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Number Theory Naoki Sato
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1 Number Theory Naoki Sato 0 Preface This set of notes on number theory was originally written in 1995 for students at the IMO level. It covers the basic background material that an IMO student should be familiar with. This text is meant to be a reference, and not a relacement but rather a sulement to a number theory textbook; several are given at the back. Proofs are given when aroriate, or when they illustrate some insight or imortant idea. The roblems are culled from various sources, many from actual contests and olymiads, and in general are very difficult. The author welcomes any corrections or suggestions. 1 Divisibility For integers a and b, we say that a divides b, or that a is a divisor (or factor) of b, or that b is a multile of a, if there exists an integer c such that b = ca, and we denote this by a b. Otherwise, a does not divide b, and we denote this by a b. A ositive integer is a rime if the only divisors of are 1 and. If k a and k+1 a where is a rime, i.e. k is the highest ower of dividing a, then we denote this by k a. Useful Facts If a, b > 0, and a b, then a b. If a b 1, a b 2,..., a b n, then for any integers c 1, c 2,..., c n, a n b i c i. i=1 Theorem 1.1. The Division Algorithm. For any ositive integer a and integer b, there exist unique integers q and r such that b = qa + r and 0 r < a, with r = 0 iff a b. 1
2 Theorem 1.2. The Fundamental Theorem of Arithmetic. Every integer greater than 1 can be written uniquely in the form e 1 1 e 2 2 e k k, where the i are distinct rimes and the e i are ositive integers. Theorem 1.3. (Euclid) There exist an infinite number of rimes. Proof. Suose that there are a finite number of rimes, say 1, 2,..., n. Let N = 1 2 n + 1. By the fundamental theorem of arithmetic, N is divisible by some rime. This rime must be among the i, since by assumtion these are all the rimes, but N is seen not to be divisible by any of the i, contradiction. Examle 1.1. Let x and y be integers. Prove that 2x + 3y is divisible by 17 iff 9x + 5y is divisible by 17. Solution. 17 (2x + 3y) 17 [13(2x + 3y)], or 17 (26x + 39y) 17 (9x + 5y), and conversely, 17 (9x + 5y) 17 [4(9x + 5y)], or 17 (36x + 20y) 17 (2x + 3y). Examle 1.2. Find all ositive integers d such that d divides both n 2 +1 and (n + 1) for some integer n. Solution. Let d (n 2 + 1) and d [(n + 1) 2 + 1], or d (n 2 + 2n + 2). Then d [(n 2 + 2n + 2) (n 2 + 1)], or d (2n + 1) d (4n 2 + 4n + 1), so d [4(n 2 +2n+2) (4n 2 +4n+1)], or d (4n+7). Then d [(4n+7) 2(2n+1)], or d 5, so d can only be 1 or 5. Taking n = 2 shows that both of these values are achieved. Examle 1.3. Suose that a 1, a 2,..., a 2n are distinct integers such that the equation (x a 1 )(x a 2 ) (x a 2n ) ( 1) n (n!) 2 = 0 has an integer solution r. Show that so (1984 IMO Short List) r = a 1 + a a 2n. 2n Solution. Clearly, r a i for all i, and the r a i are 2n distinct integers, (r a 1 )(r a 2 ) (r a 2n ) (1)(2) (n)( 1)( 2) ( n) = (n!) 2, 2
3 with equality iff {r a 1, r a 2,..., r a 2n } = {1, 2,..., n, 1, 2,..., n}. Therefore, this must be the case, so (r a 1 ) + (r a 2 ) + + (r a 2n ) = 2nr (a 1 + a a 2n ) = n + ( 1) + ( 2) + + ( n) = 0 r = a 1 + a a 2n. 2n Examle 1.4. Let 0 < a 1 < a 2 < < a mn+1 be mn + 1 integers. Prove that you can select either m + 1 of them no one of which divides any other, or n + 1 of them each dividing the following one. (1966 Putnam Mathematical Cometition) Solution. For each i, 1 i mn + 1, let n i be the length of the longest sequence starting with a i and each dividing the following one, among the integers a i, a i+1,..., a mn+1. If some n i is greater than n then the roblem is solved. Otherwise, by the igeonhole rincile, there are at least m + 1 values of n i that are equal. Then, the integers a i corresonding to these n i cannot divide each other. Useful Facts Bertrand s Postulate. For every ositive integer n, there exists a rime such that n 2n. Gauss s Lemma. If a olynomial with integer coefficients factors into two olynomials with rational coefficients, then it factors into two olynomials with integer coefficients. Problems 1. Let a and b be ositive integers such that a b 2, b 2 a 3, a 3 b 4, b 4 a 5,.... Prove that a = b. 2. Let a, b, and c denote three distinct integers, and let P denote a olynomial having all integral coefficients. Show that it is imossible that P (a) = b, P (b) = c, and P (c) = a. (1974 USAMO) 3
4 3. Show that if a and b are ositive integers, then ( a + 1 n ( + b + 2) 1 ) n 2 is an integer for only finitely many ositive integers n. (A Problem Seminar, D.J. Newman) 4. For a ositive integer n, let r(n) denote the sum of the remainders when n is divided by 1, 2,..., n resectively. Prove that r(k) = r(k 1) for infinitely many ositive integers k. (1981 Kürschák Cometition) 5. Prove that for all ositive integers n, 0 < n k=1 g(k) k 2n 3 < 2 3, where g(k) denotes the greatest odd divisor of k. (1973 Austrian Mathematics Olymiad) 6. Let d be a ositive integer, and let S be the set of all ositive integers of the form x 2 + dy 2, where x and y are non-negative integers. (a) Prove that if a S and b S, then ab S. (b) Prove that if a S and S, such that is a rime and a, then a/ S. (c) Assume that the equation x 2 + dy 2 = has a solution in nonnegative integers x and y, where is a given rime. Show that if d 2, then the solution is unique, and if d = 1, then there are exactly two solutions. 2 GCD and LCM The greatest common divisor of two ositive integers a and b is the greatest ositive integer that divides both a and b, which we denote by gcd(a, b), and similarly, the lowest common multile of a and b is the least ositive 4
5 integer that is a multile of both a and b, which we denote by lcm(a, b). We say that a and b are relatively rime if gcd(a, b) = 1. For integers a 1, a 2,..., a n, gcd(a 1, a 2,..., a n ) is the greatest ositive integer that divides all of a 1, a 2,..., a n, and lcm(a 1, a 2,..., a n ) is defined similarly. Useful Facts For all a, b, gcd(a, b) lcm(a, b) = ab. For all a, b, and m, gcd(ma, mb) = m gcd(a, b) and lcm(ma, mb) = mlcm(a, b). If d gcd(a, b), then ( a gcd d, b ) = d gcd(a, b). d In articular, if d = gcd(a, b), then gcd(a/d, b/d) = 1; that is, a/d and b/d are relatively rime. If a bc and gcd(a, c) = 1, then a b. For ositive integers a and b, if d is a ositive integer such that d a, d b, and for any d, d a and d b imlies that d d, then d = gcd(a, b). This is merely the assertion that any common divisor of a and b divides gcd(a, b). If a 1 a 2 a n is a erfect k th ower and the a i are airwise relatively rime, then each a i is a erfect k th ower. Any two consecutive integers are relatively rime. Examle 2.1. Show that for any ositive integer N, there exists a multile of N that consists only of 1s and 0s. Furthermore, show that if N is relatively rime to 10, then there exists a multile that consists only of 1s. Solution. Consider the N + 1 integers 1, 11, 111,..., (N + 1 1s). When divided by N, they leave N + 1 remainders. By the igeonhole rincile, two of these remainders are equal, so the difference in the corresonding integers, an integer of the form , is divisible by N. If N is relatively rime to 10, then we may divide out all owers of 10, to obtain an integer of the form that remains divisible by N. 5
6 Theorem 2.1. For any ositive integers a and b, there exist integers x and y such that ax + by = gcd(a, b). Furthermore, as x and y vary over all integers, ax + by attains all multiles and only multiles of gcd(a, b). Proof. Let S be the set of all integers of the form ax+by, and let d be the least ositive element of S. By the division algorithm, there exist integers q and r such that a = qd + r, 0 r < d. Then r = a qd = a q(ax + by) = (1 qx)a (qy)b, so r is also in S. But r < d, so r = 0 d a, and similarly, d b, so d gcd(a, b). However, gcd(a, b) divides all elements of S, so in articular gcd(a, b) d d = gcd(a, b). The second art of the theorem follows. Corollary 2.2. The ositive integers a and b are relatively rime iff there exist integers x and y such that ax + by = 1. Corollary 2.3. For any ositive integers a 1, a 2,..., a n, there exist integers x 1, x 2,..., x n, such that a 1 x 1 +a 2 x 2 + +a n x n = gcd(a 1, a 2,..., a n ). Corollary 2.4. Let a and b be ositive integers, and let n be an integer. Then the equation ax + by = n has a solution in integers x and y iff gcd(a, b) n. If this is the case, then all solutions are of the form ( (x, y) = x 0 + t b d, y 0 t a ), d where d = gcd(a, b), (x 0, y 0 ) is a secific solution of ax + by = n, and t is an integer. Proof. The first art follows from Theorem 2.1. For the second art, as stated, let d = gcd(a, b), and let (x 0, y 0 ) be a secific solution of ax + by = n, so that ax 0 + by 0 = n. If ax + by = n, then ax + by ax 0 by 0 = a(x x 0 ) + b(y y 0 ) = 0, or a(x x 0 ) = b(y 0 y), and hence (x x 0 ) a d = (y 0 y) b d. Since a/d and b/d are relatively rime, b/d must divide x x 0, and a/d must divide y 0 y. Let x x 0 = tb/d and y 0 y = ta/d. This gives the solutions described above. 6
7 Examle 2.2. Prove that the fraction 21n n + 3 is irreducible for every ositive integer n. (1959 IMO) Solution. For all n, 3(14n + 3) 2(21n + 4) = 1, so the numerator and denominator are relatively rime. Examle 2.3. For all ositive integers n, let T n = 2 2n + 1. Show that if m n, then T m and T n are relatively rime. Solution. We have that T n 2 = 2 2n 1 = 2 2n = (T n 1 1) 2 1 = T 2 n 1 2T n 1 = T n 1 (T n 1 2) = T n 1 T n 2 (T n 2 2) = = T n 1 T n 2 T 1 T 0 (T 0 2) = T n 1 T n 2 T 1 T 0, for all n. Therefore, any common divisor of T m and T n must divide 2. But each T n is odd, so T m and T n are relatively rime. Remark. It immediately follows from this result that there are an infinite number of rimes. The Euclidean Algorithm. By recursive use of the division algorithm, we may find the gcd of two ositive integers a and b without factoring either, and the x and y in Theorem 2.1 (and so, a secific solution in Corollary 2.4). For examle, for a = 329 and b = 182, we comute 329 = , 182 = , 147 = , 35 = 5 7, and sto when there is no remainder. The last dividend is the gcd, so in our examle, gcd(329,182) = 7. Now, working through the above equations 7
8 backwards, 7 = = ( ) = = 5 ( ) = Remark. The Euclidean algorithm also works for olynomials. Examle 2.4. Let n be a ositive integer, and let S be a subset of n + 1 elements of the set {1, 2,..., 2n}. Show that (a) There exist two elements of S that are relatively rime, and (b) There exist two elements of S, one of which divides the other. Solution. (a) There must be two elements of S that are consecutive, and thus, relatively rime. (b) Consider the greatest odd factor of each of the n + 1 elements in S. Each is among the n odd integers 1, 3,..., 2n 1. By the igeonhole rincile, two must have the same greatest odd factor, so they differ (multilication-wise) by a ower of 2, and so one divides the other. Examle 2.5. The ositive integers a 1, a 2,..., a n are such that each is less than 1000, and lcm(a i, a j ) > 1000 for all i, j, i j. Show that n i=1 1 a i < 2. (1951 Russian Mathematics Olymiad) Solution. If 1000 < a 1000, then the m multiles a, 2a,..., ma do m+1 m not exceed Let k 1 the number of a i in the interval ( 1000, 1000], k 2 2 in ( 1000, 1000], etc. Then there are k k 2 + 3k 3 + integers, no greater than 1000, that are multiles of at least one of the a i. But the multiles are distinct, so k 1 + 2k 2 + 3k 3 + < k 1 + 3k 2 + 4k 3 + = (k 1 + 2k 2 + 3k 3 + ) + (k 1 + k 2 + k 3 + ) < n <
9 Therefore, n i=1 1 2 k 1 a i k k 3 = 2k 1 + 3k 2 + 4k < Note: It can be shown that n 500 as follows: Consider the greatest odd divisor of a 1, a 2,..., a Each must be distinct; otherwise, two differ, multilication-wise, by a ower of 2, which means one divides the other, contradiction. Also, there are only 500 odd numbers between 1 and 1000, from which the result follows. It also then follows that Useful Facts n i=1 1 a i < 3 2. Dirichlet s Theorem. If a and b are relatively rime ositive integers, then the arithmetic sequence a, a + b, a + 2b,..., contains infinitely many rimes. Problems 1. The symbols (a, b,..., g) and [a, b,..., g] denote the greatest common divisor and lowest common multile, resectively of the ositive integers a, b,..., g. Prove that (1972 USAMO) [a, b, c] 2 [a, b][a, c][b, c] = (a, b, c) 2 (a, b)(a, c)(b, c). 2. Show that gcd(a m 1, a n 1) = a gcd(m,n) 1 for all ositive integers a > 1, m, n. 9
10 3. Let a, b, and c be ositive integers. Show that lcm(a, b, c) = abc gcd(a, b, c) gcd(a, b) gcd(a, c) gcd(b, c). Exress gcd(a, b, c) in terms of abc, lcm(a, b, c), lcm(a, b), lcm(a, c), and lcm(b, c). Generalize. 4. Let a, b be odd ositive integers. Define the sequence (f n ) by utting f 1 = a, f 2 = b, and by letting f n for n 3 be the greatest odd divisor of f n 1 + f n 2. Show that f n is constant for n sufficiently large and determine the eventual value as a function of a and b. (1993 USAMO) 5. Let n a 1 > a 2 > > a k be ositive integers such that lcm(a i, a j ) n for all i, j. Prove that ia i n for i = 1, 2,..., k. 3 Arithmetic Functions There are several imortant arithmetic functions, of which three are resented here. If the rime factorization of n > 1 is e 1 1 e 2 2 e k k, then the number of ositive integers less than n, relatively rime to n, is ) ) ) φ(n) = (1 ( (1 1k n the number of divisors of n is = e e e k 1 k ( 1 1)( 2 1) ( k 1), and the sum of the divisors of n is τ(n) = (e 1 + 1)(e 2 + 1) (e k + 1), σ(n) = ( e e )( e e ) ( e k + e k 1 ( e 1 +1 = k k + + 1) ) ( e ) ( e k +1 k 1 k 1 Also, φ(1), τ(1), and σ(1) are defined to be 1. We say that a function f is multilicative if f(mn) = f(m)f(n) for all relatively rime ositive 10 ).
11 integers m and n, and f(1) = 1 (otherwise, f(1) = 0, which imlies that f(n) = 0 for all n). Theorem 3.1. The functions φ, τ, and σ are multilicative. Hence, by taking the rime factorization and evaluating at each rime ower, the formula above are found easily. Examle 3.1. Find the number of solutions in ordered airs of ositive integers (x, y) of the equation where n is a ositive integer. Solution. From the given, 1 x + 1 y = 1 n, 1 x + 1 y = 1 n xy = nx + ny (x n)(y n) = n2. If n = 1, then we immediately deduce the unique solution (2,2). For n 2, let n = e 1 1 e 2 2 e k k be the rime factorization of n. Since x, y > n, there is a 1-1 corresondence between the solutions in (x, y) and the factors of n 2, so the number of solutions is τ(n 2 ) = (2e 1 + 1)(2e 2 + 1) (2e k + 1). Examle 3.2. Let n be a ositive integer. Prove that φ(d) = n. d n Solution. For a divisor d of n, let S d be the set of all a, 1 a n, such that gcd(a, n) = n/d. Then S d consists of all elements of the form b n/d, where 0 b d, and gcd(b, d) = 1, so S d contains φ(d) elements. Also, it is clear that each integer between 1 and n belongs to a unique S d. The result then follows from summing over all divisors d of n. Problems 1. Let n be a ositive integer. Prove that n n n τ(k) =. k k=1 11 k=1
12 2. Let n be a ositive integer. Prove that τ 3 (d) = τ(d) d n d n 3. Prove that if σ(n) = 2N + 1, then N is the square of an odd integer. (1976 Putnam Mathematical Cometition) 4 Modular Arithmetic For a ositive integer m and integers a and b, we say that a is congruent to b modulo m if m (a b), and we denote this by a b modulo m, or more commonly a b (mod m). Otherwise, a is not congruent to b modulo m, and we denote this by a b (mod m) (although this notation is not used often). In the above notation, m is called the modulus, and we consider the integers modulo m. Theorem 4.1. If a b and c d (mod m), then a + c b + d (mod m) and ac bd (mod m). Proof. If a b and c d (mod m), then there exist integers k and l such that a = b + km and c = d + lm. Hence, a + c = b + d + (k + l)m, so a + c b + d (mod m). Also, so ac bd (mod m). Useful Facts For all integers n, ac = bd + dkm + blm + klm 2 n 2 For all integers n, 0 n = bd + (dk + bl + klm)m, { 0 1 } { if n is even, (mod 4) if n is odd. (mod 8) 2 if n 0 (mod 4), if n 2 (mod 4), if n 1 (mod 2).. 12
13 If f is a olynomial with integer coefficients and a b (mod m), then f(a) f(b) (mod m). If f is a olynomial with integer coefficients of degree n, not identically zero, and is a rime, then the congruence f(x) 0 (mod ) has at most n solutions modulo, counting multilicity. Examle 4.1. Prove that the only solution in rational numbers of the equation x 3 + 3y 3 + 9z 3 9xyz = 0 is x = y = z = 0. (1983 Kürschák Cometition) Solution. Suose that the equation has a solution in rationals, with at least one non-zero variable. Since the equation is homogeneous, we may obtain a solution in integers (x 0, y 0, z 0 ) by multilying the equation by the cube of the lowest common multile of the denominators. Taking the equation modulo 3, we obtain x (mod 3). Therefore, x 0 must be divisible by 3, say x 0 = 3x 1. Substituting, 27x y z x 1 y 0 z 0 = 0 y z x 3 1 9x 1 y 0 z 0 = 0. Therefore, another solution is (y 0, z 0, x 1 ). We may then aly this reduction recursively, to obtain y 0 = 3y 1, z 0 = 3z 1, and another solution (x 1, y 1, z 1 ). Hence, we may divide owers of 3 out of our integer solution an arbitrary number of times, contradiction. Examle 4.2. Does one of the first Fibonacci numbers terminate with 4 zeroes? Solution. The answer is yes. Consider the sequence of airs (F k, F k+1 ) modulo Since there are only a finite number of different ossible airs (10 8 to be exact), and each air is deendent only on the revious one, this sequence is eventually eriodic. Also, by the Fibonacci relation, one can find the revious air to a given air, so this sequence is immediately eriodic. But F 0 0 (mod 10 4 ), so within 10 8 terms, another Fibonacci number divisible by 10 4 must aear. 13
14 In fact, a comuter check shows that 10 4 F 7500, and (F n ) modulo 10 4 has eriod 15000, which is much smaller than the uer bound of If ax 1 (mod m), then we say that x is the inverse of a modulo m, denoted by a 1, and it is unique modulo m. Theorem 4.2. The inverse of a modulo m exists and is unique iff a is relatively rime to m. Proof. If ax 1 (mod m), then ax = 1+km for some k ax km = 1. By Corollary 2.2, a and m are relatively rime. Now, if gcd(a, m) = 1, then by Corollary 2.2, there exist integers x and y such that ax + my = 1 ax = 1 my ax 1 (mod m). The inverse x is unique modulo m, since if x is also an inverse, then ax ax 1 xax xax x x. Corollary 4.3. If is a rime, then the inverse of a modulo exists and is unique iff does not divide a. Corollary 4.4. If ak bk (mod m) and k is relatively rime to m, then a b (mod m). Proof. Multilying both sides by k 1, which exists by Theorem 4.2, yields the result. We say that a set {a 1, a 2,..., a m } is a comlete residue system modulo m if for all i, 0 i m 1, there exists a unique j such that a j i (mod m). Examle 4.3. Find all ositive integers n such that there exist comlete residue systems {a 1, a 2,..., a n } and {b 1, b 2,..., b n } modulo n for which {a 1 + b 1, a 2 + b 2,..., a n + b n } is also a comlete residue system. Solution. The answer is all odd n. First we rove necessity. For any comlete residue system {a 1, a 2,..., a n } modulo n, we have that a 1 + a a n n(n + 1)/2 (mod n). So, if all three sets are comlete residue systems, then a 1 +a 2 + +a n +b 1 +b 2 + +b n n 2 +n 0 (mod n) and a 1 + b 1 + a 2 + b a n + b n n(n + 1)/2 (mod n), so n(n + 1)/2 0 (mod n). The quantity n(n + 1)/2 is divisible by n iff (n + 1)/2 is an integer, which imlies that n is odd. Now assume that n is odd. Let a i = b i = i for all i. Then a i + b i = 2i for all i, and n is relatively rime to 2, so by Corollary 4.4, {2, 4,..., 2n} is a comlete residue system modulo n. Theorem 4.5. Euler s Theorem. If a is relatively rime to m, then a φ(m) 1 (mod m). 14
15 Proof. Let a 1, a 2,..., a φ(m) be the ositive integers less than m that are relatively rime to m. Consider the integers aa 1, aa 2,..., aa φ(m). We claim that they are a ermutation of the original φ(m) integers a i, modulo m. For each i, aa i is also relatively rime to m, so aa i a k for some k. Since aa i aa j a i a j (mod m), each a i gets taken to a different a k under multilication by a, so indeed they are ermuted. Hence, a 1 a 2 a φ(m) (aa 1 )(aa 2 ) (aa φ(m) ) a φ(m) a 1 a 2 a φ(m) 1 a φ(m) (mod m). Remark. This gives an exlicit formula for the inverse of a modulo m: a 1 a φ(m) 2 (mod m). Alternatively, one can use the Euclidean algorithm to find a 1 x as in the roof of Theorem 4.2. Corollary 4.6. Fermat s Little Theorem (FLT). If is a rime, and does not divide a, then a 1 1 (mod ). Examle 4.4. Show that if a and b are relatively rime ositive integers, then there exist integers m and n such that a m + b n 1 (mod ab). Solution. Let S = a m + b n, where m = φ(b) and n = φ(a). Then by Euler s Theorem, S b φ(a) 1 (mod a), or S 1 0 (mod a), and S a φ(b) 1 (mod b), or S 1 0 (mod b). Therefore, S 1 0, or S 1 (mod ab). Examle 4.5. For all ositive integers i, let S i be the sum of the roducts of 1, 2,..., 1 taken i at a time, where is an odd rime. Show that S 1 S 2 S 2 0 (mod ). Solution. First, observe that (x 1)(x 2) (x ( 1)) = x 1 S 1 x 2 + S 2 x 3 S 2 x + S 1. This olynomial vanishes for x = 1, 2,..., 1. But by Fermat s Little Theorem, so does x 1 1 modulo. Taking the difference of these two olynomials, we obtain another olynomial of degree 2 with 1 roots modulo, so it must be the zero olynomial, and the result follows from comaring coefficients. 15
16 Remark. We immediately have that ( 1)! S 1 1 (mod ), which is Wilson s Theorem. Also, x x 0 (mod ) for all x, yet we cannot comare coefficients here. Why not? Theorem 4.7. If is a rime and n is an integer such that (4n 2 + 1), then 1 (mod 4). Proof. Clearly, cannot be 2, so we need only show that 3 (mod 4). Suose = 4k + 3 for some k. Let y = 2n, so by Fermat s Little Theorem, y 1 1 (mod ), since does not divide n. But, y , so y 1 y 4k+2 (y 2 ) 2k+1 ( 1) 2k+1 1 (mod ), contradiction. Therefore, 1 (mod 4). Remark. The same roof can be used to show that if is a rime and (n 2 + 1), then = 2 or 1 (mod 4). Examle 4.6. Show that there are an infinite number of rimes of the form 4k + 1 and of the form 4k + 3. Solution. Suose that there are a finite number of rimes of the form 4k + 1, say 1, 2,..., n. Let N = 4( 1 2 n ) By Theorem 4.7, N is only divisible by rimes of the form 4k + 1, but clearly N is not divisible by any of these rimes, contradiction. Similarly, suose that there are a finite number of rimes of the form 4k + 3, say q 1, q 2,..., q m. Let M = 4q 1 q 2 q m 1. Then M 3 (mod 4), so M must be divisible by a rime of the form 4k + 3, but M is not divisible by any of these rimes, contradiction. Examle 4.7. Show that if n is an integer greater than 1, then n does not divide 2 n 1. Solution. Let be the least rime divisor of n. Then gcd(n, 1) = 1, and by Corollary 2.2, there exist integers x and y such that nx+( 1)y = 1. If (2 n 1), then 2 2 nx+( 1)y (2 n ) x (2 1 ) y 1 (mod ) by Fermat s Little Theorem, contradiction. Therefore, (2 n 1) n (2 n 1). Theorem 4.8. Wilson s Theorem. If is a rime, then ( 1)! 1 (mod ). (See also Examle 4.5.) Proof. Consider the congruence x 2 1 (mod ). Then x 2 1 (x 1)(x + 1) 0, so the only solutions are x 1 and 1. Therefore, for each i, 2 i 2, there exists a unique inverse j i of i, 2 j 2, modulo 16
17 . Hence, when we grou in airs of inverses, ( 1)! 1 2 ( 2) ( 1) ( 1) 1 (mod ). Examle 4.8. Let {a 1, a 2,..., a 101 } and {b 1, b 2,..., b 101 } be comlete residue systems modulo 101. Can {a 1 b 1, a 2 b 2,..., a 101 b 101 } be a comlete residue system modulo 101? Solution. The answer is no. Suose that {a 1 b 1, a 2 b 2,..., a 101 b 101 } is a comlete residue system modulo 101. Without loss of generality, assume that a (mod 101). Then b (mod 101), because if any other b j was congruent to 0 modulo 101, then a j b j a 101 b (mod 101), contradiction. By Wilson s Theorem, a 1 a 2 a 100 b 1 b 2 b ! 1 (mod 101), so a 1 b 1 a 2 b 2 a 100 b (mod 101). But a 101 b (mod 101), so a 1 b 1 a 2 b 2 a 100 b ! 1 (mod 101), contradiction. Theorem 4.9. If is a rime, then the congruence x (mod ) has a solution iff = 2 or 1 (mod 4). (Comare to Theorem 7.1) Proof. If = 2, then x = 1 is a solution. If 3 (mod 4), then by the remark to Theorem 4.7, no solutions exist. Finally, if = 4k + 1, then let x = 1 2 (2k). Then x (2k) (2k) (2k) ( 2k) ( 2) ( 1) 1 2 (2k) ( 2k) ( 2) ( 1) ( 1)! 1 (mod ). (multilying by 2k 1s) Theorem Let be a rime such that 1 (mod 4). Then there exist ositive integers x and y such that = x 2 + y 2. Proof. By Theorem 4.9, there exists an integer a such that a 2 1 (mod ). Consider the set of integers of the form ax y, where x and y are integers, 0 x, y <. The number of ossible airs (x, y) is then ( + 1) 2 > ( ) 2 =, so by igeonhole rincile, there exist integers 0 x 1, x 2, y 1, y 2 <, such that ax 1 y 1 ax 2 y 2 (mod ). Let x = x 1 x 2 and y = y 1 y 2. At least one of x and y is non-zero, and ax y a 2 x 2 17
18 x 2 y 2 x 2 + y 2 0 (mod ). Thus, x 2 + y 2 is a multile of, and 0 < x 2 + y 2 < ( ) 2 + ( ) 2 = 2, so x 2 + y 2 =. Theorem Let n be a ositive integer. Then there exist integers x and y such that n = x 2 + y 2 iff each rime factor of n of the form 4k + 3 aears an even number of times. Theorem The Chinese Remainder Theorem (CRT). If a 1, a 2,..., a k are integers, and m 1, m 2,..., m k are airwise relatively rime integers, then the system of congruences x a 1 (mod m 1 ), x a 2 (mod m 2 ),. x a k (mod m k ) has a unique solution modulo m 1 m 2 m k. Proof. Let m = m 1 m 2 m k, and consider m/m 1. This is relatively rime to m 1, so there exists an integer t 1 such that t 1 m/m 1 1 (mod m 1 ). Accordingly, let s 1 = t 1 m/m 1. Then s 1 1 (mod m 1 ) and s 1 0 (mod m j ), j 1. Similarly, for all i, there exists an s i such that s i 1 (mod m i ) and s i 0 (mod m j ), j i. Then, x = a 1 s 1 + a 2 s a k s k is a solution to the above system. To see uniqueness, let x be another solution. Then x x 0 (mod m i ) for all i x x 0 (mod m 1 m 2 m k ). Remark. The roof shows exlicitly how to find the solution x. Examle 4.9. For a ositive integer n, find the number of solutions of the congruence x 2 1 (mod n). Solution. Let the rime factorization of n be 2 e e 1 1 e 2 2 e k k. By CRT, x 2 1 (mod n) x 2 1 (mod e i i ) for all i, and x2 1 (mod 2 e ). We consider these cases searately. We have that x 2 1 (mod e i i ) x2 1 = (x 1)(x + 1) 0 (mod e i i ). But i cannot divide both x 1 and x + 1, so it divides one of them; that is, x ±1 (mod e i i ). Hence, there are two solutions. Now, if (x 1)(x + 1) 0 (mod 2 e ), 2 can divide both x 1 and x + 1, but 4 cannot divide both. For e = 1 and e = 2, it is easily checked that there are 1 and 2 solutions resectively. For e 3, since there is at most one factor 18
19 of 2 in one of x 1 and x + 1, there must be at least e 1 in the other, for their roduct to be divisible by 2 e. Hence, the only ossibilities are x 1 or x + 1 0, 2 e 1 (mod 2 e ), which lead to the four solutions x 1, 2 e 1 1, 2 e 1 + 1, and 2 e 1. Now that we know how many solutions each rime ower factor contributes, the number of solutions modulo n is simly the roduct of these, by CRT. The following table gives the answer: e Number of solutions 0, 1 2 k 2 2 k k+2 Theorem Let m be a ositive integer, let a and b be integers, and let k = gcd(a, m). Then the congruence ax b (mod m) has k solutions or no solutions according as k b or k b. Problems 1. Prove that for each ositive integer n there exist n consecutive ositive integers, none of which is an integral ower of a rime. (1989 IMO) 2. For an odd ositive integer n > 1, let S be the set of integers x, 1 x n, such that both x and x + 1 are relatively rime to n. Show that x 1 (mod n). x S 3. Find all ositive integer solutions to 3 x + 4 y = 5 z. (1991 IMO Short List) 4. Let n be a ositive integer such that n + 1 is divisible by 24. Prove that the sum of all the divisors of n is divisible by 24. (1969 Putnam Mathematical Cometition) 5. (Wolstenholme s Theorem) Prove that if
20 is exressed as a fraction, where 5 is a rime, then 2 divides the numerator. 6. Let a be the greatest ositive root of the equation x 3 3x = 0. Show that a 1788 and a 1988 are both divisible by 17. (1988 IMO Short List) 7. Let {a 1, a 2,..., a n } and {b 1, b 2,..., b n } be comlete residue systems modulo n, such that {a 1 b 1, a 2 b 2,..., a n b n } is also a comlete residue system modulo n. Show that n = 1 or Let m, n be ositive integers. Show that 4mn m n can never be a square. (1984 IMO Proosal) 5 Binomial Coefficients For non-negative integers n and k, k n, the binomial coefficient ( n k) is defined as n! k!(n k)!, and has several imortant roerties. By convention, ( n k) = 0 if k > n. In the following results, for olynomials f and g with integer coefficients, we say that f g (mod m) if m divides every coefficient in f g. Theorem 5.1. If is a rime, then the number of factors of in n! is n n n It is also n s n 1, where s n is the sum of the digits of n when exressed in base. Theorem 5.2. If is a rime, then ( ) 0 (mod ) i 20
21 for 1 i 1. Corollary 5.3. (1 + x) 1 + x (mod ). Lemma 5.4. For all real numbers x and y, x + y x + y. Proof. x x x + y x + y Z, so x + y x + y. Theorem 5.5. If is a rime, then ( ) k 0 (mod ) i for 1 i k 1. Proof. By Lemma 5.4, k ( ) i k i + j j j=1 k k, where the LHS and RHS are the number of factors of in i!( k i)! and k i! resectively. But, = k i = 0 and k = 1, so the inequality is k k k strict, and at least one factor of divides ( ) k i. Corollary 5.6. (1 + x) k 1 + x k j=1 (mod ). Examle 5.1. Let n be a ositive integer. Show that the roduct of n consecutive ositive integers is divisible by n!. Solution. If the consecutive integers are m, m + 1,..., m + n 1, then ( ) m(m + 1) (m + n 1) m + n 1 =. n! n Examle 5.2. Let n be a ositive integer. Show that (( ) ( ) ( )) n n n (n + 1) lcm,,..., = lcm(1, 2,..., n + 1). 0 1 n (AMM E2686) Solution. Let be a rime n + 1 and let α (resectively β) be the highest ower of in the LHS (resectively RHS) of the above equality. Choose r so that r n + 1 < r+1. Then clearly β = r. We claim that ( ) m if r m < r+1, then r+1 for 0 k m. ( ) k 21 j
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Domain Name: EBISAWA-SS.COM
Registry Domain ID: 31162195_DOMAIN_COM-VRSN
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Registrant Name: JUNICHI TAMINO
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No approved endorsements
About Ms. Ruby Henderson
I have spent the last 19 years in Real Estate and love helping people with one of the biggest decision of their lives! With being in this market for over 40 years, I have come to know it very well! As a former social worker, I enjoy helping people and pets too!
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Pulsar Swarovski? crystals Mother-of-pearl Dial Womens watch #PEGF19
In stock
Pulsar Swarovski? crystals Mother-of-pearl Dial Women's watch #PEGF19 : Dress watch, Japanese quartz movement, Polished silver-tone sword hands, Applied polished silver-tone Roman numeral at twelve o'clock, Applied polished silver-tone triangular hour markers, Black indices form surrounding minute track, Pink mother-of-pearl dial with polished black logo applique below twelve o'clock, Polished silver-tone stainless steel bracelet with adjustable links and jewelry-clasp, Polished silver-tone stainless steel round case encrusted with genuine Swarovski? crystals, Textured steel crown, Stainless steel caseback, Mineral crystal, Water resistant
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Belfort Furniture offers White Glove in-home delivery and set-up by professionally trained personnel who will unwrap, place, and assemble your furniture in the appropriate room. Delivery is available Tuesday - Saturday. Once your delivery date is scheduled, you will receive a three hour delivery timeframe the day prior to your delivery.
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For delivery outside of our local metropolitan area, please contact us with your item(s) of interest for a delivery quote! 703-406-7600 or firstname.lastname@example.com
Create a beautiful display that reflects your style or worldly adventures with this bookcase. The piece offers five fixed shelves to display your favorite pieces, and is crafted from flat cut figured prima vera veneers. A Dark Mink finish provides rich, sophisticated tones, and beautifully offsets the drawer's silver eglomise glass drawer front.
The Sutton House Bookcase with 5 Shelves by Bernhardt from Belfort Furniture. We proudly serve the Washington DC, Northern Virginia, Maryland and Fairfax VA area. Product availability may vary. Contact us for the most current availability on this product.
Celebrating classic New England designs through a contemporary perspective, Sutton House's neo-traditional furniture proves that sometimes you have to look back to move forward. The collection's high end style starts with flat cut figured prima vera veneers in a rich Dark Mink finish, with Venetian mirror accents and silver leaf elements that brighten the collection and lend an element of glamour to select pieces. Silver eglomise detailing in three-dimensional diamond and cube-shaped patterns bring a sense of luxury and texture to the collection, while rosettes and sunbursts reference traditional elegance. A masterclass in scale and proportion, Sutton House reinterprets traditional style to make a modern statement, for a look that's as timeless and refined as the classic designs that inspired it.
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These images are representative of this group as a whole and may differ from this item.
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text: Ericsson and Zain Bahrain sign managed services agreement08 Oct 2013 10:44 AM Zain selects Ericsson for a four-year managed service agreement to manage, plan and optimise its network in BahrainZain Bahrain has signed a four-year managed services agreement with Ericsson, to manage its network and enhance its customer support services.The new partnership will see Ericsson managing the day-to-day operations of Zain's network across the country, enabling operational excellence. In parallel, Zain will be able to allocate more resources towards higher-end services and offerings allowing subscribers to ultimately benefit from an increased network quality and stability."At Zain, providing customers with advanced technologies that enhance their mobile experience is of a paramount importance. With that ultimate aim in mind, we selected Ericsson to manage our network operations," said Mohammed Zainalabedin, general manager, Zain in Bahrain."This latest partnership with Ericsson helps to increase our internal productivity, and the result of that efficiency will be reflected in what our subscribers will experience of superior communications that they have become accustomed to."As a telecoms operator, we are confident that this managed services solution will mean that our network and business complexity will be well-managed by Ericsson, leading to cost-effective launches of new services, dependable performance and operational competence," added Zainalabedin.Ericsson's Customer Support solutions ensure customers only receive high-quality services, supporting the overall network quality.Ray Hassan, executive vice president and head of the global customer unit Zain at Ericsson, said: "As we transition towards the networked society, when everything that benefits from being connected will be, new opportunities are opening up for communication in terms of innovation and productivity."By deploying this differentiated solution, Zain Bahrain's network will operate at upgraded productivity levels, enabling the operator to continue innovating and providing customers with a premium mobile experience."Wireless magazineLeave a CommentPlease write code to prove you're humanMore on... Critical CommunicationsFinland and Norway sign cross-border TETRA communication dealCC Europe 2017: Hytera showcasing new base station and handheld radioMore on... EricssonMWC17: Ericsson expands 5G access and transport portfolioOperators trial intercontinental 5G roaming using network slicing


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If an item you ordered from TextbookRush does not meet your expectations due to an error on our part, simply fill out a return request and then return it by mail within 30 days of ordering it for a full refund of item cost.
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Description: Their mission: To take out the scuds. Eight went out. Five came back. Their story had been closed in secrecy. Until now. They were British Special Forces, trained to be the best. In January 1991 a squad of eight men went behind the Iraqi lines on a top secret mission. It was called Bravo Two Zero. On command was Sergeant Andy McNab. "They are the true unsung heroes of the war." -- Lt. Col. Steven Turner, American F-15E commander. Dropped into "scud alley" carrying 210-pound packs, McNab and his men found themselves surrounded by Saddam's army. Their radios didn't work. The weather turned cold enough to freeze diesel fuel. And they had been spotted. Their only chance at survival was to fight their way to the Syrian border seventy-five miles to the northwest and swim the Euphrates river to freedom. Eight set out. Five came back. "I'll tell you who destroyed the scuds -- it was the British SAS. They were fabulous." -- John Major, British Prime Minister. This is their story. Filled with no-holds-barred detail about McNab's capture and excruciating torture, it tells of men tested beyond the limits of human endurance... and of the war you didn't see on CNN. Dirty, deadly, and fought outside the rules.
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List price: $7.99Copyright year: 1993Publisher: Random House Publishing GroupPublication date: 8/1/1994Binding: PaperbackPages: 432Size: 4.25" wide x 6.75" long x 1.25" tall Weight: 0.704Language: English
A former member of the crack elite force the Special Air Service (SAS), Andy McNab was involved in both covert & overt special operations on five continents, including joint ops with Delta Force, the FBI's HRT & the DEA. McNab was the British Army's most highly decorated serving soldier when he left in 1993. After the non-fiction books based on his SAS experiences, "Bravo Two Zero" (1993) & "Immediate Action" (1995), McNab turned to fiction with "Remote Control" (1997) & "Crisis Four" (1999). He lectures to security agencies & remains closely involved with the intelligence communities on both sides of the Atlantic.
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I'm a twenty-six years old graphic design student from Finland.
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High Tech Policy
- High Tech Policy
Aspen Summit: Progress and Freedom Foundation (PFF)
I just returned from several days in Aspen from PFF's 10th annual Aspen Summit…PFF is a think tank that advocatesRead More
- High Tech Policy
Trade Test Teaches Important Lesson
Last night, the U.S. House of Representatives narrowly passed, 217-215, the U.S.-Central America-Dominican Republic Free Trade Agreement (so-called CAFTA-DR), H.R.Read More
- High Tech Policy
Vlogging: the next big thing?
Read today's New York Times story on video logging or "vlogging." (Free registration required to access this site.)I, of course,Read More
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Cisco Chairman John Morgridge Testifies Before U.S. House Science Committee
Yesterday, Cisco Chairman John Morgridge testified before the House Science Committee at a hearing entitled, "U.S. Competitiveness: The Innovation Challenge."Read More
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What was life like before all these 1's and 0's…
I was talking to my wife the other day about sending her dad some information and she said she wouldRead More
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U.S. Independence Day 2005: What will life be like in another 229 years?
On July 4th, 2005, the United States celebrates its 229th birthday. As we approach Independence Day, it made me thinkRead More


text:
Hello, Dearest Accordance iOS developers:
First, just to let you know, I love the work you've done porting Accordance to iOS so much that I've dropped my laptop altogether (passed it on to my wife, actually) and I now ONLY use my iPad. I've known that I would lose some functionality that way, but I haven't really missed anything up until now� a testament to the work you've done.
But today I missed something for the first time really. A friend of mine is doing a paper for school and wanted to know if I could recommend any good resources for the life and times surrounding the second temple period. Naturally, I wanted to recommend "The Temple: Its Ministry and Services" by A. Edersheim to him. I tried to figure out a way to easily extract the bibliographical information for him so he could find a hard copy at his library. Unfortunately, there doesn't seem to be a way to do that with Accordance iOS.
Some tools have that information written in at the beginning of the module (like a normal book would� "The Complete Gathered Gold", for example), but even then you'd have to copy by handwriting it, or line-by-line as copying a "paragraph" doesn't work with all the carriage returns in those sections of the book.
Further, I found that when copying a quote from a tool, such as "The Temple...", and pasting that quote into pages or some such thing, Accordance iOS does not automatically include some kind of citation. Of course, when we copy a Bible passage, then a proper citation shows up.
All this is a rather long winded way of saying that I would really value the ability to extract bibliographical resources from Accordance iOS modules easily so that I could cite, or refer friends to the wonderful resources I enjoy.
Thanks again for all your work folks. You're doing a great job... keep it up!
Citations and Bibliographical information for iOS
Started by dgzylstra , Sep 26 2012 12:04 PMcitations bibliography copy paste
2 replies to this topic
Posted 26 September 2012 - 12:04 PM
Posted 26 September 2012 - 12:11 PM
It's definitely on the "to do" list, but I don't have an estimate for when this will show up in a release of the iOS app.
Thanks for the feedback,
Thanks for the feedback,
Posted 05 April 2013 - 03:32 PM
Thanks for posting this and for what its worth would make a similar request. I am working on a commentary for Jonah and the citation ability would be very helpful. Thanks too for a great app that I am using exclusively for this seminary project.
Joshua 1:9 "Have I not commanded you be strong and courageous, do not be terrified, do not be discouraged for the Lord your God is with you wherever you go" (ESV)
Also tagged with one or more of these keywords: citations, bibliography, copy, paste
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Started by ku.elliot , 16 May 2017 citation, bibliography
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Started by farcas , 12 May 2017 copy, citation, formating
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Started by Friarpeabody , 21 Feb 2017 copy, paste, parallels
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Generate a Bibliography of All Accordance Resources
Started by Sabra Knape , 20 Jan 2017 resource list, bibliography
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Reading mode: Light Dark
Olympus Stylus 600
Announced Aug 29, 2005 •
6 megapixels | 2.5″ screen | 35 – 105 mm (3×)
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Announced Aug 29, 2005
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Aug 29, 2005
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|Body type
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|2816 x 2112
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|6 megapixels
|Sensor size
|1/2.5" (5.744 x 4.308 mm)
|Sensor type
|Auto, 64, 100, 200, 400, 500, (2MP; 800, 1600)
|Focal length (equiv.)
|35–105 mm
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|Screen size
|Screen dots
|Max shutter speed
|1/1000 sec
|Storage types
|xD Picture Card, Internal
|USB 1.0 (1.5 Mbit/sec)
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|195 g (0.43 lb / 6.88 oz)
|97 x 55 x 27 mm (3.82 x 2.17 x 1.06″)
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A little bit about us @ Capfoto
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Heavy Industry»Materials & Chemicals»Paper & Forest Products»Sawmill
U.S. Sawmill, Woodworking, And Paper Machinery Market. Analysis And Forecast to 2020The report provides an in-depth analysis of the U.S. Sawmill, Woodworking, And Paper Machinery market. It presents the latest data of the market size and volume, domestic production, exports and imports, price dynamics and turnover in the industry. In addition, the report contains insightful information about the industry, including industry life cycle, business locations, productivity, employment and many other crucial aspects. The Company Profiles section contains relevant data on the major players in the industry. Countries coverage: the U.S. Product coverage: Woodworking machinery, including parts, attachments, and accessories, excluding for home workshops, garages, and service shops (excluding chainsaws and other power-driven handtools); Woodworking machinery for home workshops, garages, and service shops (excluding chainsaws and other power-driven handtools); Paper industry machinery; Parts and attachments for paper industry machinery (sold separately); Sawmill, woodworking, and paper machinery manufacturing, nsk, total Companies mentioned: Andritz (usa) , Valmet, Alliance Machine Systems International , Kadant Inc., Paper Converting Machine Company, Simonds International , Curt G. Joa, Maxcess International Corp, Westervelt Company, Barry-Wehmiller Papersystems, Paper Machinery Corporation, Wood-Mizer Holdings, Andritz, C.G. Bretting Manufacturing Co., The Ward Machinery Company, Tecnau, GL&v USA , ASC Machine Tools, G L & V USA , Globe Machine Manufacturing Company, Kirk-Rudy, Metso Usa, Selway Corporation, Voith Paper Inc., Marquip , Voith Paper Fabric and Rolls Systems, Glv US Holding , Maxcess International Holding Corp., Will-Pemco, W + D Machinery Co. Data coverage:
Learn how to effectively navigate the market research process to help guide your organization on the journey to success.


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Our brewery is located at the intersection NE Sandy & Couch, if you blink, you'll miss it.
We are just a production facility and currently do not have a tasting room. Check our "On Tap!" page to see where we're being poured around town!
Ian McGuinness – owner, brewer, all around okay guy.
– Ian McGuinness
First brewery in Portland, Oregon to package beer in aluminum cans! We are now operating on a 10bbl system after upgrading from our little 1bbl nano-brewery. After 3 years of churning out tiny, little, mini "nano" batches on a non-stop basis, we actually get to have a little time left over for events and minor necessities like sleep.
Located near the NE Sandy Blvd/Burnside intersection we're brewing away in the big bay windows you pass right when Sandy turns into Couch St. We are monkeys on display through the brewery windows & are always willing to put on a show for anyone that stops by to peer/leer/stare in.
Due to our recently increased size, we are being found in more and more locations around Portland and the surrounding areas, including Vancouver WA. But wait, there's more! We are also available statewide in Pennsylvania!
To keep updated on what's offered and where, please check out the "On Tap!" page in which we try to keep as updated as humanly possible.
Even though you may have to go out of your way to grab a pint, we hope you enjoy it!


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We have a new Chobani Yogurt printable coupon! Both this week and next, the 5.3 oz greek yogurt cups are on sale for $1. We also have a Target Cartwheel offer to use as well. Combine everything to pay just $0.62 at Target.
$1/3 Chobani Simply 100 Yogurt 4.2-5.3 oz printable manufacturer's coupon
At Target through 1/16:
BUY THREE CHOBANI SIMPLY 100 GREEK YOGURT 5.3 OZ, on sale $1
Use the 5% OFF Chobani Greek Yogurt & Simply 100 5.3 oz, 32 oz or 4-pack Target Cartwheel Offer (exp 2/13; deducts $0.15)
and use one $1/3 Chobani Simply 100 Yogurt 4.2-5.3 oz printable manufacturer's coupon
NET PRICE FOR ALL: $1.85 (or 62 cents each!)
Thanks Totally Target.Extreme Couponing Tip: You can usually get two prints of most internet coupons. To find them in the easiest possible way, use our preclipped links that take you right to the coupon. Then come back and click the same link to get your 2nd print. This will save you the time spent searching for the coupon.
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Red artificial vertical garden highlights are created with Garden Beet's freshly harvested fake berry and currents. We create rich colours and texture. Each square meter is hand finished by Melbourne artists who artfully arrange each and every feature stem. No two gardens are the same and we can also make to order.
These artificial vertical gardens are ideal for indoor locations such as a hospital, kindergarten, retail outlet restaurant or you very own bedroom.
Our artificial vertical gardens can be located outdoors however we do not recommend full sun. A UV resistant finish can be applied nevertheless we seriously question the longevity of synthetic plants in the Australian sun.
Pricing is per square meter.


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speckled belly goose sounds download 1dFI free file
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Duck hunters hunt Specklebelly Geese due to the birds' responsiveness to calling, decoy placement and the appeal of obtaining a trophy bird.Product Description. Why do Specklebelly hunters rave about this goose call? 1. Pure Sound – This call is designed to create the agressive, clear, piercing sound of ...5/30/2011 · Ricelandcustomcalls.com, Specklebelly Calling TipsOur Specklebelly call was designed to be a simple, versatile, easy blowing, goose fool'n, "Katy bar the door", Speck call. We know, because we hunt them ourselves.It's awfully quiet in here, you fellas killin' birds? Have seen people on other forums advise new speck hunters not to call too much. Seen ...Shop RNT Goose Calls at MacksPW.com. Large Selection and Quality Customer Service! We are Americas Premier Waterfowl Outfitter!Shop Goose Calls at MacksPW.com. Large Selection and Quality Customer Service! We are Americas Premier Waterfowl Outfitter!9/4/2007 · How to Use a Goose Call. Along with selecting a prime location and arranging decoys in a natural pattern, calling is one of the essential components of ...Rotary Machines / Bouncers | Rotary Machines - Hunting Snows / Snow Goose Decoy Shop Tamarack, MN 218-341-4477We offer a wide selection of cheap goose calls, cheap goose decoys and goose layout blinds for sale. Including name brands like Foiles, Tim Grounds, Avery, Big Foot ...
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The Lewis & Clark Motel, 824 West Main Street, Bozeman, MT., is evidence that local ownership (not just a local franchise of a national chain) makes a difference. Rates are reasonable, accommodations are fine, but the real value lies in the home-like feeling of the place. The theme is tradition; the music is old fashioned (e.g., Sinatra); the fitness center actually is a "rec room" in the basement (complete with fooseball); the lobby has a jig-saw puzzle in progress for any to work on, a collection of antique Life magazines, and the proprietor's doll collection. The proprietor herself, Lisa, the daughter of the founder, succeeds in warmth, customer service, and competence. I am especially indebted to her for single-handedly managing a fouled up FedEx delivery of an important package. If you're ever in Bozeman, stay at the Lewis & Clark Motel and enjoy the local ambiance!
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Conde Naste Lease Finalizes 1WTC Lease
I don't typically comment much on the New York City office market. However, the events of September 11, 2001, will never be forgotten and the World Trade Center is such an iconic piece of real estate for our country. Today the New York Times is reporting that Conde Naste Publications has finalized a much anticipated lease for one million square feet of office space at 1 World Trade Center in a lease transaction estimated to be valued at $2 billion, over the 25 year lease term. Read more at The New York Times


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Lightning Jack Kane likes to think of himself as the meanest cowboy in the Wild West, sadly no-one else does. So to rectify his reputation he recruits a sidekick, mute cowboy Ben Doyle (Gooding Jr.) and the two perform various stunts across the West.
Hitching a ride on the current cowboy bandwagon Paul Hogan is Lightning Jack Kane, allegedly the meanest, fastest gun in the West. Unfortunately this reputation is all in Jack's mind and he can't seem to get his picture on a wanted poster much less a price on his head. After failing to carry out a bank robbery on his own with no lookout, the customers keep coming through the door Jack finds himself a partner in the shape of the mute Ben Doyle (Gooding Jr.) and the duo spend the rest of the film trying to improve their outlaw image.
Writing one half of the film's double act as a mute was an adventurous move which would have paid off had Hogan's script been snappier. But with Ben unable to provide quick-witted retorts in the best buddy movie tradition, the majority of Jack's lines fall flat. And while Gooding's facial contortions make his a fairly amusing performance to watch, their relationship means Hogan gets centre stage, grabbing the best (and indeed the only) lines for himself.
The story also suffers from a distinct lack of excitement: when Jack and Ben are pursued by Comanche, the chase ends with the Indians getting bored and giving up, rather than with any thrilling heroics. It's a shame too that D'Angelo has to put up with a rather lame role as Jack's saloon sweetheart, always ready with a soothing bedside manner. After the abysmal Almost An Angel, Hogan clearly needed a boost. Unfortunately this isn't it. Perhaps next time he should let someone else write the script.
Paul Hogan had a successful turn as Crocodile Dundee but it appears that was a one hit wonder as he has appeared in flop after flop ever since and that includes Crocodile Dundee II and III. Gooding Jr. gives impressive support as a mute but sadly gets too few of the jokes.


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Books > Other
THE SCIENCE OF GETTING RICH: Financial Success Through Creative Thought(SKU :
Wallace D. WattlesPublisher:
Cosimo ClassicsISBN:
All books will be dispatched the next working day after receipt of the order.
Refunds & Exchange:
We offer a 'No Quibble' exchange or refund policy if the item you purchased does not meet your expectations in any way. In the unlikely event that you do need to return an item please contact us for details as to the easiest way to do this.


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Black Aggie is the folkloric name given to a statue formerly placed on the grave of General Felix Agnus in Druid Ridge Cemetery in Pikesville, Maryland. It is an unauthorized replica, rendered by Edward L. A. Pausch, of Augustus St. Gaudens' allegorical figure, popularly called Grief, at the Adams Memorial in Rock Creek Cemetery in Washington, D.C.[1]
The statue is of a seated figure in a cowl or shroud.
Beginning with its installation in 1926, the statue was surrounded by many urban legends.
Principally that someone spending a night in its lap would be haunted by the ghosts of those buried there; that the spirits of individuals buried at Druid Ridge would annually convene at the statue; that no grass would grow on the ground where the statue's shadow would lay during the daytime; or that the statue would animate itself during the night, whether by physically moving or by showing glowing red eyes.
These urban legends led to much unwelcome attention toward the statue; many people were caught breaking into the cemetery at night to visit it, and the pedestal was frequently vandalized. The Agnus family, disturbed by the attention the statue received, donated it to the Smithsonian in 1967. It sat for many years in storage at the National Museum of American Art (later named the Smithsonian American Art Museum) where an authorized recasting of the original Adams Memorial statue now sits.[2] Black Aggie was moved from her previous home at the museum to a courtyard behind the Dolley Madison House on Lafayette Square in Washington, D.C. where she currently stands. The bare, blank pedestal remains at the statue's former home at Druid Ridge Cemetery.[1]
The Pasta
When Felix Agnus put up the life-sized shrouded bronze statue of a grieving angel, seated on a pedestal, in the Agnus family plot in the Druid Ridge Cemetery, he had no idea what he had started. The statue was a rather eerie figure by day, frozen in a moment of grief and terrible pain. At night, the figure was almost unbelievably creepy; the shroud over its head obscuring the face until you were up close to it. There was a living air about the grieving angel, as if its arms could really reach out and grab you if you weren't careful.
It didn't take long for rumors to sweep through the town and surrounding countryside. They said that the statue - nicknamed Black Aggie - was haunted by the spirit of a mistreated wife who lay beneath her feet. The statue's eyes would glow red at the stroke of midnight, and any living person who returned the statues gaze would instantly be struck blind. Any pregnant woman who passed through her shadow would miscarry. If you sat on her lap at night, the statue would come to life and crush you to death in her dark embrace. If you spoke Black Aggie's name three times at midnight in front of a dark mirror, the evil angel would appear and pull you down to hell. They also said that spirits of the dead would rise from their graves on dark nights to gather around the statue at night.
People began visiting the cemetery just to see the statue, and it was then that the local fraternity decided to make the statue of Grief part of their initiation rites. "Black Aggie" sitting, where candidates for membership had to spend the night crouched beneath the statue with their backs to the grave of General Agnus, became popular.
One dark night, two fraternity members accompanied new hopeful to the cemetery and watched while he took his place underneath the creepy statue. The clouds had obscured the moon that night, and the whole area surrounding the dark statue was filled with a sense of anger and malice. It felt as if a storm were brewing in that part of the cemetery, and to their chagrin, the two fraternity members noticed that gray shadows seemed to be clustering around the body of the frightened fraternity candidate crouching in front of the statue.
What had been a funny initiation rite suddenly took on an air of danger. One of the fraternity brothers stepped forward in alarm to call out to the initiate. As he did, the statue above the boy stirred ominously. The two fraternity brothers froze in shock as the shrouded head turned toward the new candidate. They saw the gleam of glowing red eyes beneath the concealing hood as the statue's arms reached out toward the cowering boy.
With shouts of alarm, the fraternity brothers leapt forward to rescue the new initiate. But it was too late. The initiate gave one horrified yell, and then his body disappeared into the embrace of the dark angel. The fraternity brothers skidded to a halt as the statue thoughtfully rested its glowing eyes upon them. With gasps of terror, the boys fled from the cemetery before the statue could grab them too.
Hearing the screams, a night watchman hurried to the Agnus plot. To his chagrin, he discovered the body of a young man lying at the foot of the statue. The young man had apparently died of fright.
The disruption caused by the statue grew so acute that the Agnus family finally donated it to the Smithsonian museum in Washington D.C.. The grieving angel sat for many years in storage there, never again to plague the citizens visiting the Druid Hill Park Cemetery.