# Concentration of points on two and three dimensional modular hyperbolas and applications

Save this PDF as:

Size: px
Start display at page:

Download "Concentration of points on two and three dimensional modular hyperbolas and applications"

## Transcription

1 Concentration of points on two and three dimensional modular hyperbolas and applications J. Cilleruelo Instituto de Ciencias Matemáticas (CSIC-UAM-UC3M-UCM) and Departamento de Matemáticas Universidad Autónoma de Madrid Madrid-28049, Spain M. Z. Garaev Instituto de Matemáticas Universidad Nacional Autónoma de México C.P , Morelia, Michoacán, México Abstract Let p be a large prime number, K, L, M, λ be integers with 1 M p and gcd(λ, p) = 1. The aim of our paper is to obtain sharp upper bound estimates for the number I 2 (M; K, L) of solutions of the congruence xy λ (mod p), K + 1 x K + M, L + 1 y L + M and for the number I 3 (M; L) of solutions of the congruence xyz λ (mod p), L + 1 x, y, z L + M. Using the idea of Heath-Brown from [6], we obtain a bound for I 2 (M; K, L), which improves several recent results of Chan and Shparlinski [3]. For instance, we prove that if M < p 1/4, then I 2 (M; K, L) M o(1). The problem with I 3 (M; L) is more difficult and requires a different approach. Here, we connect this problem with the Pell diophantine equation and prove that for M < p 1/8 one has I 3 (M; L) M o(1). Our results have applications to some other problems as well. For instance, it follows that if I 1, I 2, I 3 are intervals in F p of length I i < p 1/8, then I 1 I 2 I 3 = ( I 1 I 2 I 3 ) 1 o(1). MSC Classification: 11A07, 11B75 1

2 1 Introduction In what follows, p denotes a large prime number, K, L, M, λ are integers with 1 M p and gcd(λ, p) = 1. By x, y, z we denote variables that take integer values. The notation o(1) in the exponent on M means a function that tends to 0 as M. Let I 2 (M; K, L) be the number of solutions of the congruence xy λ (mod p), K + 1 x K + M, L + 1 y L + M and let I 3 (M; L) be the number of solutions of the congruence xyz λ (mod p), L + 1 x, y, z L + M. Estimates of incomplete Kloosterman sums implies that I 2 (M; K, L) = M 2 In particular, if M/(p 3/4 (log p) 2 ) as p, one gets that p + O(p1/2 (log p) 2 ). (1) I 2 (M; K, L) = (1 + o(1)) M 2 p. This asymptotic formula also holds when M/p 3/4 as p (see [5]). The problem of upper bound estimates of I 2 (M; K, L) for smaller values of M has been a subject of the work of Chan and Shparlinski [3]. Using Bourgain s sum-product estimate [1], they have shown that there exists an effectively computable constant η > 0 such that for any positive integer M < p, uniformly over arbitrary integers K and L, the following bound holds: I 2 (M; K, L) M 2 p + M 1 η. In the present paper we obtain the following upper bound estimates for I 2 (M; K, L). Theorem 1. Uniformly over arbitrary integers K and L, we have When K = L, we have I 2 (M; K, L) < M 4/3+o(1) I 2 (M; L, L) < M 3/2+o(1) In other words, for any fixed ε > 0 one has I 2 (M; K, L) M 4/3+ε p 1/3 + M o(1). (2) p 1/2 + M o(1). (3) p 1/3 + M ε, I 2 (M; L, L) M 3/2+ε p 1/2 + M ε, where the implied constant in Vinogradov s symbol may depend only on ε. In particular, from Theorem 1 it follows that if M < p 1/4 then I 2 (M; K, L) < M o(1). Theorem 1 together with (1) easily implies the following consequence, which improves upon the mentioned result of Chan and Shparlinski. 2

3 Corollary 1. Uniformly over arbitrary integers K and L, we have If K = L, then I 2 (M; K, L) M 2 I 2 (M; L, L) M 2 p + M 4/5+o(1). p + M 3/4+o(1). The proof of Theorem 1 is based on an idea of Heath-Brown [6]. The problem with I 3 (M; L) is more difficult and requires a different approach. Here, we shall connect this problem with the Pell diophantine equation and establish the following statement. Theorem 2. Let M p 1/8. Then, uniformly over arbitrary integer L, we have I 3 (M; L) M o(1). (4) From Theorem 2 we can easily derive a sharp bound for the cardinality of product of three small intervals in F p. Corollary 2. Let I 1, I 2, I 3 be intervals in F p of length I i < p 1/8. Then I 1 I 2 I 3 = ( I 1 I 2 I 3 ) 1 o(1). We remark that Corollary 2 is equivalent to saying that for any fixed ε > 0 there exists c = c(ε) > 0 such that I 1 I 2 I 3 > c( I 1 I 2 I 3 ) 1 ε. Theorems 1 and 2 have also applications to the problem on concentration points on exponential curves as well. Let g 2 be an integer of multiplicative order t, and let M < t. Denote by J a (M; K, L) the number of solutions of the congruence y ag x (mod p); x [K + 1, K + M], y [L + 1, L + M]. Chan and Shparlinski [3] used a sum product estimate of Bourgain and Garaev [2] to prove that J a (M; K, L) < max{m 10/11+o(1), M 9/8+o(1) p 1/8 } as M. From our Theorem 1 we shall derive the following improvement on this result. Corollary 3. Let M < t. Uniformly over arbitrary integers K and L, we have J a (M; K, L) < (1 + M 3/4 p 1/4 )M 1/2+o(1). In particular, if M p 1/3, then we have J a (M; K, L) < M 1/2+o(1). Theorem 2 allows us to strengthen Corollary 3 when M p 3/20. Corollary 4. The following bound holds: J a (M; K, L) < (1 + Mp 1/8 )M 1/3+o(1). In particular, if M p 1/8, then we have J a (M; K, L) < M 1/3+o(1). 3

4 2 Proof of Theorem 1 We will need the following lemma which is a simple version of a more precise result about divisors in short intervals, see, for example, [4]. Lemma 1. For every positive integer n and every integer m n, the interval [m, m+n 1/6 ] contains at most two divisors of n. Proof. Suppose that d 1, d 2, d 3 [m, m+l] are three divisors of n. We claim that the number r = d 1 d 2 d 3 (d 1, d 2 )(d 1, d 3 )(d 2, d 3 ) is also a divisor of n. To see this, for a given prime q, let α 1, α 2, α 3, α such that q α i d i, i = 1, 2, 3 and q α n. Assume that α 1 α 2 α 3 α. The exponent of q in the rational number r is α 1 + α 2 + α 3 (min(α 1, α 2 ) + min(α 1, α 3 ) + min(α 2, α 3 )) = α 3 α 1. Since 0 α 3 α 1 α we have that r is an integer divisor of n. On the other hand, since (d i, d j ) d i d j L we have and the result follows. n r > m3 L 3 n3/2 L 3, Now we proceed to prove Theorem 1. Our approach is based on Heath-Brown s idea from [6]. We can assume that M is a sufficiently large integer. The congruence xy λ (mod p), K + 1 x K + M, L + 1 y L + M is equivalent to xy + Lx + Ky b (mod p), 1 x, y M, (5) where b = λ KL. From the pigeon-hole principle it follows that for any positive integer T < p there exists a positive integer t T 2 and integers u 0, v 0 such that From (5) we get that tl u 0 (mod p), tk v 0 (mod p), u 0 p/t, v 0 p/t. txy + u 0 x + v 0 y b 0 (mod p), 1 x, y M, for some b 0 < p/2. We write this congruence as an equation txy + u 0 x + v 0 y = b 0 + zp, 1 x, y M, z Z. (6) Comparing the minimum and maximum value of the left hand side we can see that z txy + u 0x + v 0 y b 0 < T 2 M 2 + 2M p p T We observe that for each given z the equation (6) is equivalent to the equation (tx + u 0 )(ty + v 0 ) = n z, 1 x, y M (7) for certain integer n z. If n z = 0, then either tx + u 0 = 0 or ty + v 0 = 0. Since λ 0 (mod p), in either case x and y are both determined uniquely. So, we need only consider those z for which n z 0. 4

5 Case M < p 1/4 /4. In this case we take T = 8M. Then z < 1 and we have to consider only the integer n z = n 0 in (7). Each solution of (7) produces two divisors of n 0, tx + u 0 and ty + v 0, one of them is greater than or equal to n 0. If n M 18 the number of solutions of (7) is bounded by the number of divisors of n 0, which is M o(1). If n 0 > 2 36 M 18 the positive integers tx + u 0 and ty + v 0 lie in two intervals I 1 and I 2 of length T 2 M 2 6 M 3 < n 0 1/6. If there were five solutions, we would have three divisors greater of equal to n 0 in an interval of length n 0 1/6. We apply Lemma 1 to conclude that there are at most four solutions. Hence, in this case we have I 2 (M; K, L) < M o(1). Case M p 1/4 /4. In this case we take T (p/m) 1/3. Thus z M 4/3 /p 1/3. For each z the number of solutions of (7) is bounded by the number of divisors of n z which is p o(1) = M o(1). Hence, in this case we get Thus, we have proved that I 2 (M; K, L) < M 4/3+o(1) p 1/3. I 2 (M; K, L) < M 4/3+o(1) p 1/3 + M o(1) which proves the first part of Theorem 1. The proof of the second part of Theorem 1 (corresponding to the case K = L) is similar, with the only difference that we simply take t T (instead t T 2 ) satisfying 3 An auxiliary statement tk u 0 (mod p), u 0 p/t. To prove Theorem 2 we need the following auxiliary statement. Proposition 1. Let A, B, C, D, E, F M O(1) and assume that = B 2 4AC is not a perfect square (in particular, 0). Then the diophantine equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 (8) has at most M o(1) solutions in integers x, y with 1 x, y M O(1). The proof of Proposition 1 given here can be shortened by a direct appeal to Lemma 3.5 of Vaughan and Wooley [8]. Their proof uses results from Chapter 11 of Hua [7] and is somewhat sketchy. Here we give a self-contained proof of Proposition 1. The following lemma is well-known from the classical theory of Pell s equation. Lemma 2. Let A be a positive integer that is not a perfect square and let (x 0, y 0 ) be a solution of the equation x 2 Ay 2 = 1 in positive integers with the smallest value of x 0. Then for any other integer solution (x, y) there exists a non-negative integer n such that x + A y = (x 0 + Ay 0 ) n. 5

6 Lemma 3. Let A be a squarefree integer, N is a positive integer. Then the congruence z 2 A (mod N), 0 z N 1, has at most N o(1) solutions. Proof. Let J(N) be the number of solutions of the congruence in question and let N = p α 1 1 p α k k be a canonical factorization of N. Clearly, J(N) = J(p α 1 1 ) J(p α k k ), where J(pα ) is the number of solutions of the congruence z 2 A (mod p α ), 0 z p α 1. Since A is squarefree, we have J(2 α ) 4 and J(p α ) 2 for odd primes p. The result follows. Lemma 4. Let A, E be integers with A, E < M O(1) such that A is not a perfect square. Then the equation x 2 Ay 2 = E, 1 x, y < M O(1) has at most M o(1) solutions. Proof. (1) We can assume that A is also a squarefree number. Indeed, let A = A 1 B 2 1, where A 1, B 1 are nonzero integers, A 1 is squarefree and is not a perfect square. Then our equation takes the form x 2 A 1 (B 1 y) 2 = E, 1 x, y < M O(1). Since B 1 y < M O(1), it follows that indeed we can assume that A is squarefree. (2) We can assume that in our equation gcd(x, y) = 1. Indeed, if d = gcd(x, y), then d 2 E. In particular, since E has M o(1) divisors, we have M o(1) possible values for d. Besides, (x/d) 2 A(y/d) 2 = E/d 2, where we have now gcd(x/d, y/d) = 1. Thus, without loss of generality, we can assume that gcd(x, y) = 1. In particular, it follows that gcd(y, E) = 1. (3) Since A is not a perfect square, we have, in particular, that E 0. (4) For any x, y Z + with (y, E) = 1 there exists 1 z E such that x zy (mod E). Given 1 z E, let K z be the set of all pairs (x, y) with x 2 Ay 2 = E, 1 x, y < M O(1), (x, y) = 1 such that x zy (mod E). If (x, y) K z, then (zy) 2 Ay 2 0 (mod E). Since (y, E) = 1, it follows that z 2 A (mod E). Due to Lemma 3, the number of solutions of this congruence is at most E o(1) = M o(1). Thus, we have at most M o(1) possible values for z. Therefore, it suffices to show that K z = M o(1) for any such z. Let x 0 be the smallest positive integer such that x 2 0 Ay 2 0 = E, (x 0, y 0 ) K z. Let (x, y) be any other solution from K z. Then, From this we derive that x 2 0 Ay 2 0 = E, x 2 Ay 2 = E. (x 0 x Ayy 0 ) 2 A(xy 0 x 0 y) 2 = (x 2 0 Ay 2 0)(x 2 Ay 2 ) = E 2. (9) On the other hand, from (x 0, y 0 ), (x, y) K z it follows that x 0 zy 0 (mod E), x zy (mod E) 6

7 Since z 2 A (mod E), we get xx 0 z 2 yy 0 (mod E) Ayy 0 (mod E). We also have x 0 y xy 0 (mod E), as both hand sides are zyy 0 (mod E). Therefore, x 0 x Ay 0 y 0 (mod E), xy 0 x 0 y (mod E). (10) From (9) and (10) we get that ( ) 2 ( ) 2 x0 x Ay 0 y xy0 x 0 y A = 1 E E and the numbers inside of parenthesis are integers. Now there are two cases to consider: (1) A > 0. In view of Lemma 2, x 0 x Ay 0 y E + A xy 0 x 0 y E = (u 0 + A v 0 ) n, where (u 0, v 0 ) is the smallest solution to X 2 AY 2 = 1 in positive integers, and n is some non-negative integer. Since the left hand side is of the order of magnitude M O(1), we have that n log M = M o(1). Thus, there are M o(1) possible values for n and, each given n produces at most 4 pairs (x, y). This proves the statement in the first case. (2) A < 0. Then we get that x 0 x Ay 0 y E { 1, 0, 1}, xy 0 x 0 y E { 1, 0, 1}, and the result follows. The proof of Proposition 1. Now we can deduce Proposition 1 from Lemma 4. Multiplying (8) by 4A, we get (2Ax + By + D) 2 y 2 + (4EA 2BD)y + 4AF D 2 = 0, where = B 2 4AC. Multiplying by we get, ( y + BD 2EA) 2 (2Ax + By + D) 2 = T, where T = (BD 2EA) 2 + (4AF D 2 ). Now, since is not a full square, and since T, M O(1), we have, by Lemma 4 and the condition A, B, C, D, E, F M O(1), that there are at most M o(1) possible pairs ( y + BD 2EA, 2Ax + By + D). Each such pair uniquely determines y (as 0) and x (as is not a full square, therefore A 0). This finishes the proof of Proposition 1. 4 Proof of Theorem 2 In what follows, by v we denote the least positive integer such that vv 1 (mod p). We rewrite our congruence in the form (L + x)(l + y)(l + z) λ (mod p), 1 x, y, z M 7

8 which, in turn, is equivalent to the congruence L 2 (x + y + z) + L(xy + xz + yz) + xyz λ L 3 (mod p), 1 x, y, z M. (11) Assume that M p 1/8 and that p is large enough to satisfy several inequalities throughout the proof. Let k = max{1, 2M 2 /p 1/4 }. (12) Lemma 5. If L = uv for some integers u, v with u M 3 /k and 1 v M 2 /k, then the number of solutions of the congruence (11) is at most M o(1). Proof. The congruence (11) is equivalent to v 2 xyz + uv(xy + xz + yz) + u 2 (x + y + z) µ (mod p), where µ < p/2 and µ λv 2 u 3 v (mod p). The absolute value of the left hand side is bounded by (M 2 /k) 2 M 3 + (M 3 /k)(m 2 /k)(3m 2 ) + (M 3 /k) 2 (3M) 7M 7 /k 2 7M 7 /(2M 2 /p 1/4 ) 2 Hence, the congruence (11) is equivalent to the equality Multiplying by v, we get = 7 4 M 3 p 1/2 < p/2. v 2 xyz + uv(xy + xz + yz) + u 2 (x + y + z) = µ. (vx + u)(vy + u)(vz + u) = vµ + u 3 The absolute value of the right and the left hand sides is M O(1), and besides it is distinct from zero (since vµ + u 3 λv 3 (mod p), and λv 3 0 (mod p). Therefore, the number of solutions of the latter equation is bounded by M o(1) and the lemma follows. Due to this conclusion, from now on we can assume that L does not satisfy the condition of Lemma 5, that is we shall assume that L uv, u M 3 /k, v M 2 /k. (13) For 0 r, s 3k 1 and 0 t k 1 let S r,s,t be the set of solutions (x, y, z) such that x + y + z ( rm, (r+1)m ] k k xy + xz + yz ( sm 2 xyz ( tm 3 k, (t+1)m 3, (s+1)m 2 ] k k ] k Clearly, the number of solutions I 3 (M; L) of our congruence satisfies I 3 (M; L) 9k 3 max S rst. We fix one solution (x 0, y 0, z 0 ) S rst. Any other solution (x i, y i, z i ) S rst satisfies the congruence A i L 2 + B i L + C i 0 (mod p) (14) 8

9 where We have A i = x i + y i + z i (x 0 + y 0 + z 0 ), B i = x i y i + x i z i + y i z i (x 0 y 0 + x 0 z 0 + y 0 z 0 ), C i = x i y i z i x 0 y 0 z 0. A i M/k, B i M 2 /k, C i M 3 /k. (15) A solution (x i, y i, z i ) (x 0, y 0, z 0 ) we call degenerated if A i = 0, and non-degenerated otherwise. The set of non-degenerated solutions. We shall show that there are at most M o(1) non-degenerated solutions. We can assume that there are at least several non-degenerated solutions (otherwise we are done). With this set of solutions we shall form a system of congruence with respect to L, L 2. Let us fix one solution (A 1, B 1, C 1 ). Note that the condition A i 0 and the inequalities (15) imply that A i 0 (mod p). Case (1). If A i B 1 A 1 B i for some i, then in view of (15) we also have that A i B 1 A 1 B i (mod p). Solving the system of equations (14) corresponding to the indices i and 1, we obtain that L (C i A 1 A i C 1 )(A i B 1 A 1 B i ) uv (mod p), where From this we derive that L 2 (B i C 1 C i B 1 )(A i B 1 A 1 B i ) u v (mod p), u = C i A 1 A i C 1, v = A i B 1 A 1 B i, u = B i C 1 C i B 1. u 2M 4 /k 2, u 2M 5 /k 2, v 2M 3 /k 2 (16) and (uv ) 2 L 2 u v (mod p). Hence, u 2 u v (mod p) and, using (16), (12), we get u 2, u v 4M 8 /k 4 p/4, so that we actually have the equality u 2 = u v. Multiplying (11) by v, we get vxyz + u(xy + xz + yz) + u (x + y + z) v(λ L 3 ) (mod p) (17) Since 1 x, y, z M, the inequalities (16) give vxyz + u(xy + xz + yz) + u (x + y + z) 14M 6 k 2 14M 6 (2M 2 p 1/4 ) = 7M 2 p 1/2 2 2 < p/2. This converts the congruence (17) into the equality vxyz + u(xy + xz + yz) + u (x + y + z) = µ for some µ M O(1) and µ v(λ L 3 ) (mod p). We multiply this equality by v 2 and use u v = u 2 ; we get that (vx + u)(vy + u)(vz + u) = µv 2 + u 3. (18) 9

10 Since µv 2 + u 3 0, the total number of solutions of the latter equation is M o(1). Case (2). If we are not in case (1), then for any index i one has A 1 B i = A i B 1, which, in turn, implies that we also have A 1 C i A i C 1 (mod p). In view of inequalities (15), we get that the latter congruence is also an equality, so that we have A 1 B i = A i B 1, A 1 C i = A i C 1. (19) From the first equation and the definition of A i, B i, C i, we get from the second equation we get where z i (A 1 (x i + y i ) B 1 ) = B 1 (x i + y i a 0 ) A 1 x i y i + b 0 A 1, (20) z i (A 1 x i y i C 1 ) = C 1 (x i + y i a 0 ) + c 0 A 1, (21) a 0 = x 0 + y 0 + z 0, b 0 = x 0 y 0 + y 0 z 0 + z 0 x 0, c 0 = x 0 y 0 z 0. Multiplying (20) by A 1 x i y i C 1, and (21) by A 1 (x i + y i ) B 1, subtracting the resulting equalities, and making the change of variables x i + y i = u i, x i y i = v i, we obtain (B 1 (u i a 0 ) A 1 v i + b 0 A 1 ) (A 1 v i C 1 ) = (C 1 (u i a 0 ) + c 0 A 1 ) (A 1 u i B 1 ). We rewrite this equation in the form A 1 v 2 i + C 1 u 2 i B 1 u i v i (a 0 C 1 c 0 A 1 )u i (b 0 A 1 a 0 B 1 + C 1 )v i + b 0 C 1 c 0 B 1 = 0. If B 2 1 4A 1 C 1 is a full square (as a number), say R 2 1, then from (14) we obtain that L ( B 1 ± R 1 )(2A 1 ) = uv with u B 1 + B 1 + 4A 1 C 1 4M 2 /k, v 2M/k, which contradicts our condition (13). If B 2 1 4A 1 C 1 is not a full square, then we are at the conditions of Proposition 1 and we can claim that the number of pairs (u i, v i ) is at most M o(1). We now conclude the proof observing that each pair u i, v i produces at most two pairs x i, y i, which, in turn, determines z i. Therefore, the number of non-degenerated solutions counted in S rst is at most M o(1). The set of degenerated solutions. We now consider the set of solutions for which A i = 0. If B i 0, then B i 0 (mod p) and thus we get L = C i B i with C i M 3 /k, B i M 2 /k, which contradicts condition (13). If B i = 0 then together with A i = 0 this implies that C i = 0. Thus, Hence, x i + y i + z i = a 0 = x 0 + y 0 + z 0, x i y i + x i z i + y i z i = b 0 = x 0 y 0 + y 0 z 0 + z 0 x 0, x i y i z i = c 0 = x 0 y 0 z 0. (L + x i )(L + y i )(L + z i ) = (L + x 0 )(L + y 0 )(L + z 0 ). The right hand side is not zero (since it is congruent to λ (mod p) and gcd(λ, p) = 1). Thus, the number of solutions of this equation is at most M o(1). The result follows. 10

11 5 Proof of Corollaries If M < p 5/8 then M 4/3+o(1) p 1/3 + M o(1) < M 4/5+o(1) and the statement of Corollary 1 for I 2 (M; K, L) follows from Theorem 1. If M > p 5/8 then, p 1/2 (log p) 2 < M 4/5+o(1) and the statement of Corollary 1 for I 2 (M; K, L) follows from (5). Analogously we deal with I 2 (M; K, K) considering the cases M > p 2/3 and M < p 2/3. In order to prove Corollary 3, let k = J a (M; K, L) and let (x i, y i ), i = 1,..., k, be all solutions of the congruence y ag x (mod p) with x i [K+1, K+M] and y i [L+1, L+M]. Since M < t, the numbers y 1,..., y k are distinct. Since y i y j ag z (mod p) for some z [2K + 2, 2K + 2M], there exists a value λ such that for at least k 2 /2M pairs (y i, y j ) we have y i y j λ (mod p). Hence, theorem 1 implies that k 2 2M < M 3/2+o(1) p 1/2 + M o(1), and the result follows. Corollary 4 is proved similar to Corollary 3. For any triple (i, j, l) we have y i y j y l ag z (mod p) for some z [3K + 3, 3K + 3M]. Hence, there exists λ 0 (mod p) such that the congruence y i y j y l λ (mod p) has at least k 3 /3M solutions. Thus, k 3 3M < M o(1), and the result follows in this case. If M > p 1/8, then in the interval [L + 1, L + M] we can find a subinterval of length p 1/8 which would contain at least k/(2mp 1/8 ) members from y 1,..., y k. Thus, the preceding argument gives that ( ) 3 k Mp 1/8 3M < M o(1), and the result follows. Now we prove Corollary 2. Let W be the number of solutions of the congruence Then, xyz x y z (mod p), (x, x, y, y, z, z ) I 1 I 1 I 2 I 2 I 3 I 3. W = 1 p 1 χ(x) 2 χ(y) 2 χ(z) 2, χ x I 1 y I 1 z I 1 where χ runs through the set of Dirichlet s characters modulo p. Applying the Holder s inequality, we obtain W ( 1 p 1 χ(x) 6 ) 1/3 ( 1 χ(y) 6 ) 1/3 ( 1 χ(z) 6 ) 1/3. p 1 p 1 χ x I 1 χ y I 2 χ z I 3 Thus, W W 1/3 1 W 1/3 2 W 1/3 3, 11

12 where W j is the number of solutions of the congruence xyz x y z (mod p), x, y, z, x, y, z I j. According to Theorem 2, for each given triple (x, y, z ) there are at most I j o(1) possibilities for (x, y, z). Thus, we have that W i I j 3+o(1). Therefore, W ( I 1 I 2 I 3 ) 1+o(1). Now, using the well known relationship between the cardinality of a product set and the number of solutions of the corresponding equation, we get and the result follows. I 1 I 2 I 3 I 1 2 I 2 2 I 3 2 W ( I 1 I 2 I 3 ) 1 o(1) 6 Conjectures and Open Problems It is very interesting to us the problem of obtaining similar results in four and higher dimensional cases. Let I 4 (M; L) be the number of solutions of the congruence xyzt λ (mod p), L + 1 x, y, z, t L + M. When we estimated I 3 (M; L) we used its connection with the Pell s diophantine equation. If we apply the same method for I 4 (M; L) at the first sight it may look that the method works, however we have not been able to handle diophantine equations that appear in the course of the argument. Thus, the problem we are interested is to prove that there exists an absolute constant δ > 0 such that if M < p δ, then I 4 (M; L) < M o(1). We conclude our paper with several conjectures and open problems related to I 2 (M; K, L), I 3 (M; L), J a (M; K, L) and the product of three small intervals in F p. Conjecture 1. For M < p 1/2 one has I 2 (M; K, L) < M o(1) Conjecture 2. For M < p 1/3 one has I 3 (M; L) < M o(1) Conjecture 3. For M < p 1/2 one has J a (M; K, L) < M o(1). Conjecture 4. Let I 1, I 2, I 3 be intervals in F p of length I i < p 1/3. Then I 1 I 2 I 3 = ( I 1 I 2 I 3 ) 1 o(1). Problem 1. From Theorem 1 it follows that if M < p 1/4, then I 2 (M; K, L) < M o(1). Improve the exponent 1/4 to a larger constant. Problem 2. From Theorem 1 it follows that if M < p 1/3, then I 2 (M; L, L) < M o(1). Improve the exponent 1/3 to a larger constant. Problem 3. Theorem 2 claims that if M < p 1/8, then I 3 (M; L) < M o(1). Improve the exponent 1/8 to a larger constant. 12

13 References [1] J. Bourgain, More on the sum-product phenomenon in prime fields and its applications, Int. J. Number Theory 1 (2005), no.1, [2] J. Bourgain and M. Z. Garaev, On a variant of sum-product estimates and explicit exponential sum bounds in prime fields, Math. Proc. Cambridge Philos. Soc. 146 (2009), no. 1, [3] T. H. Chan and I. Shparlinski, On the concentration of points on modular hyperbolas and exponential curves, Acta Arith. 142 (2010), no. 1, [4] J. Cilleruelo and J. Jiménez, Divisors in a Dedekind domain, Acta Arith. 85 (1998), no. 3, [5] M. Z. Garaev, On the logarithmic factor in error term estimates in certain additive congruence problems, Acta Arith. 124 (2006), no. 1, [6] D. R. Heath-Brown, Almost-primes in arithmetic progressions and short intervals, Math. Proc. Cambridge Philos. Soc. 83 (1978), no. 3, [7] L. K. Hua, Introduction to Number Theory, Springer Verlag, Berlin, [8] R. C. Vaughan and T. D. Wooley, Further improvements in Waring s problem, Acta Math. 174 (1995),

### MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES

MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES 2016 47 4. Diophantine Equations A Diophantine Equation is simply an equation in one or more variables for which integer (or sometimes rational) solutions

### On the greatest and least prime factors of n! + 1, II

Publ. Math. Debrecen Manuscript May 7, 004 On the greatest and least prime factors of n! + 1, II By C.L. Stewart In memory of Béla Brindza Abstract. Let ε be a positive real number. We prove that 145 1

### APPLICATIONS OF THE ORDER FUNCTION

APPLICATIONS OF THE ORDER FUNCTION LECTURE NOTES: MATH 432, CSUSM, SPRING 2009. PROF. WAYNE AITKEN In this lecture we will explore several applications of order functions including formulas for GCDs and

### CHAPTER 6: RATIONAL NUMBERS AND ORDERED FIELDS

CHAPTER 6: RATIONAL NUMBERS AND ORDERED FIELDS LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we construct the set of rational numbers Q using equivalence

### SUM OF TWO SQUARES JAHNAVI BHASKAR

SUM OF TWO SQUARES JAHNAVI BHASKAR Abstract. I will investigate which numbers can be written as the sum of two squares and in how many ways, providing enough basic number theory so even the unacquainted

### Prime divisors of Lucas sequences and a conjecture of Ska lba

Prime divisors of Lucas sequences and a conjecture of Ska lba Florian Luca, Instituto de Matemáticas Universidad Nacional Autónoma de México C.P. 58089, Morelia, Michoacán, México fluca@matmor.unam.mx

### On the largest prime factor of x 2 1

On the largest prime factor of x 2 1 Florian Luca and Filip Najman Abstract In this paper, we find all integers x such that x 2 1 has only prime factors smaller than 100. This gives some interesting numerical

### Characterizing the Sum of Two Cubes

1 3 47 6 3 11 Journal of Integer Sequences, Vol. 6 (003), Article 03.4.6 Characterizing the Sum of Two Cubes Kevin A. Broughan University of Waikato Hamilton 001 New Zealand kab@waikato.ac.nz Abstract

### Applications of Fermat s Little Theorem and Congruences

Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4

### Chapter 6. Number Theory. 6.1 The Division Algorithm

Chapter 6 Number Theory The material in this chapter offers a small glimpse of why a lot of facts that you ve probably nown and used for a long time are true. It also offers some exposure to generalization,

### On the Exponential Diophantine Equation (12m 2 + 1) x + (13m 2 1) y = (5m) z

International Journal of Algebra, Vol. 9, 2015, no. 6, 261-272 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ija.2015.5529 On the Exponential Diophantine Equation (12m 2 + 1) x + (13m 2 1) y

### Pythagorean Triples Pythagorean triple similar primitive

Pythagorean Triples One of the most far-reaching problems to appear in Diophantus Arithmetica was his Problem II-8: To divide a given square into two squares. Namely, find integers x, y, z, so that x 2

### Further linear algebra. Chapter I. Integers.

Further linear algebra. Chapter I. Integers. Andrei Yafaev Number theory is the theory of Z = {0, ±1, ±2,...}. 1 Euclid s algorithm, Bézout s identity and the greatest common divisor. We say that a Z divides

### CHAPTER 5: MODULAR ARITHMETIC

CHAPTER 5: MODULAR ARITHMETIC LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we will consider congruence modulo m, and explore the associated arithmetic called

### MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on

### Divisors in a Dedekind domain

ACTA ARITHMETICA LXXXV.3 (1998) Divisors in a Dedekind domain by Javier Cilleruelo and Jorge Jiménez-Urroz (Madrid) 1. Introduction. Let A be a Dedekind domain in which we can define a notion of distance.

### k, then n = p2α 1 1 pα k

Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

### ON INTEGERS EXPRESSIBLE BY SOME SPECIAL LINEAR FORM

Acta Math. Univ. Comenianae Vol. LXXXI, (01), pp. 03 09 03 ON INTEGERS EXPRESSIBLE BY SOME SPECIAL LINEAR FORM A. DUBICKAS and A. NOVIKAS Abstract. Let E(4) be the set of positive integers expressible

### Topics in Number Theory

Chapter 8 Topics in Number Theory 8.1 The Greatest Common Divisor Preview Activity 1 (The Greatest Common Divisor) 1. Explain what it means to say that a nonzero integer m divides an integer n. Recall

### Homework until Test #2

MATH31: Number Theory Homework until Test # Philipp BRAUN Section 3.1 page 43, 1. It has been conjectured that there are infinitely many primes of the form n. Exhibit five such primes. Solution. Five such

### Solutions to Homework Set 3 (Solutions to Homework Problems from Chapter 2)

Solutions to Homework Set 3 (Solutions to Homework Problems from Chapter 2) Problems from 21 211 Prove that a b (mod n) if and only if a and b leave the same remainder when divided by n Proof Suppose a

### The minimal number with a given number of. divisors

The minimal number with a given number of divisors Ron Brown Department of Mathematics, 2565 McCarthy Mall, University of Hawaii, Honolulu, HI 96822, USA Phone (808) 956-4665 Fax (808) 956-9139 Email address:

### Math 319 Problem Set #3 Solution 21 February 2002

Math 319 Problem Set #3 Solution 21 February 2002 1. ( 2.1, problem 15) Find integers a 1, a 2, a 3, a 4, a 5 such that every integer x satisfies at least one of the congruences x a 1 (mod 2), x a 2 (mod

### Simultaneous equal sums of three powers

Simultaneous equal sums of three powers T.D. Browning 1 and D.R. Heath-Brown 2 1 School of Mathematics, Bristol University, Bristol BS8 1TW 2 Mathematical Institute, 24 29 St. Giles, Oxford OX1 3LB 1 t.d.browning@bristol.ac.uk,

### On the Union of Arithmetic Progressions

On the Union of Arithmetic Progressions Shoni Gilboa Rom Pinchasi August, 04 Abstract We show that for any integer n and real ɛ > 0, the union of n arithmetic progressions with pairwise distinct differences,

### Prime Numbers. Chapter Primes and Composites

Chapter 2 Prime Numbers The term factoring or factorization refers to the process of expressing an integer as the product of two or more integers in a nontrivial way, e.g., 42 = 6 7. Prime numbers are

### THE CONGRUENT NUMBER PROBLEM

THE CONGRUENT NUMBER PROBLEM KEITH CONRAD 1. Introduction A right triangle is called rational when its legs and hypotenuse are all rational numbers. Examples of rational right triangles include Pythagorean

### Tangent and normal lines to conics

4.B. Tangent and normal lines to conics Apollonius work on conics includes a study of tangent and normal lines to these curves. The purpose of this document is to relate his approaches to the modern viewpoints

### Define the set of rational numbers to be the set of equivalence classes under #.

Rational Numbers There are four standard arithmetic operations: addition, subtraction, multiplication, and division. Just as we took differences of natural numbers to represent integers, here the essence

### CHAPTER 5. Number Theory. 1. Integers and Division. Discussion

CHAPTER 5 Number Theory 1. Integers and Division 1.1. Divisibility. Definition 1.1.1. Given two integers a and b we say a divides b if there is an integer c such that b = ac. If a divides b, we write a

### PYTHAGOREAN TRIPLES PETE L. CLARK

PYTHAGOREAN TRIPLES PETE L. CLARK 1. Parameterization of Pythagorean Triples 1.1. Introduction to Pythagorean triples. By a Pythagorean triple we mean an ordered triple (x, y, z) Z 3 such that x + y =

### The square-free kernel of x 2n a 2n

ACTA ARITHMETICA 101.2 (2002) The square-free kernel of x 2n a 2n by Paulo Ribenboim (Kingston, Ont.) Dedicated to my long-time friend and collaborator Wayne McDaniel, at the occasion of his retirement

### ARITHMETIC PROGRESSIONS OF FOUR SQUARES

ARITHMETIC PROGRESSIONS OF FOUR SQUARES KEITH CONRAD 1. Introduction Suppose a, b, c, and d are rational numbers such that a 2, b 2, c 2, and d 2 form an arithmetic progression: the differences b 2 a 2,

### On the largest prime factor of the Mersenne numbers

On the largest prime factor of the Mersenne numbers Kevin Ford Department of Mathematics The University of Illinois at Urbana-Champaign Urbana Champaign, IL 61801, USA ford@math.uiuc.edu Florian Luca Instituto

### Congruent Numbers, the Rank of Elliptic Curves and the Birch and Swinnerton-Dyer Conjecture. Brad Groff

Congruent Numbers, the Rank of Elliptic Curves and the Birch and Swinnerton-Dyer Conjecture Brad Groff Contents 1 Congruent Numbers... 1.1 Basic Facts............................... and Elliptic Curves.1

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### ON TOPOLOGICAL SEQUENCE ENTROPY AND CHAOTIC MAPS ON INVERSE LIMIT SPACES. 1. Introduction

Acta Math. Univ. Comenianae Vol. LXVIII, 2(1999), pp. 205 211 205 ON TOPOLOGICAL SEQUENCE ENTROPY AND CHAOTIC MAPS ON INVERSE LIMIT SPACES J. S. CANOVAS Abstract. The aim of this paper is to prove the

### SUBGROUPS OF CYCLIC GROUPS. 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by

SUBGROUPS OF CYCLIC GROUPS KEITH CONRAD 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by g = {g k : k Z}. If G = g, then G itself is cyclic, with g as a generator. Examples

### FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization

### Geometry and Arithmetic

Geometry and Arithmetic Alex Tao 10 June 2008 1 Rational Points on Conics We begin by stating a few definitions: A rational number is a quotient of two integers and the whole set of rational numbers is

MODULAR ARITHMETIC KEITH CONRAD. Introduction We will define the notion of congruent integers (with respect to a modulus) and develop some basic ideas of modular arithmetic. Applications of modular arithmetic

### ON THE FIBONACCI NUMBERS

ON THE FIBONACCI NUMBERS Prepared by Kei Nakamura The Fibonacci numbers are terms of the sequence defined in a quite simple recursive fashion. However, despite its simplicity, they have some curious properties

### MATH REVIEW KIT. Reproduced with permission of the Certified General Accountant Association of Canada.

MATH REVIEW KIT Reproduced with permission of the Certified General Accountant Association of Canada. Copyright 00 by the Certified General Accountant Association of Canada and the UBC Real Estate Division.

### Chapter Three. Functions. In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics.

Chapter Three Functions 3.1 INTRODUCTION In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics. Definition 3.1: Given sets X and Y, a function from X to

### THEORY OF SIMPLEX METHOD

Chapter THEORY OF SIMPLEX METHOD Mathematical Programming Problems A mathematical programming problem is an optimization problem of finding the values of the unknown variables x, x,, x n that maximize

### Every Positive Integer is the Sum of Four Squares! (and other exciting problems)

Every Positive Integer is the Sum of Four Squares! (and other exciting problems) Sophex University of Texas at Austin October 18th, 00 Matilde N. Lalín 1. Lagrange s Theorem Theorem 1 Every positive integer

### PRIME FACTORS OF CONSECUTIVE INTEGERS

PRIME FACTORS OF CONSECUTIVE INTEGERS MARK BAUER AND MICHAEL A. BENNETT Abstract. This note contains a new algorithm for computing a function f(k) introduced by Erdős to measure the minimal gap size in

### A Non-Existence Property of Pythagorean Triangles with a 3-D Application

A Non-Existence Property of Pythagorean Triangles with a 3-D Application Konstantine Zelator Department of Mathematics and Computer Science Rhode Island College 600 Mount Pleasant Avenue Providence, RI

### CHAPTER I THE REAL AND COMPLEX NUMBERS DEFINITION OF THE NUMBERS 1, i, AND 2

CHAPTER I THE REAL AND COMPLEX NUMBERS DEFINITION OF THE NUMBERS 1, i, AND 2 In order to mae precise sense out of the concepts we study in mathematical analysis, we must first come to terms with what the

### Theorem 2. If x Q and y R \ Q, then. (a) x + y R \ Q, and. (b) xy Q.

Math 305 Fall 011 The Density of Q in R The following two theorems tell us what happens when we add and multiply by rational numbers. For the first one, we see that if we add or multiply two rational numbers

### Quotient Rings and Field Extensions

Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

### ARITHMETIC PROGRESSIONS OF THREE SQUARES

ARITHMETIC PROGRESSIONS OF THREE SQUARES KEITH CONRAD 1 Introduction Here are the first 10 perfect squares (ignoring 0): 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 In this list is the arithmetic progression

### DIVISORS IN A DEDEKIND DOMAIN. Javier Cilleruelo and Jorge Jiménez-Urroz. 1 Introduction

DIVISORS IN A DEDEKIND DOMAIN Javier Cilleruelo and Jorge Jiménez-Urroz 1 Introduction Let A be a Dedekind domain in which we can define a notion of distance. We are interested in the number of divisors

### arxiv:math/ v2 [math.nt] 6 Aug 2006

PROBLEMS IN ADDITIVE NUMBER THEORY, I arxiv:math/0604340v2 [math.nt] 6 Aug 2006 MELVYN B. NATHANSON Abstract. Talk at the Atelier en combinatoire additive (Workshop on Arithmetic Combinatorics) at the

### Homework 5 Solutions

Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which

### Groups, Rings, and Fields. I. Sets Let S be a set. The Cartesian product S S is the set of ordered pairs of elements of S, S S = {(x, y) x, y S}.

Groups, Rings, and Fields I. Sets Let S be a set. The Cartesian product S S is the set of ordered pairs of elements of S, A binary operation φ is a function, S S = {(x, y) x, y S}. φ : S S S. A binary

### The Dirichlet Unit Theorem

Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if

### ON FIBONACCI NUMBERS WITH FEW PRIME DIVISORS

ON FIBONACCI NUMBERS WITH FEW PRIME DIVISORS YANN BUGEAUD, FLORIAN LUCA, MAURICE MIGNOTTE, SAMIR SIKSEK Abstract If n is a positive integer, write F n for the nth Fibonacci number, and ω(n) for the number

### U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra

U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory

### Practice Problems for First Test

Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the human mind will never penetrate.-

### INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS

INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS STEVEN HEILMAN Contents 1. Homework 1 1 2. Homework 2 6 3. Homework 3 10 4. Homework 4 16 5. Homework 5 19 6. Homework 6 21 7. Homework 7 25 8. Homework 8 28

### Notes on the second moment method, Erdős multiplication tables

Notes on the second moment method, Erdős multiplication tables January 25, 20 Erdős multiplication table theorem Suppose we form the N N multiplication table, containing all the N 2 products ab, where

### Collinear Points in Permutations

Collinear Points in Permutations Joshua N. Cooper Courant Institute of Mathematics New York University, New York, NY József Solymosi Department of Mathematics University of British Columbia, Vancouver,

### Elementary Number Theory

Elementary Number Theory Ahto Buldas December 3, 2016 Ahto Buldas Elementary Number Theory December 3, 2016 1 / 1 Division For any m > 0, we define Z m = {0, 1,... m 1} For any n, m Z (m > 0), there are

### 1.3 Induction and Other Proof Techniques

4CHAPTER 1. INTRODUCTORY MATERIAL: SETS, FUNCTIONS AND MATHEMATICAL INDU 1.3 Induction and Other Proof Techniques The purpose of this section is to study the proof technique known as mathematical induction.

### 11 Ideals. 11.1 Revisiting Z

11 Ideals The presentation here is somewhat different than the text. In particular, the sections do not match up. We have seen issues with the failure of unique factorization already, e.g., Z[ 5] = O Q(

### On the number-theoretic functions ν(n) and Ω(n)

ACTA ARITHMETICA LXXVIII.1 (1996) On the number-theoretic functions ν(n) and Ω(n) by Jiahai Kan (Nanjing) 1. Introduction. Let d(n) denote the divisor function, ν(n) the number of distinct prime factors,

### (Positive) Rational Numbers

(Positive) Rational Numbers Definition Definition 1 Consider all ordered pairs (, ) with, N Define the equivalance (, ) (y 1, ) = y 1 (1) Theorem 2 is indeed an equivalence relation, that is 1 (, ) (,

### Theorem 5. The composition of any two symmetries in a point is a translation. More precisely, S B S A = T 2

Isometries. Congruence mappings as isometries. The notion of isometry is a general notion commonly accepted in mathematics. It means a mapping which preserves distances. The word metric is a synonym to

### CONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12

CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.

### Notes on Factoring. MA 206 Kurt Bryan

The General Approach Notes on Factoring MA 26 Kurt Bryan Suppose I hand you n, a 2 digit integer and tell you that n is composite, with smallest prime factor around 5 digits. Finding a nontrivial factor

### Section 5.1 Number Theory: Prime & Composite Numbers

Section 5.1 Number Theory: Prime & Composite Numbers Objectives 1. Determine divisibility. 2. Write the prime factorization of a composite number. 3. Find the greatest common divisor of two numbers. 4.

### Congruences. Robert Friedman

Congruences Robert Friedman Definition of congruence mod n Congruences are a very handy way to work with the information of divisibility and remainders, and their use permeates number theory. Definition

### (x + a) n = x n + a Z n [x]. Proof. If n is prime then the map

22. A quick primality test Prime numbers are one of the most basic objects in mathematics and one of the most basic questions is to decide which numbers are prime (a clearly related problem is to find

PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient

### THE CONGRUENT NUMBER PROBLEM

THE CONGRUENT NUMBER PROBLEM KEITH CONRAD 1. Introduction A right triangle is called rational when its legs and hypotenuse are all rational numbers. Examples of rational right triangles include Pythagorean

### TRIANGLES ON THE LATTICE OF INTEGERS. Department of Mathematics Rowan University Glassboro, NJ Andrew Roibal and Abdulkadir Hassen

TRIANGLES ON THE LATTICE OF INTEGERS Andrew Roibal and Abdulkadir Hassen Department of Mathematics Rowan University Glassboro, NJ 08028 I. Introduction In this article we will be studying triangles whose

### Lesson 6: Proofs of Laws of Exponents

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 8 Student Outcomes Students extend the previous laws of exponents to include all integer exponents. Students base symbolic proofs on concrete examples to

### Solving a family of quartic Thue inequalities using continued fractions

Solving a family of quartic Thue inequalities using continued fractions Andrej Dujella, Bernadin Ibrahimpašić and Borka Jadrijević Abstract In this paper we find all primitive solutions of the Thue inequality

### PROBLEM SET 7: PIGEON HOLE PRINCIPLE

PROBLEM SET 7: PIGEON HOLE PRINCIPLE The pigeonhole principle is the following observation: Theorem. Suppose that > kn marbles are distributed over n jars, then one jar will contain at least k + marbles.

### PRIMITIVE EVEN 2-REGULAR POSITIVE QUATERNARY QUADRATIC FORMS

J. Korean Math. Soc. 45 (2008), No. 3, pp. 621 630 PRIMITIVE EVEN 2-REGULAR POSITIVE QUATERNARY QUADRATIC FORMS Byeong-Kweon Oh Reprinted from the Journal of the Korean Mathematical Society Vol. 45, No.

### Number Theory Hungarian Style. Cameron Byerley s interpretation of Csaba Szabó s lectures

Number Theory Hungarian Style Cameron Byerley s interpretation of Csaba Szabó s lectures August 20, 2005 2 0.1 introduction Number theory is a beautiful subject and even cooler when you learn about it

### Completeness I. Chapter Rational Numbers

Chapter 5 Completeness I Completeness is the key property of the real numbers that the rational numbers lack. Before examining this property we explore the rational and irrational numbers, discovering

### Theorem (The division theorem) Suppose that a and b are integers with b > 0. There exist unique integers q and r so that. a = bq + r and 0 r < b.

Theorem (The division theorem) Suppose that a and b are integers with b > 0. There exist unique integers q and r so that a = bq + r and 0 r < b. We re dividing a by b: q is the quotient and r is the remainder,

### 1. R In this and the next section we are going to study the properties of sequences of real numbers.

+a 1. R In this and the next section we are going to study the properties of sequences of real numbers. Definition 1.1. (Sequence) A sequence is a function with domain N. Example 1.2. A sequence of real

### Prime Numbers and Irreducible Polynomials

Prime Numbers and Irreducible Polynomials M. Ram Murty The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry.

### MATH 361: NUMBER THEORY FIRST LECTURE

MATH 361: NUMBER THEORY FIRST LECTURE 1. Introduction As a provisional definition, view number theory as the study of the properties of the positive integers, Z + = {1, 2, 3, }. Of particular interest,

### HTP in Positive Characteristic

HTP in Positive Characteristic Alexandra Shlapentokh East Carolina University October 2007 Table of Contents 1 A Brief History of Diophantine Undecidability over Number Fields The Original Problem Extensions

### EULER S THEOREM. 1. Introduction Fermat s little theorem is an important property of integers to a prime modulus. a p 1 1 mod p.

EULER S THEOREM KEITH CONRAD. Introduction Fermat s little theorem is an important property of integers to a prime modulus. Theorem. (Fermat). For prime p and any a Z such that a 0 mod p, a p mod p. If

### PERIMETERS OF PRIMITIVE PYTHAGOREAN TRIANGLES. 1. Introduction

PERIMETERS OF PRIMITIVE PYTHAGOREAN TRIANGLES LINDSEY WITCOSKY ADVISOR: DR. RUSS GORDON Abstract. This paper examines two methods of determining whether a positive integer can correspond to the semiperimeter

### Instructor: Bobby Kleinberg Lecture Notes, 5 May The Miller-Rabin Randomized Primality Test

Introduction to Algorithms (CS 482) Cornell University Instructor: Bobby Kleinberg Lecture Notes, 5 May 2010 The Miller-Rabin Randomized Primality Test 1 Introduction Primality testing is an important

### Chapter 1 Basic Number Concepts

Draft of September 2014 Chapter 1 Basic Number Concepts 1.1. Introduction No problems in this section. 1.2. Factors and Multiples 1. Determine whether the following numbers are divisible by 3, 9, and 11:

### 1 Die hard, once and for all

ENGG 2440A: Discrete Mathematics for Engineers Lecture 4 The Chinese University of Hong Kong, Fall 2014 6 and 7 October 2014 Number theory is the branch of mathematics that studies properties of the integers.

### Finite Fields and Error-Correcting Codes

Lecture Notes in Mathematics Finite Fields and Error-Correcting Codes Karl-Gustav Andersson (Lund University) (version 1.013-16 September 2015) Translated from Swedish by Sigmundur Gudmundsson Contents

### QUADRATIC RECIPROCITY IN CHARACTERISTIC 2

QUADRATIC RECIPROCITY IN CHARACTERISTIC 2 KEITH CONRAD 1. Introduction Let F be a finite field. When F has odd characteristic, the quadratic reciprocity law in F[T ] (see [4, Section 3.2.2] or [5]) lets

### Integer Factorization using the Quadratic Sieve

Integer Factorization using the Quadratic Sieve Chad Seibert* Division of Science and Mathematics University of Minnesota, Morris Morris, MN 56567 seib0060@morris.umn.edu March 16, 2011 Abstract We give

### The minimum number of monochromatic 4-term progressions

The minimum number of monochromatic 4-term progressions Rutgers University May 28, 2009 Monochromatic 3-term progressions Monochromatic 4-term progressions 1 Introduction Monochromatic 3-term progressions