The Hasse Minkowski Theorem Lee Dicker University of Minnesota, REU Summer 2001
|
|
- Lucas Weaver
- 5 years ago
- Views:
Transcription
1 The Hasse Minkowski Theorem Lee Dicker University of Minnesota, REU Summer 2001 The Hasse-Minkowski Theorem rovides a characterization of the rational quadratic forms. What follows is a roof of the Hasse-Minkowski Theorem arahrased from the book, Number Theory by Z.I. Borevich and I.R. Shafarevich [1]. Throughout this aer, some familiarity with the -adic numbers and the Hilbert symbol is assumed and some basic facts about quadratic forms are stated without roof. Also, the roof of the Hasse Minkowski theorem given here uses the Dirichlet theorem on rimes in arithmetic rogressions. A roof of Dirichlet s theorem will not be given here (see [1], for a roof of the theorem) due to its length, but the result is stated resently. Theorem 0 (Dirichlet s theorem). Every residue class modulo m which consists of numbers relatively rime to m contains an infinite number of rime numbers. To begin, we state the definition of a quadratic form over a field K: Definition 1. A quadratic form, f, over the field K is a homogeneous olynomial of degree 2 with coefficients in K where f can be written n f = a ij x i x j, and a ij = a ji. i,j=1 Using the same notation as above, the symmetric matrix A = (a ij ) comletely determines the quadratic form f. We call A the matrix of f and we sometimes refer to det A as the determinant of f. Let X be the column vector of the variables x 1,..., x n, then we can write f = X t AX. The quadratic form g is equivalent to f = X t AX if the matrix of g, A 1, can be written as A 1 = C t AC, where C is an n n invertible matrix with entries in K. The quadratic form f is said to reresent zero in Kif there exist values α i K, not all zero such that f(α 1,..., α n ) = 0. Likewise, f reresents γ K if there are values α i K such that f(α 1,..., α n ) = γ. It is easily seen that two equivalent quadratic reresent the same elements in K. Furthermore, if a nonsingular quadratic form f (ı.e. the matrix of f is invertible) reresents zero in a field K, then f reresents all elements of K. Additionally, we note that every quadratic form is equivalent to a diagonal quadratic form. That is, if f is a quadratic form in n variables and A is the matrix of f, there exists an invertible matrix C such that the matrix C t AC is diagonal. This follows immediately from the Gram-Schmidt rocess from linear algebra. Also, it can be shown that if a diagonal quadratic form reresents zero in a field K and the number of elements of K is greater than five then there is a reresentation of zero where all the variables take on nonzero values in K. We are now reared to state the Hasse Minkowski Theorem: 1
2 Theorem 1 (Hasse Minkowski). A quadratic form with rational coefficients reresents zero in the field of rational numbers if and only if it reresents zero in the field of real numbers and in all fields of -adic numbers, Q (for all rimes ). The necessity of the condition is clear so we must show its sufficiency. In light of the facts that every quadratic form is equivalent to a diagonal quadratic form, and that reresenting zero is reserved when assing between equivalent quadratic forms, only diagonal quadratic forms need be considered in our roof. Essentially, the roof deends on the number of variables, n, of the quadratic form. For n = 1 the theorem is trivial. We divide the roof into four remaining cases. We consider searately when n = 2, n = 3, n = 4, or n 5. When n = 2, we first need a lemma ertaining to binary quadratic forms: Lemma 1. A binary quadratic form f(x, y) = ax 2 + 2bxy + cy 2 with determinant d = ac b 2 reresents zero in a field K if and only if d = α 2 for some α K. Sketch of roof. For the necessity of the condition, when d = 0 the roof is trivial. When d 0 we note the following two facts: A nonsingular binary quadratic form that reresents zero is equivalent to the form g = y 1 y 2 (see, for examle, [1]). Also, the determinants of equivalent quadratic forms differ by a nonzero factor which is a square in K. For the sufficiency, if f = ax 2 + by 2 and d = ab = α 2, then f(α, a) = 0. If the binary quadratic form f reresents zero in R and d is the determinant of f, it follows from Lemma 1 that d > 0, hence, d = k1 1 k2 2 ks s where k i is a rational integer. If f also reresents zero in Q for all rimes, d must a square in Q i (i = 1,..., s). Thus, k i must be even for i = 1,..., s and d is a square in Q. By Lemma 1, f reresents zero in Q. Before roving the cases where n 3 some things should be noted. The first of these observations follow from the fact that a quadratic form with rational integer coefficients f reresents zero if and only if cf reresents zero, where c is a nonzero constant in Q: throughout the roof Theorem 1, we may assume that the coefficients of the quadratic form f(x 1,..., x n ) are rational integers (if necessary, multily f by the least common multile of the denominators of the coefficients). Also, it is clear the equation f(x 1,..., x n ) = 0 (1) is solvable in Q (or in Q ) if and only if it is solvable in the rational integers, Z (resectively, in the -adic integers, Z ). We note that (1) is solvable in R if and only if the form f is indefinite. Moving on to the case n = 3, let f = a 1 x 2 + a 2 y 2 + a 3 z 2. Since f reresents zero in R, the coefficients a 1, a 2, a 3 do not all have the same sign. We may assume that two coefficients of f are ositive and one is negative (if necessary, we may multily f by 1). We may also assume that a 1, a 2, a 3 are rational integers, square-free (making a change of variables if necessary), and relatively rime. Suose a 1 and a 2 have a common rime factor, then multilying f by and setting x/ and y/ as new variables, we get a form with coefficients a 1 /, a 2 /, a 3. By reeating this rocess we obtain a quadratic form whose coefficients are ositive rational integers, airwise relatively rime and square-free: ax 2 + by 2 cz 2. (2) We now rove a theorem that is relevant to the case at hand and will also be needed again, later, in our roof of the Hasse Minkowski theorem. We begin with a lemma. 2
3 Lemma 2 (Hensel s lemma). Let F (x 1,..., x n ) be a olynomial whose coefficients are -adic integers. Let γ 1,..., γ n be -adic integers such that for some i (1 i n) we have then there exist -adic integers θ 1,..., θ n such that F (γ 1,..., γ n ) 0 (mod ), F x i (γ 1,..., γ n ) 0 (mod ), F (θ 1,..., θ n ) = 0 and θ i γ i (mod ) (i = 1,..., n). Proof. Consider the olynomial in x, f(x) = F (γ 1,..., γ i 1, x, γ i+1,..., γ n ). We will find α Z such that f(α) = 0 and α γ i (mod ). Then set θ j = γ j for i j, and θ i = α. It is easily seen that this suffices to rove the lemma. Let γ i = γ. We construct a sequence of -adic integers α 0, α 1,..., α m,... (3) congruent to γ modulo, such that f(α m ) 0 (mod m+1 ) for all m 0. For m = 0 take α 0 = γ. We roceed by induction. Suose for some m 1, α 0,..., α m 1 have been determined. Then, α m 1 γ (mod ) and f(α m 1 ) 0 (mod m ). We exand the olynomial f(x) in owers of x α m 1 : f(x) = β 0 + β 1 (x α m 1 ) + β 2 (x α m 1 ) (β i Z ). By the induction hyothesis, β 0 = f(α m 1 ) = m A where A Z. Also, since α m 1 γ (mod ) and F x i (γ 1,..., γ n ) 0 (mod ), β 1 = f (α m 1 ) = B, where B Z and B is not divisible by. Set x = α m 1 + ξ m and we get f(α m 1 + ξ m ) = m (A + Bξ) + β 2 2m ξ Since B 0 (mod ) we may choose ξ = ξ 0 Z such that A + Bξ 0 0 (mod ). Furthermore, since km 1 + m for k 2, we have f(α m 1 + ξ 0 m ) 0 (mod m+1 ). In addition m 1, so α m 1 + ξ 0 m γ (mod ) and we see that we may take α m = α m 1 + ξ 0 m. We now check that the sequence (3) converges -adically. By construction, v (α m α m 1 ) m (here v (ξ), ξ Q is the -adic value of ξ) and since the -adic numbers are comlete, (3) converges to a -adic integer, call it α. Clearly, α γ (mod ). Since f(α m ) 0 (mod m+1 ), lim m f(α m ) = 0. By the continuity of the olynomial f, lim m f(α m ) = f(α). It follows f(α) = 0. Now, for the theorem mentioned above: Theorem 2. Let 2 and 0 < r < n. The quadratic form F = F 0 + F 1 = ɛ 1 x ɛ r x 2 r + (ɛ r+1 x 2 r ɛ n x 2 n), where ɛ i (1 i n), is a -adic unit, reresents zero in Q if and only if at least one of the forms F 0 or F 1 reresents zero. Proof. The sufficiency of the condition is clear, we rove the necessity. Suose F reresents zero: ɛ 1 ξ ɛ r ξ 2 r + (ɛ r+1 ξ 2 r ɛ n ξ 2 n) = 0. (4) 3
4 Without loss of generality, assume that ξ i Z (1 i n) and that at least one ξ i is not divisible by. Suose ξ 1, or some ξ i among i = 1,..., r, is not divisble by. Consider equation (4) modulo, then and F 0 (ξ 1,..., ξ r ) 0 (mod ), F 0 x 1 (ξ 1,..., ξ r ) = 2ɛ 1 ξ 1 0 (mod ). By Hensel s lemma, F 0 reresents zero. Now assume ξ 1,..., ξ r are all divisible by, then some ξ i (r + 1 i n) is not divisible by and ɛ 1 ξ ɛ r ξ 2 r 0 (mod 2 ). When looking at equation (4) modulo 2 we have F 1 (ξ r+1,..., ξ n ) 0 (mod 2 ), or after dividing by F 1 (ξ r+1,..., ξ n ) 0 (mod ). We roceed as above, alying Hensel s lemma, and conclude that F 1 reresents zero. Two useful corollaries follow: Corollary 1. If ɛ 1,..., ɛ r are -adic units and 2, then the quadratic form f = ɛ 1 x ɛ r x 2 r reresents zero in Q if and only if the congruence f(x 1,..., x r ) 0 (mod ) has a nontrivial solution in Z. Corollary 2. Let f be the quadratic form from Corollary 1. If r 3, then f always reresents zero in Q. Proof. Corollary 1 is an immediate result of Theorem 2. Corollary 2 follows from Chevaley s theorem, which states that if F (x 1,..., x n ) is a form of degree less than n, then the congruence F (x 1,..., x n ) 0 (mod ), a rime, has a nontrivial solution. Getting back to the Hasse Minkowski theorem, we are considering the quadratic form (2) where a, b, c are ositive, square-free, rational integers that are airwise relatively rime. Let 2 be a rime divisor of c. Since (2) reresents zero in Q by assumtion, we may aly Theorem 2 and Corollary 1 and conclude that the congruence ax 2 + by 2 0 (mod ) has a nontrivial solution, (x 0, y 0 ). It follows that the form ax 2 + by 2 factors linearly, modulo : if we assume y 0 is not divisible by, we have ax 2 + by 2 ay 2 0 (xy 0 + yx 0 )(xy 0 yx 0 ) (mod ). Thus, since divides c, (2) factors into linear factors modulo : ax 2 + by 2 cz 2 L () (x, y, z)m () (x, y, z) (mod ), where L () and M () are integral linear forms. Similarly, for,, an odd rime divisor of a or b, the quadratic form (2) factors into linear integral forms modulo. Also, when = 2 we note that the quadratic form (2) factors linearly modulo 2 since ax 2 + by 2 cz 2 (ax + by cz) 2 (mod 2). By the Chinese Remainder theorem we may find integral linear forms L(x, y, z) and M(x, y, z) such that L(x, y, z) L () (x, y, z) (mod ), M(x, y, z) M () (x, y, z) (mod ) 4
5 for all rime divisors of a, b, and c. Since a, b, c are square-free and airwise relatively rime, we have the congruence ax 2 + by 2 cz 2 L(x, y, z)m(x, y, z) (mod abc). (5) Now give rational integer values to the variables x, y, z satisfying the inequalities 0 x < bc, 0 y < ac, 0 z < ab. (6) Without loss of generality, we may exclude the case a = b = c = 1, then, since a, b, c are airwise relatively rime and square-free, bc, ac, and ab are not integers. It follows that the number of triles (x, y, z) satisfying the inequalities (6) will be strictly greater than bc ac ab = abc. Since the number of triles is greater than the number of resiue classes modulo abc, there exist distinct triles (x 1, y 1, z 1 ) and (x 2, y 2, z 2 ) such that L(x 1, y 1, z 1 ) L(x 2, y 2, z 2 ) (mod abc). If we set it follows from the linearity of L that Looking at the congruence (5) we see that x 0 = x 1 x 2, y 0 = y 1 y 2, z 0 = z 1 z 2, L(x 0, y 0, z 0 ) 0 (mod abc). ax by 2 0 cz (mod abc). (7) Since the triles (x 1, y 1, z 1 ) and (x 2, y 2, z 2 ) satisfy the inequalities (6), we have Thus, Combining (7) and (8) we see that either or x 0 < bc, y 0 < ac, z 0 < ab. abc < ax by 2 0 cz 2 0 < 2abc. (8) ax by 2 0 cz 2 0 = 0, (9) ax by 2 0 cz 2 0 = abc. (10) In the first case, (x 0, y 0, z 0 ) is a nontrivial reresentation of zero in Q. If the second case holds, we aeal to the following. Since b and c are square-free and relatively rime, bc is not a square. We show that (2) reresents zero if and ony if ac is the norm of some element from the field Q( bc). Suosing (9) holds, we assume without loss of generality that x 0 0, it follows that ( cz0 ) 2 ( y0 ) 2 ( cz0 ac = bc = N + y ) 0 bc. x 0 x 0 x 0 x 0 For the converse, if ac = N(u + v bc), then ac 2 + b(cv) 2 cu 2 = 0. Now suose (10) holds. Multilying by c, we get ac(x 2 0 bc) = (cz 0 ) 2 bcy 2 0. Setting α = x 0 + bc, β = cz 0 + y 0 bc, it follows that and acn(α) = N(β), ( β ac = N. α) Thus ac is the norm of β α Q( bc) and (2) reresents zero in Q. In rearation for the last two cases to be roved, that is n = 4 and n 5, we need another lemma. Before that, we review some roerties of the Hilbert symbol. 5
6 Definition 2. For any air α 0, β 0 of -adic numbers, the Hilbert symbol (α, β) is equal to +1 or 1 if the form αx 2 + βy 2 z 2 reresents zero in the field Q or not, accordingly. We extend the Hilbert symbol to the real numbers by calling R the field Q. Whereas the field Q is the comletion of Q with resect to a finite rime, we say the real numbers are the comletion of Q with resect to the infinite rime. What follows are some basic roerties of the Hilbert symbol (see, for examle, [1]): (α, β 1 β 2 ) = (α, β 1 ) (α, β 2 ), (α, β) = (β, α), (α, α) = (α, 1). For -adic units ɛ and η, and real numbers a and b, ( ɛ (, ɛ) =, (ɛ, η) = 1 for 2,, ) where ( ɛ ) is the Legendre symbol. (2, ɛ) 2 = ( 1) (ɛ2 1)/8, (ɛ, η) 2 = ( 1) [(ɛ 1)/2][(η 1)/2], (a, b) = 1, if a > 0 or b > 0, (a, b) = 1, if a < 0 and b < 0, Lemma 3. If a rational quadratic form in three variables reresents zero in all fields Q, where runs through all rimes and, excet ossibly for Q q, then it reresents zero in Q q. Proof. Consider the rational quadratic form ax 2 + by 2 z 2. It follows from Corollary 2 that (a, b) = 1 for all, excet ossibly when = 2, or is in the rime factorization of a or b. Thus, there are only finitely many values of for which (a, b) = 1 and the roduct (a, b), where takes on the value of all rimes and the symbol, makes sense. To rove the lemma it suffices to show that (a, b) = 1, (11) that is, the number of for which (a, b) = 1 is even. The basic roerties of the Hilbert symbol allow us to reduce the roof of (11) to three cases: (1) a = 1, b = 1, (2) a = q, b = 1 (q a rime), (3) a = q, b = q (q and q rimes). All of the following comutations follow from the basic roerties of the Hilbert symbol and, in one lace, the law of quadratic recirocity. Case (1): ( 1, 1) = ( 1, 1) 2 ( 1, 1) = ( 1) ( 1) = 1. Case (2): (q, 1) = (q, 1) q (q, 1) 2 = (2, 1) = (2, 1) 2 (2, 1) = 1 1 = 1, ( 1 ) ( 1) [(q 1)/2][( 1 1)/2] = ( 1) (q 1)/2 ( 1) [(q 1)/2][( 1 1)/2] = 1. q 6
7 Case (3): (2, q) = (2, q) q (2, q) 2 = ( 2 q ) ( 1) (q2 1)/8 = ( 1) (q2 1)/8 ( 1) (q2 1)/8 = 1, ( q (q, q ) = (q, q ) q (q, q ) q (q, q )( q ) ) 2 = q q ( 1) [(q 1)/2][(q 1)/2] = ( 1) [(q 1)/2][(q 1)/2] ( 1) [(q 1)/2][(q 1)/2] = 1. This roves (11), hence, the lemma. For the case n = 4 of the Hasse Minkowski theorem we will consider the quadratic form a 1 x a 2 x a 3 x a 4 x 2 4 (12) where a i (1 i 4) is a square-free integer. The form (12) reresents zero in R, thus, we may assume a 1 > 0, a 4 < 0. Furthermore, let g = a 1 x a 2 x 2 2 and h = a 3 x 2 3 a 4 x 2 4. To rove the theorem for n = 4, we show that there exists a rational number a, such that both g and h reresent a in Q. Let 1,..., s be all the distinct odd rimes that divide a 1, a 2, a 3, a 4. For each of these rimes and the rime = 2 choose a reresentation of zero in Q : a 1 ξ a 2 ξ a 3 ξ a 4 ξ 2 4 = 0. We ick the ξ i so that ξ i 0 (1 i 4) (see the review of quadratic forms, above). Set b = a 1 ξ a 2 ξ 2 2 = a 3 ξ 2 3 a 4 ξ 2 4. Choose our ξ i so that b is divisible by at most the first ower of (if b = 0, then g, h reresent zero and thus reresent all elements of Q ). The congruences a b 2 (mod 16) a b 1 (mod 2 1) (13). a b s (mod 2 s) determine a rational integer a unique modulo m = s. Since b i is divisble by at most the first ower of i, following congruence holds: b i a 1 1 (mod i ). Any -adic unit congruent to 1 modulo is a square in Q (see [1]), thus b i a 1 is a square in Q i. Similarly, b 2 a 1 1 (mod 8), thus b 2 a 1 is a square in Q 2 (a 2-adic unit ɛ is a square in Q 2 if and only if ɛ 1 (mod 8), see [1]). Note that b and a differ by a square in Q. Thus, for = 2, 1,..., s the quadratic forms reresent zero in Q. ax g and ax h (14) Choosing a > 0, we see that the forms (14) reresent zero in R since a 1 > 0 and a 4 < 0. Suose now that 2, 1,..., s and a, then by Corollary 2 the forms (14) reresent zero in Q. To summarize our current osition, the forms (14) reresent zero in R and all -adic fields excet ossibly when divides 7
8 a and 2, 1,..., s. Thus, if we can choose a ositive rational integer a that satisfies the congruences (13) and is divisible by only rimes among 2, 1,..., s and ossibly one other rime q, we may aeal to Lemma 3 and conclude that (14) reresents zero in Q for all rimes including = q. We use Dirichlet s theorem on rimes in arithmetic rogressions. Pick a rational integer a > 0 that satisfies the congruences (13). Set d = gcd(a, m), then gcd(a /d, m/d) = 1. By Dirichlet s theorem, there exists a k N such that a d + k m d = q is rime. Take a = a + km = dq. Hence, the forms (14) reresent zero in R and Q for all rimes. By the Hasse Minkowski theorem for three variables, the forms (14) reresent zero in Q. It clearly follows that g and h both reresent a in Q. This roves the Hasse Minkowski theorem for quadratic forms in four variables. Lastly we deal with case n 5. Consider the quadratic form Since the form is indefinite, we may assume a 1 > 0 and a 5 < 0. We set a 1 x a 2 x a 3 x a 4 x a 5 x 2 5. (15) g = a 1 x a 2 x 2 2 and h = a 3 x 2 3 a 4 x 2 4 a 5 x 2 5. Proceeding exactly as in the case n = 4, we find a rational integer a > 0 such that g and h reresent a in R and Q for all rimes with the ossible excetion of one rime q. By Lemma 3, g reresents a in all -adic fields including when = q. To show that h reresents a in all -adic fields we note that h reresents zero in Q q by Corollary 2 (as we saw in the roof of the case n = 4, q a i (3 i 5)). It follows that h reresents all elements of Q q, namely h reresents a. Thus g and h reresent a in R and Q for all rimes. Clearly then, ax g and ax h reresent zero in R and Q for all rimes. By the Hasse Minkowski theorem for forms in three and four variables, ax g and ax h both reresent zero in Q. Hence, g and h both reresent a in Q and the form (15) reresents zero in Q. In order to generalize to n > 5 we first note that any quadratic form over Q with five or more variables always reresents zero in Q (see [1]). So the Hasse Minkowski theorem reads: a quadratic form in five or more variables with rational coefficients reresents zero in Q if and only if it reresents zero in R. An indefinite quadratic form in more than five variable is equivalent to an indefinite diagonal quadratic form f, which may be written as f = f 0 + f 1 where f 0 is an indefinite quadratic form in five variables. Since we are assured that f 0 reresents zero in Q for all (and in R), by what was just roved f 0 reresents zero in Q. It clearly follows that f reresents zero in Q. With that, Theorem 1 has been roved. References: [1] Borevich, Z. I. and Shafarevich, I. R. (translated by Greenleaf, Newcomb), Number Theory, Academic Press Inc.,
Introduction to Arithmetic Geometry Fall 2013 Lecture #10 10/8/2013
18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #10 10/8/2013 In this lecture we lay the groundwork needed to rove the Hasse-Minkowski theorem for Q, which states that a quadratic form over
More informationMATH 3240Q Introduction to Number Theory Homework 7
As long as algebra and geometry have been searated, their rogress have been slow and their uses limited; but when these two sciences have been united, they have lent each mutual forces, and have marched
More informationRepresenting Integers as the Sum of Two Squares in the Ring Z n
1 2 3 47 6 23 11 Journal of Integer Sequences, Vol. 17 (2014), Article 14.7.4 Reresenting Integers as the Sum of Two Squares in the Ring Z n Joshua Harrington, Lenny Jones, and Alicia Lamarche Deartment
More informationMath 4400/6400 Homework #8 solutions. 1. Let P be an odd integer (not necessarily prime). Show that modulo 2,
MATH 4400 roblems. Math 4400/6400 Homework # solutions 1. Let P be an odd integer not necessarily rime. Show that modulo, { P 1 0 if P 1, 7 mod, 1 if P 3, mod. Proof. Suose that P 1 mod. Then we can write
More informationMA3H1 TOPICS IN NUMBER THEORY PART III
MA3H1 TOPICS IN NUMBER THEORY PART III SAMIR SIKSEK 1. Congruences Modulo m In quadratic recirocity we studied congruences of the form x 2 a (mod ). We now turn our attention to situations where is relaced
More informationPOINTS ON CONICS MODULO p
POINTS ON CONICS MODULO TEAM 2: JONGMIN BAEK, ANAND DEOPURKAR, AND KATHERINE REDFIELD Abstract. We comute the number of integer oints on conics modulo, where is an odd rime. We extend our results to conics
More informationSolvability and Number of Roots of Bi-Quadratic Equations over p adic Fields
Malaysian Journal of Mathematical Sciences 10(S February: 15-35 (016 Secial Issue: The 3 rd International Conference on Mathematical Alications in Engineering 014 (ICMAE 14 MALAYSIAN JOURNAL OF MATHEMATICAL
More information6 Binary Quadratic forms
6 Binary Quadratic forms 6.1 Fermat-Euler Theorem A binary quadratic form is an exression of the form f(x,y) = ax 2 +bxy +cy 2 where a,b,c Z. Reresentation of an integer by a binary quadratic form has
More informationA CONCRETE EXAMPLE OF PRIME BEHAVIOR IN QUADRATIC FIELDS. 1. Abstract
A CONCRETE EXAMPLE OF PRIME BEHAVIOR IN QUADRATIC FIELDS CASEY BRUCK 1. Abstract The goal of this aer is to rovide a concise way for undergraduate mathematics students to learn about how rime numbers behave
More informationMAT 311 Solutions to Final Exam Practice
MAT 311 Solutions to Final Exam Practice Remark. If you are comfortable with all of the following roblems, you will be very well reared for the midterm. Some of the roblems below are more difficult than
More informationPrimes - Problem Sheet 5 - Solutions
Primes - Problem Sheet 5 - Solutions Class number, and reduction of quadratic forms Positive-definite Q1) Aly the roof of Theorem 5.5 to find reduced forms equivalent to the following, also give matrices
More informationx 2 a mod m. has a solution. Theorem 13.2 (Euler s Criterion). Let p be an odd prime. The congruence x 2 1 mod p,
13. Quadratic Residues We now turn to the question of when a quadratic equation has a solution modulo m. The general quadratic equation looks like ax + bx + c 0 mod m. Assuming that m is odd or that b
More informationHENSEL S LEMMA KEITH CONRAD
HENSEL S LEMMA KEITH CONRAD 1. Introduction In the -adic integers, congruences are aroximations: for a and b in Z, a b mod n is the same as a b 1/ n. Turning information modulo one ower of into similar
More informationMobius Functions, Legendre Symbols, and Discriminants
Mobius Functions, Legendre Symbols, and Discriminants 1 Introduction Zev Chonoles, Erick Knight, Tim Kunisky Over the integers, there are two key number-theoretic functions that take on values of 1, 1,
More informationMath 104B: Number Theory II (Winter 2012)
Math 104B: Number Theory II (Winter 01) Alina Bucur Contents 1 Review 11 Prime numbers 1 Euclidean algorithm 13 Multilicative functions 14 Linear diohantine equations 3 15 Congruences 3 Primes as sums
More informationON THE LEAST SIGNIFICANT p ADIC DIGITS OF CERTAIN LUCAS NUMBERS
#A13 INTEGERS 14 (014) ON THE LEAST SIGNIFICANT ADIC DIGITS OF CERTAIN LUCAS NUMBERS Tamás Lengyel Deartment of Mathematics, Occidental College, Los Angeles, California lengyel@oxy.edu Received: 6/13/13,
More informationHASSE-MINKOWSKI THEOREM
HASSE-MINKOWSKI THEOREM KIM, SUNGJIN 1. Introduction In rough terms, a local-global principle is a statement that asserts that a certain property is true globally if and only if it is true everywhere locally.
More informationMATH 361: NUMBER THEORY EIGHTH LECTURE
MATH 361: NUMBER THEORY EIGHTH LECTURE 1. Quadratic Recirocity: Introduction Quadratic recirocity is the first result of modern number theory. Lagrange conjectured it in the late 1700 s, but it was first
More informationElementary Analysis in Q p
Elementary Analysis in Q Hannah Hutter, May Szedlák, Phili Wirth November 17, 2011 This reort follows very closely the book of Svetlana Katok 1. 1 Sequences and Series In this section we will see some
More informationt s (p). An Introduction
Notes 6. Quadratic Gauss Sums Definition. Let a, b Z. Then we denote a b if a divides b. Definition. Let a and b be elements of Z. Then c Z s.t. a, b c, where c gcda, b max{x Z x a and x b }. 5, Chater1
More informationDiophantine Equations and Congruences
International Journal of Algebra, Vol. 1, 2007, no. 6, 293-302 Diohantine Equations and Congruences R. A. Mollin Deartment of Mathematics and Statistics University of Calgary, Calgary, Alberta, Canada,
More informationp-adic Measures and Bernoulli Numbers
-Adic Measures and Bernoulli Numbers Adam Bowers Introduction The constants B k in the Taylor series exansion t e t = t k B k k! k=0 are known as the Bernoulli numbers. The first few are,, 6, 0, 30, 0,
More informationBy Evan Chen OTIS, Internal Use
Solutions Notes for DNY-NTCONSTRUCT Evan Chen January 17, 018 1 Solution Notes to TSTST 015/5 Let ϕ(n) denote the number of ositive integers less than n that are relatively rime to n. Prove that there
More informationRECIPROCITY LAWS JEREMY BOOHER
RECIPROCITY LAWS JEREMY BOOHER 1 Introduction The law of uadratic recirocity gives a beautiful descrition of which rimes are suares modulo Secial cases of this law going back to Fermat, and Euler and Legendre
More informationarxiv: v1 [math.nt] 9 Sep 2015
REPRESENTATION OF INTEGERS BY TERNARY QUADRATIC FORMS: A GEOMETRIC APPROACH GABRIEL DURHAM arxiv:5090590v [mathnt] 9 Se 05 Abstract In957NCAnkenyrovidedanewroofofthethreesuarestheorem using geometry of
More informationJacobi symbols and application to primality
Jacobi symbols and alication to rimality Setember 19, 018 1 The grou Z/Z We review the structure of the abelian grou Z/Z. Using Chinese remainder theorem, we can restrict to the case when = k is a rime
More information1 Integers and the Euclidean algorithm
1 1 Integers and the Euclidean algorithm Exercise 1.1 Prove, n N : induction on n) 1 3 + 2 3 + + n 3 = (1 + 2 + + n) 2 (use Exercise 1.2 Prove, 2 n 1 is rime n is rime. (The converse is not true, as shown
More informationMersenne and Fermat Numbers
NUMBER THEORY CHARLES LEYTEM Mersenne and Fermat Numbers CONTENTS 1. The Little Fermat theorem 2 2. Mersenne numbers 2 3. Fermat numbers 4 4. An IMO roblem 5 1 2 CHARLES LEYTEM 1. THE LITTLE FERMAT THEOREM
More informationMA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES 2018
MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES 2018 J. E. CREMONA Contents 0. Introduction: What is Number Theory? 2 Basic Notation 3 1. Factorization 4 1.1. Divisibility in Z 4 1.2. Greatest Common
More informationCONGRUENCES CONCERNING LUCAS SEQUENCES ZHI-HONG SUN
Int. J. Number Theory 004, no., 79-85. CONGRUENCES CONCERNING LUCAS SEQUENCES ZHI-HONG SUN School of Mathematical Sciences Huaiyin Normal University Huaian, Jiangsu 00, P.R. China zhihongsun@yahoo.com
More informationWe collect some results that might be covered in a first course in algebraic number theory.
1 Aendices We collect some results that might be covered in a first course in algebraic number theory. A. uadratic Recirocity Via Gauss Sums A1. Introduction In this aendix, is an odd rime unless otherwise
More informationMATH 2710: NOTES FOR ANALYSIS
MATH 270: NOTES FOR ANALYSIS The main ideas we will learn from analysis center around the idea of a limit. Limits occurs in several settings. We will start with finite limits of sequences, then cover infinite
More informationNumber Theory Naoki Sato
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
More informationGAUSSIAN INTEGERS HUNG HO
GAUSSIAN INTEGERS HUNG HO Abstract. We will investigate the ring of Gaussian integers Z[i] = {a + bi a, b Z}. First we will show that this ring shares an imortant roerty with the ring of integers: every
More informationHOMEWORK # 4 MARIA SIMBIRSKY SANDY ROGERS MATTHEW WELSH
HOMEWORK # 4 MARIA SIMBIRSKY SANDY ROGERS MATTHEW WELSH 1. Section 2.1, Problems 5, 8, 28, and 48 Problem. 2.1.5 Write a single congruence that is equivalent to the air of congruences x 1 mod 4 and x 2
More informationLEGENDRE S THEOREM, LEGRANGE S DESCENT
LEGENDRE S THEOREM, LEGRANGE S DESCENT SUPPLEMENT FOR MATH 370: NUMBER THEORY Abstract. Legendre gave simple necessary and sufficient conditions for the solvablility of the diophantine equation ax 2 +
More information3 Properties of Dedekind domains
18.785 Number theory I Fall 2016 Lecture #3 09/15/2016 3 Proerties of Dedekind domains In the revious lecture we defined a Dedekind domain as a noetherian domain A that satisfies either of the following
More informationMA3H1 Topics in Number Theory. Samir Siksek
MA3H1 Toics in Number Theory Samir Siksek Samir Siksek, Mathematics Institute, University of Warwick, Coventry, CV4 7AL, United Kingdom E-mail address: samir.siksek@gmail.com Contents Chater 0. Prologue
More informationPractice Final Solutions
Practice Final Solutions 1. True or false: (a) If a is a sum of three squares, and b is a sum of three squares, then so is ab. False: Consider a 14, b 2. (b) No number of the form 4 m (8n + 7) can be written
More informationNUMBER SYSTEMS. Number theory is the study of the integers. We denote the set of integers by Z:
NUMBER SYSTEMS Number theory is the study of the integers. We denote the set of integers by Z: Z = {..., 3, 2, 1, 0, 1, 2, 3,... }. The integers have two oerations defined on them, addition and multilication,
More informationMATH342 Practice Exam
MATH342 Practice Exam This exam is intended to be in a similar style to the examination in May/June 2012. It is not imlied that all questions on the real examination will follow the content of the ractice
More informationAlgebraic number theory LTCC Solutions to Problem Sheet 2
Algebraic number theory LTCC 008 Solutions to Problem Sheet ) Let m be a square-free integer and K = Q m). The embeddings K C are given by σ a + b m) = a + b m and σ a + b m) = a b m. If m mod 4) then
More informationApplicable Analysis and Discrete Mathematics available online at HENSEL CODES OF SQUARE ROOTS OF P-ADIC NUMBERS
Alicable Analysis and Discrete Mathematics available online at htt://efmath.etf.rs Al. Anal. Discrete Math. 4 (010), 3 44. doi:10.98/aadm1000009m HENSEL CODES OF SQUARE ROOTS OF P-ADIC NUMBERS Zerzaihi
More informationFactorability in the ring Z[ 5]
University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Dissertations, Theses, and Student Research Paers in Mathematics Mathematics, Deartment of 4-2004 Factorability in the ring
More informationQuadratic Reciprocity
Quadratic Recirocity 5-7-011 Quadratic recirocity relates solutions to x = (mod to solutions to x = (mod, where and are distinct odd rimes. The euations are oth solvale or oth unsolvale if either or has
More informationSmall Zeros of Quadratic Forms Mod P m
International Mathematical Forum, Vol. 8, 2013, no. 8, 357-367 Small Zeros of Quadratic Forms Mod P m Ali H. Hakami Deartment of Mathematics, Faculty of Science, Jazan University P.O. Box 277, Jazan, Postal
More informationInfinitely Many Insolvable Diophantine Equations
ACKNOWLEDGMENT. After this aer was submitted, the author received messages from G. D. Anderson and M. Vuorinen that concerned [10] and informed him about references [1] [7]. He is leased to thank them
More informationf(r) = a d n) d + + a0 = 0
Math 400-00/Foundations of Algebra/Fall 07 Polynomials at the Foundations: Roots Next, we turn to the notion of a root of a olynomial in Q[x]. Definition 8.. r Q is a rational root of fx) Q[x] if fr) 0.
More informationAlgebraic Number Theory
Algebraic Number Theory Joseh R. Mileti May 11, 2012 2 Contents 1 Introduction 5 1.1 Sums of Squares........................................... 5 1.2 Pythagorean Triles.........................................
More informationOn the Multiplicative Order of a n Modulo n
1 2 3 47 6 23 11 Journal of Integer Sequences, Vol. 13 2010), Article 10.2.1 On the Multilicative Order of a n Modulo n Jonathan Chaelo Université Lille Nord de France F-59000 Lille France jonathan.chaelon@lma.univ-littoral.fr
More information#A47 INTEGERS 15 (2015) QUADRATIC DIOPHANTINE EQUATIONS WITH INFINITELY MANY SOLUTIONS IN POSITIVE INTEGERS
#A47 INTEGERS 15 (015) QUADRATIC DIOPHANTINE EQUATIONS WITH INFINITELY MANY SOLUTIONS IN POSITIVE INTEGERS Mihai Ciu Simion Stoilow Institute of Mathematics of the Romanian Academy, Research Unit No. 5,
More informationQUADRATIC RECIPROCITY
QUADRATIC RECIPROCIT POOJA PATEL Abstract. This aer is an self-contained exosition of the law of uadratic recirocity. We will give two roofs of the Chinese remainder theorem and a roof of uadratic recirocity.
More informationMAS 4203 Number Theory. M. Yotov
MAS 4203 Number Theory M. Yotov June 15, 2017 These Notes were comiled by the author with the intent to be used by his students as a main text for the course MAS 4203 Number Theory taught at the Deartment
More informationArithmetic Consequences of Jacobi s Two-Squares Theorem
THE RAMANUJAN JOURNAL 4, 51 57, 2000 c 2000 Kluwer Academic Publishers. Manufactured in The Netherlands. Arithmetic Consequences of Jacobi s Two-Squares Theorem MICHAEL D. HIRSCHHORN School of Mathematics,
More informationSets of Real Numbers
Chater 4 Sets of Real Numbers 4. The Integers Z and their Proerties In our revious discussions about sets and functions the set of integers Z served as a key examle. Its ubiquitousness comes from the fact
More informationDIRICHLET S THEOREM ABOUT PRIMES IN ARITHMETIC PROGRESSIONS. Contents. 1. Dirichlet s theorem on arithmetic progressions
DIRICHLET S THEOREM ABOUT PRIMES IN ARITHMETIC PROGRESSIONS ANG LI Abstract. Dirichlet s theorem states that if q and l are two relatively rime ositive integers, there are infinitely many rimes of the
More informationOn the Rank of the Elliptic Curve y 2 = x(x p)(x 2)
On the Rank of the Ellitic Curve y = x(x )(x ) Jeffrey Hatley Aril 9, 009 Abstract An ellitic curve E defined over Q is an algebraic variety which forms a finitely generated abelian grou, and the structure
More informationNew weighing matrices and orthogonal designs constructed using two sequences with zero autocorrelation function - a review
University of Wollongong Research Online Faculty of Informatics - Paers (Archive) Faculty of Engineering and Information Sciences 1999 New weighing matrices and orthogonal designs constructed using two
More informationSQUAREFREE VALUES OF QUADRATIC POLYNOMIALS COURSE NOTES, 2015
SQUAREFREE VALUES OF QUADRATIC POLYNOMIALS COURSE NOTES, 2015 1. Squarefree values of olynomials: History In this section we study the roblem of reresenting square-free integers by integer olynomials.
More informationANALYTIC NUMBER THEORY AND DIRICHLET S THEOREM
ANALYTIC NUMBER THEORY AND DIRICHLET S THEOREM JOHN BINDER Abstract. In this aer, we rove Dirichlet s theorem that, given any air h, k with h, k) =, there are infinitely many rime numbers congruent to
More informationA CRITERION FOR POLYNOMIALS TO BE CONGRUENT TO THE PRODUCT OF LINEAR POLYNOMIALS (mod p) ZHI-HONG SUN
A CRITERION FOR POLYNOMIALS TO BE CONGRUENT TO THE PRODUCT OF LINEAR POLYNOMIALS (mod ) ZHI-HONG SUN Deartment of Mathematics, Huaiyin Teachers College, Huaian 223001, Jiangsu, P. R. China e-mail: hyzhsun@ublic.hy.js.cn
More informationAn Overview of Witt Vectors
An Overview of Witt Vectors Daniel Finkel December 7, 2007 Abstract This aer offers a brief overview of the basics of Witt vectors. As an alication, we summarize work of Bartolo and Falcone to rove that
More informationBOUNDS FOR THE SIZE OF SETS WITH THE PROPERTY D(n) Andrej Dujella University of Zagreb, Croatia
GLASNIK MATMATIČKI Vol. 39(59(2004, 199 205 BOUNDS FOR TH SIZ OF STS WITH TH PROPRTY D(n Andrej Dujella University of Zagreb, Croatia Abstract. Let n be a nonzero integer and a 1 < a 2 < < a m ositive
More informationNumber Theory. Lectured by V. Neale Michaelmas Term 2011
Number Theory Lectured by V Neale Michaelmas Term 0 NUMBER THEORY C 4 lectures, Michaelmas term Page Page 5 Page Page 5 Page 9 Page 3 Page 4 Page 50 Page 54 Review from Part IA Numbers and Sets: Euclid
More informationAlmost All Palindromes Are Composite
Almost All Palindromes Are Comosite William D Banks Det of Mathematics, University of Missouri Columbia, MO 65211, USA bbanks@mathmissouriedu Derrick N Hart Det of Mathematics, University of Missouri Columbia,
More informationERRATA AND SUPPLEMENTARY MATERIAL FOR A FRIENDLY INTRODUCTION TO NUMBER THEORY FOURTH EDITION
ERRATA AND SUPPLEMENTARY MATERIAL FOR A FRIENDLY INTRODUCTION TO NUMBER THEORY FOURTH EDITION JOSEPH H. SILVERMAN Acknowledgements Page vii Thanks to the following eole who have sent me comments and corrections
More information= 1 2x. x 2 a ) 0 (mod p n ), (x 2 + 2a + a2. x a ) 2
8. p-adic numbers 8.1. Motivation: Solving x 2 a (mod p n ). Take an odd prime p, and ( an) integer a coprime to p. Then, as we know, x 2 a (mod p) has a solution x Z iff = 1. In this case we can suppose
More informationCharacteristics of Fibonacci-type Sequences
Characteristics of Fibonacci-tye Sequences Yarden Blausa May 018 Abstract This aer resents an exloration of the Fibonacci sequence, as well as multi-nacci sequences and the Lucas sequence. We comare and
More informationPractice Final Solutions
Practice Final Solutions 1. Find integers x and y such that 13x + 1y 1 SOLUTION: By the Euclidean algorithm: One can work backwards to obtain 1 1 13 + 2 13 6 2 + 1 1 13 6 2 13 6 (1 1 13) 7 13 6 1 Hence
More informationTHUE-VINOGRADOV AND INTEGERS OF THE FORM x 2 + Dy 2. Contents. Introduction Study of an Elementary Proof
THUE-VINOGRADOV AND INTEGERS OF THE FORM x 2 + Dy 2 PETE L. CLARK Contents Introduction Study of an Elementary Proof 1 1. The Lemmas of Thue and Vinogradov 4 2. Preliminaries on Quadratic Recirocity and
More informationAn Inverse Problem for Two Spectra of Complex Finite Jacobi Matrices
Coyright 202 Tech Science Press CMES, vol.86, no.4,.30-39, 202 An Inverse Problem for Two Sectra of Comlex Finite Jacobi Matrices Gusein Sh. Guseinov Abstract: This aer deals with the inverse sectral roblem
More informationGOOD MODELS FOR CUBIC SURFACES. 1. Introduction
GOOD MODELS FOR CUBIC SURFACES ANDREAS-STEPHAN ELSENHANS Abstract. This article describes an algorithm for finding a model of a hyersurface with small coefficients. It is shown that the aroach works in
More informationarxiv: v5 [math.nt] 22 Aug 2013
Prerint, arxiv:1308900 ON SOME DETERMINANTS WITH LEGENDRE SYMBOL ENTRIES arxiv:1308900v5 [mathnt] Aug 013 Zhi-Wei Sun Deartment of Mathematics, Nanjing University Nanjing 10093, Peole s Reublic of China
More informationChapter 3. Number Theory. Part of G12ALN. Contents
Chater 3 Number Theory Part of G12ALN Contents 0 Review of basic concets and theorems The contents of this first section well zeroth section, really is mostly reetition of material from last year. Notations:
More informationInfinitely Many Quadratic Diophantine Equations Solvable Everywhere Locally, But Not Solvable Globally
Infinitely Many Quadratic Diohantine Equations Solvable Everywhere Locally, But Not Solvable Globally R.A. Mollin Abstract We resent an infinite class of integers 2c, which turn out to be Richaud-Degert
More informationRINGS OF INTEGERS WITHOUT A POWER BASIS
RINGS OF INTEGERS WITHOUT A POWER BASIS KEITH CONRAD Let K be a number field, with degree n and ring of integers O K. When O K = Z[α] for some α O K, the set {1, α,..., α n 1 } is a Z-basis of O K. We
More informationCERIAS Tech Report The period of the Bell numbers modulo a prime by Peter Montgomery, Sangil Nahm, Samuel Wagstaff Jr Center for Education
CERIAS Tech Reort 2010-01 The eriod of the Bell numbers modulo a rime by Peter Montgomery, Sangil Nahm, Samuel Wagstaff Jr Center for Education and Research Information Assurance and Security Purdue University,
More informationDIRICHLET S THEOREM ON PRIMES IN ARITHMETIC PROGRESSIONS. 1. Introduction
DIRICHLET S THEOREM ON PRIMES IN ARITHMETIC PROGRESSIONS INNA ZAKHAREVICH. Introduction It is a well-known fact that there are infinitely many rimes. However, it is less clear how the rimes are distributed
More informationQUADRATIC RESIDUES AND DIFFERENCE SETS
QUADRATIC RESIDUES AND DIFFERENCE SETS VSEVOLOD F. LEV AND JACK SONN Abstract. It has been conjectured by Sárközy that with finitely many excetions, the set of quadratic residues modulo a rime cannot be
More informationCONGRUENCE PROPERTIES OF TAYLOR COEFFICIENTS OF MODULAR FORMS
CONGRUENCE PROPERTIES OF TAYLOR COEFFICIENTS OF MODULAR FORMS HANNAH LARSON AND GEOFFREY SMITH Abstract. In their work, Serre and Swinnerton-Dyer study the congruence roerties of the Fourier coefficients
More informationSQUARES IN Z/NZ. q = ( 1) (p 1)(q 1)
SQUARES I Z/Z We study squares in the ring Z/Z from a theoretical and comutational oint of view. We resent two related crytograhic schemes. 1. SQUARES I Z/Z Consider for eamle the rime = 13. Write the
More informationOn the smallest point on a diagonal quartic threefold
On the smallest oint on a diagonal quartic threefold Andreas-Stehan Elsenhans and Jörg Jahnel Abstract For the family x = a y +a 2 z +a 3 v + w,,, > 0, of diagonal quartic threefolds, we study the behaviour
More informationOn generalizing happy numbers to fractional base number systems
On generalizing hay numbers to fractional base number systems Enriue Treviño, Mikita Zhylinski October 17, 018 Abstract Let n be a ositive integer and S (n) be the sum of the suares of its digits. It is
More information(Workshop on Harmonic Analysis on symmetric spaces I.S.I. Bangalore : 9th July 2004) B.Sury
Is e π 163 odd or even? (Worksho on Harmonic Analysis on symmetric saces I.S.I. Bangalore : 9th July 004) B.Sury e π 163 = 653741640768743.999999999999.... The object of this talk is to exlain this amazing
More informationRational Points on Conics, and Local-Global Relations in Number Theory
Rational Points on Conics, and Local-Global Relations in Number Theory Joseph Lipman Purdue University Department of Mathematics lipman@math.purdue.edu http://www.math.purdue.edu/ lipman November 26, 2007
More information#A45 INTEGERS 12 (2012) SUPERCONGRUENCES FOR A TRUNCATED HYPERGEOMETRIC SERIES
#A45 INTEGERS 2 (202) SUPERCONGRUENCES FOR A TRUNCATED HYPERGEOMETRIC SERIES Roberto Tauraso Diartimento di Matematica, Università di Roma Tor Vergata, Italy tauraso@mat.uniroma2.it Received: /7/, Acceted:
More informationThe Euler Phi Function
The Euler Phi Function 7-3-2006 An arithmetic function takes ositive integers as inuts and roduces real or comlex numbers as oututs. If f is an arithmetic function, the divisor sum Dfn) is the sum of the
More informationOn Character Sums of Binary Quadratic Forms 1 2. Mei-Chu Chang 3. Abstract. We establish character sum bounds of the form.
On Character Sums of Binary Quadratic Forms 2 Mei-Chu Chang 3 Abstract. We establish character sum bounds of the form χ(x 2 + ky 2 ) < τ H 2, a x a+h b y b+h where χ is a nontrivial character (mod ), 4
More informationQUADRATIC FORMS, BASED ON (A COURSE IN ARITHMETIC BY SERRE)
QUADRATIC FORMS, BASED ON (A COURSE IN ARITHMETIC BY SERRE) HEE OH 1. Lecture 1:Introduction and Finite fields Let f be a olynomial with integer coefficients. One of the basic roblem is to understand if
More information19th Bay Area Mathematical Olympiad. Problems and Solutions. February 28, 2017
th Bay Area Mathematical Olymiad February, 07 Problems and Solutions BAMO- and BAMO- are each 5-question essay-roof exams, for middle- and high-school students, resectively. The roblems in each exam are
More informationQuadratic Residues, Quadratic Reciprocity. 2 4 So we may as well start with x 2 a mod p. p 1 1 mod p a 2 ±1 mod p
Lecture 9 Quadratic Residues, Quadratic Recirocity Quadratic Congruence - Consider congruence ax + bx + c 0 mod, with a 0 mod. This can be reduced to x + ax + b 0, if we assume that is odd ( is trivial
More informationPartII Number Theory
PartII Number Theory zc3 This is based on the lecture notes given by Dr.T.A.Fisher, with some other toics in number theory (ossibly not covered in the lecture). Some of the theorems here are non-examinable.
More information2 Asymptotic density and Dirichlet density
8.785: Analytic Number Theory, MIT, sring 2007 (K.S. Kedlaya) Primes in arithmetic rogressions In this unit, we first rove Dirichlet s theorem on rimes in arithmetic rogressions. We then rove the rime
More informationMATH 433 Applied Algebra Lecture 4: Modular arithmetic (continued). Linear congruences.
MATH 433 Applied Algebra Lecture 4: Modular arithmetic (continued). Linear congruences. Congruences Let n be a postive integer. The integers a and b are called congruent modulo n if they have the same
More informationA FEW EQUIVALENCES OF WALL-SUN-SUN PRIME CONJECTURE
International Journal of Mathematics & Alications Vol 4, No 1, (June 2011), 77-86 A FEW EQUIVALENCES OF WALL-SUN-SUN PRIME CONJECTURE ARPAN SAHA AND KARTHIK C S ABSTRACT: In this aer, we rove a few lemmas
More informationp-adic Properties of Lengyel s Numbers
1 3 47 6 3 11 Journal of Integer Sequences, Vol. 17 (014), Article 14.7.3 -adic Proerties of Lengyel s Numbers D. Barsky 7 rue La Condamine 75017 Paris France barsky.daniel@orange.fr J.-P. Bézivin 1, Allée
More information2 Asymptotic density and Dirichlet density
8.785: Analytic Number Theory, MIT, sring 2007 (K.S. Kedlaya) Primes in arithmetic rogressions In this unit, we first rove Dirichlet s theorem on rimes in arithmetic rogressions. We then rove the rime
More informationElementary Number Theory
Elementary Number Theory WISB321 = F.Beukers 2012 Deartment of Mathematics UU ELEMENTARY NUMBER THEORY Frits Beukers Fall semester 2013 Contents 1 Integers and the Euclidean algorithm 4 1.1 Integers................................
More informationDISCRIMINANTS IN TOWERS
DISCRIMINANTS IN TOWERS JOSEPH RABINOFF Let A be a Dedekind domain with fraction field F, let K/F be a finite searable extension field, and let B be the integral closure of A in K. In this note, we will
More informationMath 261 Exam 2. November 7, The use of notes and books is NOT allowed.
Math 261 Eam 2 ovember 7, 2018 The use of notes and books is OT allowed Eercise 1: Polynomials mod 691 (30 ts In this eercise, you may freely use the fact that 691 is rime Consider the olynomials f( 4
More information