, a, b) such that ab = n, <x<y,andax + by (mod xy). (b) The solutions (, x, Y, y) to Equation ( ), with fixed value = n Y > 0, x, y >, Y,andx<y if = Y, are in a one-to-one correspondence with the set of the quadruples (, Y, a, b) satisfying n = ab > n = Y, gcd(a +, b + Y ), ( ) where a + >,b + Y >, Y, and a < b if = Y. Moreover, for every solution (, x, Y, y) to Equation ( ), x = b+y and y = a+. Proof: (a) As noted in the introduction, a solution (, x, Y, y) to Equation ( ) with <x<ygives the yy + congruence ax + by (mod xy), where a = x and b = x+ y, ab = n. Conversely, if (x, y) is a solution to ax + by (mod xy), where ab = n and <x<y, then we easily chec that (, x, Y, y) with = by x and Y = ax y is a solution to Equation ( ). (b) Let (, x, Y, y) be a solution to Equation ( ) with = n Y, x, y >, Y,andx<yif = Y. Then with a, b as above, we get a quadruple (, Y, a, b). According to Lemma., n = ab > n = Y, gcd(a +, b + Y ), and x = b+y, y = a+. Hence, x, y > imply a + >and b + Y >. Moreover, if = Y, then x<ygives a<b. Conversely, if (, Y, a, b) is any quadruple satisfying the conditions in (b), then we get (, x, Y, y), where x = b+y and y = a+, which is easily seen to be a solution of Equation ( ) satisfying all the conditions in (b).

4 94 Experimental Mathematics, Vol. 4 (005), No. 4 Remar.. Notice that the condition gcd(a +, b + Y ) is equivalent to gcd(a +, b + Y )=, since = ab Y =(a + )b (b + Y ) implies that gcd(a +, b + Y ). Moreover, if gcd(n, ) =, then the conditions a + and b + Y are equivalent. In fact, gcd(n, ) = implies gcd(, ) = gcd(b, ) =, so the identity above implies the equivalence of both conditions. Thus, if gcd(n, ) =, then in order to find a solution to Equation ( ), it is sufficient to find factors a of n and of n such that a + with a + >and n a + n >. Then, (, x = n a + n,y = n,y = a + is a solution. In particular, if a =, we obtain solutions for every, such that gcd(, n) =, n, +, and +>. ( ) On the other hand, if a = n, we get solutions for, such that gcd(, n) =, n, n +, and + n >. ( ) We shall use these observations frequently in Section 6. Theorem.(a) implies that in order to estimate the number of solutions to Equation ( ), we have to estimate the number f(n) = ab=n ρ(a, b) of solutions with <x<yto all the congruences ax + by (mod xy) when ab = n. It is well nown that for every ε > 0 there is a constant C ε only depending on ε such that τ(n) C ε n ε. Applying this fact and Theorem., we get a bound on f(n) depending on n,, andε. However, we can get a somewhat sharper estimate if we use only one of the congruences ax + by (mod xy) and bx + ay (mod xy), and instead, count all solutions with x, y > (that is, we disregard the condition x<y). In fact, it is clear that (x, y) solves the first congruence if and only if (y, x) solves the second one. In such a way, we can use the estimate from Theorem., but only for the pairs a, b with ab = n and a b. The number of such pairs is τ(n)+ε n, where ε n =0ifn is not a square and ε n = when n is a square. This gives the following result: Theorem.. Let f(n) denote the number of solutions to Equation ( ) and let g(n, ) = n log(n)+ (+ 0.6 ) n (n ) + n n. ) Then, for every ε>0 and any real n, there is a constant C ε such that f(n) τ(n)g(n, ) C ε n ε when n is not a square, and f(n) (τ(n)+)g(n, ) (C ε n ε +) n log(n)+ ( (n ) n n log(n)+ ( ) n (n ) n, ) n, when n is a square. In particular, if n is sufficiently big then f(n) <n. 4. AN ALGORITHM We can now construct a reasonably efficient algorithm for computing the number of solutions (, x, Y, y) to Equation ( ) following their description in Theorem.(b). First of all, write down the list of divisors of n. For each divisor a of n and for all integers such that < n, repeat the following: compute all the divisors of a + ; foreach, chec whether or not Y = n and x = b+y, where b = n a+ a, are integers, let y =, in the affirmative case. If = Y and x>y replace (x, y) by(y, x). Chec whether x>, y>and accept the quadruple (, x, Y, y) as a solution if all these conditions are satisfied. Theorem.(b) easily implies that this algorithm gives all the solutions to Equation ( ) and every solution exactly once. We are now ready for the numerical computations proving that the number f(n) of solutions to Equation ( ) is always less than n. As we noted before, for each ε>0 there is a constant C ε depending only on ε such that τ(n) C ε n ε for all n. For simplicity, let ε = 4 and denote by C the least constant corresponding to this value of ε. Itiseasy to show that for the positive integers the quotient x = b+y C(n) = τ(n) n 4 attains its maximum value for n =, 6, 600, which gives C <C 0 =

5 Brzezińsi et al.: On the Congruence ax + by Modulo xy 95 According to Theorem., if n is not a square, we want to decide when Let f(n) τ(n)g(n, ) C n 4 log(n) ( <n. ) C n 4 + (n ) n C n 4 h(n,, C) =n 4 C log(n) ( ) (n ) C n n C. Choose =.95. Then, it is easy to chec that h(n,, C ) > h(n,, C 0 ) > 0 when n, 6, 000. By the definition of C, this shows that f(n) <nfor all n, 6, 000, and it remains to chec this inequality for all n<, 6, 000. In order to carry out the numerical computation, we find all the numbers n for which τ(n)g(n, ) n. This happens when τ(n) is big, which occurs when n has many small prime factors. The computations give 6,5 numbers in the interval [ 0 4,, 6, 000],,00 in [ 0 4, 0 5 ],,48 in [0 5, ], and in [5 0 6,, 6, 000]. The numbers in the last interval are 5,045,040 (4,559), 5,66,800 (4,05), 5,405,400 (5,069), 5,569,00 (4,494), 5,654,880 (4,54), 5,765,760 (5,86), 6,6,0 (5,), 6,0,60 (5,407), 6,486,480 (4,), 7,07,00 (6,09), 8,648,640 (5,0), where the number in the parenthesis is the corresponding value of f(n). If n is a square, then we repeat the same procedure as above taing into account the extra term on the righthand side in the second inequality in Theorem.. The bound,6,000 wors in this case as well, so we have to consider all squares less than this bound (,408 numbers). Short computations show that there are 8 such squares for which the expression in the second inequality in Theorem. is not less than n (the biggest one is,587,600). For these 8 numbers, we chec by computer calculations that f(n) < n. 5. REDUCED SOLUTIONS In this section, we loo at some special solutions to Equation ( ), and we also give a direct proof of the inequality f(n) <nfor the case when n = p is a prime number. Let, x, Y, y be a solution to Equation ( ), which in this section will be denoted by n =[, x, Y, y]. Recall that a, b denote integers such that ax = yy + and by = x+. We say that a solution, x, Y, y is reduced if < y and Y < x. The reduced solutions are characterized in the following way: Proposition 5.. Let n =[, x, Y, y]. Then, x, Y, y is reduced if and only if Y = n. Proof: If < y and Y < x, then Lemma. gives xy = ax+by = x+yy+ <x(y )+y(x )+ = xy+ x y < xy. Thus = n Y =. Conversely, if Y = n, then by Lemma., = n Y =. This implies <yand Y < x, since otherwise, xy = x + yy +>xy,thatis,>. Corollary 5.. The number of reduced solutions to Equation ( ) is τ(n)τ(n ). Proof: If, y, Y, y is a reduced solution, then ab = n, Y = n, and = according to Lemma.. Thus, each pair of divisors of n and n defines a solution and every solution gives such a pair of divisors. Of course, we have to divide the total number of such pairs by in order to obtain each desymmetrized solution exactly once. Proposition 5.. If p is a prime, then f(p) <p. Proof: According to Corollary 5., the number of reduced solutions to p =[, x, Y, y] equals τ(p ). Assume that the solution, x, Y, y is not reduced. Without loss of generality, we may assume that x +=py and yy +=x. The second equation gives Y < yy + = x. Since the solution is not reduced, we have >y(the equality is, of course, impossible by the first equation). The second equation gives y x, so y<x. We also have x<p, since otherwise py = x + >pgives a contradiction. Thus x belongs to the set {,...,p } with p elements. Moreover, py (mod x) andy<x, so the congruence allows at most one y that gives a solution to the equation. If now p (mod x), then y must be equal to, which is impossible. Thus x> can not assume values that divide p. The number of such x is τ(p ). Thus, x assumes at most (p ) (τ(p ) ) = p τ(p ) different values that give nonreduced solutions. According to Corollary 5., the number of reduced solutions

6 96 Experimental Mathematics, Vol. 4 (005), No. 4 is τ(p ), so the total number of solutions is at most p. Every solution, x, Y, y to Equation ( ) has the corresponding value of = n Y. By Proposition 5., = corresponds to the reduced solutions. For these solutions, and Y must be the least positive solutions to the congruences x (mod y) andyy (mod x) when x, y are fixed. All other positive solutions to these congruences with x, y fixed are given by + ry, Y + sx, where r, s 0. Thus, starting from n = [, x, Y, y] with a fixed pair x, y, weget N =[ + ry,x,y + sx, y], where N =(rx + b)(sy + a). The number n = ab is the least number for which such a (reduced) solution with fixed x, y exists. We have N ( + ry)(y + sx) = + r + s. In particular, if r =,s =0orr =0,s =,we get quadruples for which the corresponding parameter decreases by : [, x, Y, y] [ + y, x, Y, y], [, x, Y, y] [, x, Y + x, y]. (5 ) We shall say that these two transformations are elementary. Thus we can describe the solutions for a given n in the following way: Proposition 5.4. Every solution to n =[, x, Y, y] with = n Y > can be obtained from a reduced solution to m =[ 0,x,Y 0,y], forsomem<n, by successive use of elementary transformations (5 ). Proof: If we have a solution n =[, x, Y, y] with = n Y and >, then the solution is not reduced, which means that > y or Y > x, since Lemma. implies immediately that the equalities are impossible. If >y, then we get n (yy +)=[ y, x, Y, y], while Y>xgives n (x +) = [, x, Y x, y], both with the corresponding value of =[n (x + )] (Y x) =. This reduction process eventually leads to a reduced solution for a natural m<nand the same x, y. Starting from such a reduced solution and reversing the process, we get the given solution n =[, x, Y, y] after steps. By Lemma.(d), n+. Observe, that for t and n =t, we have n =[,, t, ] and in this case, = n Y = n+. 6. ON THE RATE OF GROWTH OF f(n) What is the true rate of growth of f(n)? As Figure shows, f(n) oscillates rather wildly FIGURE. The numbers of solutions to ( ) for 0, 000 n 0, 500. There are a number of natural questions regarding the behavior of f(n). What is the correct upper bound for f(n)? Does f(n) tend to infinity as n tends to infinity? If so, how fast? As Figure indicates, not only are the large values of f quite far from the mean value of f, but so are the small values e+006 "f(n)" "max" "mean" "min" FIGURE. The x-coordinate for each plotted point corresponds to a window of the form [00l +, 00l + 00], and the y-coordinate corresponds to the maximum (respectively, minimum) value taen by f(n) for n in that window. 6. The Average Rate of Growth of f(n) Figure indicates that the average of f(n) behaves quite regularly, and we can in fact show that f(n) on average is of size log n. For simplicity, if g, h are positive functions,

7 Brzezińsi et al.: On the Congruence ax + by Modulo xy 97 we write g(n) h(n) if there is a positive constant C such that g(n) Ch(n) for all sufficiently big natural n. Theorem 6.. There exist positive constants C,C such that for T, T n= C f(n) T log T C. In the proof we need the following result: Lemma 6.. n T τ(n)τ(n ) = O(T log T ). Proof: If m T,wehave τ(m), l m l T and thus τ(n)τ(n ) 4 n T n T =4,l T =4,l T =4 4,l T,l T l n l T (n ) T {n T : l n, (n )} {d T/l : (dl )} {d T/l : d l ( ) T l + mod } = O(T log T )+O(T )=O(T log T ). Proof of Theorem 6.: Using the notation in the introduction, given relatively prime x, y, let us choose 0 and Y 0 such that x 0 (mod y), 0 < 0 <y, yy 0 (mod x), and 0 <Y 0 <x. We want to count the number of integers, Y such that 0 (mod y), Y Y 0 (mod x), and ( + )( Y + ) = n T, x y when x, y > are fixed. Noting that (x + )(Yy+) Y < < 4Y, xy we will obtain lower bounds by estimating from below the number of, Y such that 4Y T. The congruences 0 (mod y), Y Y 0 (mod x) are equivalent to, Y being of the form = 0 + ry, Y = Y 0 + sx for r, s nonnegative integers. Thus, it is enough to estimate {r, s 0:( 0 + ry)(y 0 + sx) T/4}, which, since 0 <yand Y 0 <x, we may bound from below by {r, s 0:(r + )(s +)xy T/4}. This, in turn, is greater than { r, s 0:rs T 6xy } T 6xy log T 6xy. Summing over relatively prime x, y T /, we then find that the number of ways of finding x, y,, Y such that x,y T / (x,y)= n = is strictly greater than T 6xy log (x + )(Yy+) xy T 6xy log T T log T T T log T. x,y T / (x,y)= x,y T / (x,y)= T 6xy xy In other words, on average, there are at least C log T solutions for some C > 0. In order to prove the existence of an upper bound, we note first that if r = s = 0, then the solution ( 0,Y 0,x,y) is reduced. For n T the number of reduced solutions, according to Corollary 5. and Lemma 6., is τ(n)τ(n ) = O(T log T ). n T Assume now that r ands = 0. Then, the number of solutions (, Y, x, y) to Equation ( ) such that n T is less than {r, x, y > 0:Y =( 0 + ry)y 0 T } {r, x, y > 0:ryY 0 T }.

8 98 Experimental Mathematics, Vol. 4 (005), No. 4 Since r, we have M := yy 0 T, and since yy 0 (mod x), we have {r, x, y > 0:ryY 0 T } {r, x, y > 0:x M +,y M, rm T } M T M T τ(m)τ(m +)T/M, which, by partial summation and Lemma 6., is O(T log T ). The case r =0,s > 0 follows in a similar way to the previous one. Finally, if r, s > 0 and (, Y, x, y) is a solution to ( ) such that n T, then since Y < T,weget {r, s, x, y > 0:Y =( 0 + ry)(y 0 + sx) T } {r, s, x, y > 0:rysx T } = O(T log T ). In other words, on average, there are at most C log T solutions for some C > 0. Remar 6.. With a more careful analysis of the cases r =0,s>0andr>0,s= P 0, it is possible to prove that n T lim f(n) T exists and equals T log T π. 6. Large Values of f(n) Since there are τ(n)τ(n ) reduced solutions (see Corollary 5.), it is clear that the order of magnitude of f(n) is sometimes larger than any power of log n, and thus f(n) can deviate quite a bit from its average value. However, there are other sources of large oscillations. Let M(n, ) denote the number of solutions, x, Y, y to Equation ( ) such that n Y =. Then, f(n) = (n+)/ = M(n, ), (6 ) since (n +)/ by Lemma.(d). Taing into account the contribution to f(n) from the number of solutions with = and a similar contribution for = (see Lemma 6.7), one might expect that the most significant fluctuations of f(n) depend on M(n, ) for small values of. However, this is not the case as shown by the following construction (we than Andrew Granville for pointing this out to us). Fix an arbitrary and let M>be a large integer. Choose n = + p i, where all p i are primes such that p i (mod ) andp i M. Denote the number of such primes p i by π(m,, ). By the prime number theorem for arithmetic progressions (see [Davenport 00, Chapters 0 and ]), c M φ() log M π(m,, ) C M φ() log M for suitable positive constants c,c only depending on. Now, half of the divisors of n are congruent to modulo, so taing into account ( ), we get f(n) π(m,, ). Hence log f(n) π(m,, ) M φ()logm. On the other hand, since p i M π(m,, ),weget and similarly, log n log log n. Thus Hence log f(n) log n log p i M φ(), M φ(), which implies log M M φ() log M log n log M log n log log n. ( ) c log n f(n) exp log log n for some constant c>0 only depending on. Thus for each there exists a suitable n such that M(n, ) gives a big contribution to f(n). It is also possible to show that the contribution to f(n) may come from many different values of. If n + has many different divisors, then according to ( ), where we choose =, each such divisor gives a solution to Equation ( ). However, according to the following heuristic reasoning it is reasonable to expect that M(n, ) will be zero for most. We note that each solution to ax+by =+xy, ab = n, for fixed, corresponds to divisors a n and n such that a + 0 mod. Since there are τ(n) possible values for a and τ(n ) possible values for, it is natural to expect that M(n, ) should be of size τ(n)τ(n )/. In particular, for n ε and ε>0, M(n, ) should rarely be nonzero. Thus it seems reasonable to mae the following conjecture: Conjecture 6.4. For all ε>0, f(n) ε n ε. 6. Small Values of f(n) Given the heuristic that M(n, ) should be of size τ(n)τ(n )/, one would expect that f(n) tends to be small when n is a prime number, and Figure seems to confirm this.

9 Brzezińsi et al.: On the Congruence ax + by Modulo xy "min(f(n))" "min(f(p))-min(f(n))" "min(g(n))" e+006 FIGURE. The x-coordinate for each plotted point corresponds to a window of the form [00l +, 00l + 00], and the y-coordinate corresponds to the minimum value taen by f(n) forn integer. In most of the windows, f assumes its minimum at a prime. In each window, the difference of the minimum at the primes and the minimum is also graphed e+006 FIGURE 5. The x-coordinate for each plotted point corresponds to a window of the form [000l +, 000l + 000], and the y-coordinate corresponds to the minimum value taen by g(n) =f(n)/(τ(n)log n) in that window "f(n) (smoothed)" 00 "f(p)" e+008 e+008 e+008 4e+008 5e+008 6e+008 7e+008 8e+008 9e+008 e+009 FIGURE 4. f(p) for p the first prime in the interval [0007l, 0007(l +)]. Thus, in order to try to understand small values of f(n), we will concentrate on f(n) forn taing values in a sparse subset of the primes. (Note that our algorithm for calculating f(n) is quite a bit faster when τ(n) is small.) From the data in Figure 4, we are led to mae the following conjecture: Conjecture 6.5. f(n) tends to infinity when n. In fact, as Figure 5 indicates, it might be true that f(n) τ(n)(log n) o(), but the evidence for this is perhaps not so convincing FIGURE 6. The function f(n) in[ , ] smoothed by removing the 0 smallest and 0 largest values of f(n) forn in the windows [00(l ) +, 00l], where l =5 0 +,..., Following a suggestion by the referee, we tried to investigate the normal order of f(n) (in the sense of [Hardy and Wright 79, page 56]), and doing this we looed at several windows of length 00 (for small ) plotting the middle 80 examples in each window for n up to 0 6. Unfortunately, the oscillation of the values in the windows is very strong (see Figure 6). We also looed at the quotient max(f(n))/ min(f(n)), where the maximum and minimum are taen among the 80 middle values in windows of length 00, for n up to 0 6. This also shows considerable oscillation (see Figure 7). Our conclusion is that f(n) does not have a (reasonably simple) normal order and the same can be said about f(n)/τ(n) Numerics seem to indicate that log f(n) has normal order log log n

10 400 Experimental Mathematics, Vol. 4 (005), No "max divided by min (smoothed)" be odd. Thus all the possibilities for the sums a + and b + Y are given by all the choices of the odd factors of n and n. Only one such choice gives a + =or b + Y =. This proves the second case. Now we prove that if n>, then M(n, ) + M(n, ) 7. (6 ) First let n be odd. Then, M(n, ) + M(n, ) = τ(n)(τ(n ) + τ(n )) FIGURE 7. The smoothed quotient max(f(n)) in each window [00(l ) +, 00l] forl min(f(n)) =,...,0000. (up to a constant), but the evidence is wea and it is not computationally feasible to study this for sufficiently large n. As for rigorous lower bounds, with some effort, we can prove the following: Proposition 6.6. If n 9, then f(n) 8. Let τ odd (n) denote the number of odd divisors of n. Then for =, we have the following lemma: Lemma 6.7. For n, we have { M(n, ) = τ(n) τ(n ), if n is odd, τ odd (n) τ odd (n ), if n is even. Proof: In fact, if n is odd, then according to Remar., we get all solutions to ( ) taing any divisor a of n (b = n n a ) and any divisor of n (Y = ) such that a + > andb + Y >. The number of pairs of such divisors giving different quadruples (, x, Y, y) with x<yis τ(n) τ(n ), and the only case when a+ = or b + Y = corresponds to the choice of a = =or a = n, = n, which gives only one quadruple with x<y. This proves the first case. If n is even, let n = r m, where m is odd. One of the numbers n, n must be divisible by 4, so let us assume that r (the case with n divisible by 4 is considered in similar way with the roles of n, n interchanged). Thus, n = ( r m ), and n is odd. If n =Y, then exactly one of the factors, Y is even and the other one is odd. Since a + and b + Y are even, exactly one of the factors a, b of n = ab must Since n > 4 is even, τ(n ) 4. Assume that τ(n) =. Then n is a prime. If also τ(n ) =, then 6 n. Since n > 6, we have τ(n ) 6, so M(n, ) + M(n, ) 7. Assume now that τ(n ) =, that is, n = p, where p > is a prime. Then p +=n, which is impossible. Thus, τ(n ) 4, which gives M(n, ) + M(n, ) 7. Notice that if n is a prime, n twice a prime, and n is a product of two different primes, then M(n, ) + M(n, ) = 7. By Schinzel s conjecture (see [Schinzel and Sierpińsi 58]), this situation happens for infinitely many n. Ifτ(n) >, then it is easy to chec that M(n, ) + M(n, ) 8. Assume now that n is even, so M(n, )+M(n, ) = τ(n)τ(n )+τ odd(n)τ odd (n ). We have τ(n) >, since n > 4. Assume τ(n) = 4. Since n > 8, we have n = p, where p is an odd prime. If n is a prime, then n =(p ), so τ odd (n)τ odd (n ) 4andM(n, ) + M(n, ) 7. If τ(n) = 5, then n = 6, and the claim follows by a direct computation. If τ(n) = 6, then n =orn = p q for two different primes p, q. Ifp =, then n = (q ) has at least two odd factors, so M(n, ) + M(n, ) 7. If q =andp =, we chec the claim directly, and when p>, then n =(p ) is divisible by, so τ odd (n)τ odd (n ) 6. If finally, τ(n) 7, then of course, the inequality holds. Now we prove that if n>, then M(n, ). (6 ) Assume first that n (so n ) and let n be even. Then n =m and n =m. If for a prime p (mod ), p n, then p 7, so = p, Y = n p, x = +Y >, and y = m+ > (see Remar.) give a solution to Equation. If for a prime p (mod ), p n, then p 5, so =,Y = n p, x = p+y >,

11 Brzezińsi et al.: On the Congruence ax + by Modulo xy 40 and y = n+ > give such a solution. If n is odd, then n is even, and we repeat the same arguments looing instead at the prime factors p of n. Now let n. Letn be even. Then n = s m, where m, andn = ( s m ) = r. Ifr has a prime divisor p or (mod ), we proceed exactly as in the previous case when n. Otherwise, m is a power of, so n = s+ and n = ( s + ). In this case, n must have a prime factor p> congruent to modulo and we get a solution to Equation ( ) as before. If n is odd, then n is even and divisible by, so the considerations are similar with the role of n and n interchanged. Now the proof of Proposition 6.6 follows immediately from (6 ), (6 ), (6 ), and by direct inspection of the cases n = 9, 0,,. Still more elaborate arguments show that f(n) if n 0 (we than Jerzy Browin for sending us his proof of this result and, in particular, for the proof of Lemma 6.7). Remar 6.8. It is no longer true that M(n, 4) for all sufficiently large n. If all primes dividing both n and n 4 are congruent to modulo 4, then by Remar., there are no solutions to Equation ( ) with =4. In fact, this happens for infinitely many n by the following argument, for which we than Mariusz Sa lba. Let m be a natural number such that m (mod ) and put n =m +m +5. Then, n =(m ) +(m +) and n 4=m +(m +) are only divisible by primes congruent to modulo 4. ACKNOWLEDGMENTS We than Jerzy Browin for many valuable suggestions and helpful comments on an earlier version of this paper. We also express our thans to the referee for many interesting comments and suggestions on how to improve the experimental part of this paper. The third author is partially supported by the Royal Swedish Academy of Sciences and the Swedish Research Council. REFERENCES [Davenport 00] H. Davenport. Multiplicative Number Theory, Third edition. Berlin: Springer Verlag, 000. [Hardy and Wright 79] G. H. Hardy and E. M. Wright. An Introduction to the Theory of Numbers, Fifth edition. Oxford, UK: Oxford Science Publications, 979. [Schinzel and Sierpińsi 58] A. Schinzel and W. Sierpińsi. Sur certaines hypothèses concernant les nombres premiers. Acta Arithmetica 4 (958), J. Brzezińsi, Department of Mathematics, Chalmers University of Technology and Göteborg University, S 496 Göteborg, Sweden (jub@math.chalmers.se) W. Holsztyńsi, 400 E. Remington Dr, #D Sunnyvale, CA (wlod@earthlin.net) P. Kurlberg, Department of Mathematics, Royal Institute of Technology, S 0044 Stocholm, Sweden (urlberg@math.th.se) Received December, 00; accepted in revised form June, 005.