Math 104B: Number Theory II (Winter 2012)

Transcription

1 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 of squares 4 1 Recirocity ste 5 Descent ste 5 3 Pythagorean triles 7 4 More descent 8 5 Diohantine equations and congruences 10 6 Fermat-Pell equations Diohantine aroximation 1 6 Continued fractions 18 7 Primes of the form = x + ny 4 8 Quadratic recirocity 7 81 Legendre symbol 7 8 Jacobi symbol 36 9 Quadratic forms Elementary genus theory 54 1

2 1 Review 11 Prime numbers A rime number has the following roerties: has no other divisors than 1 and ; ab = a or b There are infinitely many rimes Every ositive integer can be written uniquely as a roduct of rimes 1 Euclidean algorithm The algorithm is used to find the greatest common divisor d = (a, b) of two ositive integers a and b It also can be used to find integers r, s such that 13 Multilicative functions d = ar + bs A function f : Z C is multilicative if f(mn) = f(m)f(n) whenever (m, n) = 1 It is comletely multilicative if f(mn) = f(m)f(n) for all integers m, n regardless of their gcd Fact: If f is a multilicative function than so is g : Z C defined by g(n) = d n f(d) Examles of multilicative, but not comletely multilicative functions: d(n) = the number of divisors of n (divisor function) σ(n) = the sum of the divisors of n φ(n) = the number of ositive integers < n and relatively rime to n (Euler s φ function) Assume that we know that f(n) is a multilicative function Then in order to be able to evaluate it at any ositive integer, it is enough to know its value at rime owers That is because f( a 1 1 ar r ) = f( a 1 1 ) f( ar r ) for any distinct rimes 1,, r We can take advantage of this feature to deduce formulas for various multilicative functions For instance, for n = a 1 1 ar r,

3 d(n) = (a 1 + 1) (a r + 1) σ(n) = a ar r 1 r 1 ) ) φ(n) = a ( 1 1) ar 1 ( r 1) = n (1 11 (1 1r r 14 Linear diohantine equations These are equations of the form a 1 x a k x k = c with integer coefficients and for which we want to find integer solutions We know that such solutions exist if and only if the gcd of the coefficients a 1,, a n divides c 15 Congruences Definition We say that two integers a and b are congruent modulo some integer n and write a b (mod n) if n a b (That is to say, a and b give the same remainder when divided by n) Here a few roerties of congruences: a a (mod n) a b (mod n) b a (mod n) a b (mod n) and b c (mod n) = a c (mod n) a b (mod n) and c d (mod n) = a ± c b ± d (mod n), ac bd (mod n) (a, n) = d and ab ac (mod n) = b c (mod n d ) Given integers a and n, the equation ax b (mod n) has solutions if (a, n) b Therefore it has solutions for all b iff (a, n) = 1 That is to say, if there exists an integer c such that ac 1 (mod n) If such a c exists, it is unique modulo n, and the solution is x = bc (mod n) is also unique modulo n a φ(n) 1 (mod n) 3

4 In addition to all these similarities to normal arithmetic oerations (addition, subtraction, multilication, division), there are similarities to linear algebra as well For instance, the system of linear congruences has unique solution a 11 x a 1r x r b 1 (mod n) a r1 x a rr x r b r (mod n) (mod n) iff det(a ij ) and n are corime Theorem 11 (Chinese Remainder Theorem) Assume that m 1,, m r are ositive integer with the roerty that any two of them are relatively rime Then, for any a 1,, a r Z, the system of equations x a 1 (mod m 1 ) x a r (mod m r ) has a unique solution (mod m 1 m r ) Primes as sums of squares Our goal in the next coule of lectures is to rove the following result formulated by Fermat Theorem 1 A rime can be written as the sum of two squares if and only if = or 1 (mod 4) Proof One of the direction is easy Assume = a + b Since a and b are each either congruent to 0 or 1 modulo 4, it follows that 0, 1 or (mod 4) But let s not forget that is a rime, so it cannot ossible be divisible by 4, and the only way it can be (mod 4) is for it to equal The other direction is much harder It s clear to do when =, but we also have to show that any rime 1 (mod 4) can be written as the sum of two squares For that, we will follow Euler s roof It might not be the shortest roof one can write down, but it has the advantage that it illustrates the concet of descent (which was the idea Fermat used in his sketch of the roof) and recirocity that we will encounter again later in the course Recirocity ste: A rime 1 (mod 4), then it divides N = a + b with a and b relatively rime integers Descent ste: If a rime divides a number N of the form N = a +b, where (a, b) = 1, then itself can be written as = x + y for some (x, y) = 1 Clearly the these two claims imly our result We are going to deviate from the historical order and rove first the recirocity ste (Euler first found the roof for the descent ste) 4

5 1 Recirocity ste The recirocity ste follows immediately from the following result Lemma The equation has solutions = or 1 (mod 4) Proof If =, then x = 1 is a solution x 1 (mod ) If 1 (mod 4), then 4 1 = φ() and therefore there exists an integer a with ord a = 4 This means that a 4 1 (mod ) and a, a, a 3 1 (mod 4) We have a 4 1 = (a 1)(a + 1) 0 (mod ) But a 1 0 (mod ), hence a equation 1 (mod ), and x = a is a solution of our If 3 (mod 4), assume that x = a is a solution, ie a 1 (mod ) Then a 4 1 (mod ), so ord a 4 But we also know that ord a φ() = 1 Hence ord a ( 1, 4) =, which means that a 1 (mod ) The ushot is that 1 1 (mod ), so The only way this will haen is for =, and we reached a contradiction Descent ste Fermat s idea (which he used on a number of other occasions), formalized in this case by Euler in this case, is to show that if we have a solution to a diohantine equation, then we can find a smaller (in some sense) solution Iterating this rocess means that we can find smaller and smaller ositive integers Hence the rocess needs to terminate at some oint, or we reach a contradiction Lemma 3 If N is an integer of the form N = a +b for some (a, b) = 1 and q = x +y is a rime divisor of N, then there exist relatively rime integers c and d such that N/q = c +d Proof First note that since q has no trivial divisors, x and y are forced to be relatively rime We have x N a q = x (a + b ) a (x + y ) = x b a y = (xb ay)(xb + ay) Since q N, it follows that x N a q 0 (mod q), and so (xb ay)(xb + ay) 0 (mod q) Since q is a rime, this can haen only if one of the factors is divisible by q Since we can change the sign of a without affecting our theorem, we can assume that q xb ay, that is xb ay = dq for some integer d 5

6 We would like to show that x a + dy Since (x, y) = 1, this is equivalent to showing that x y(a + dy) But y(a + dy) = ay + dy = xb dq + dy = xb d(x + y ) + dy = xb dx which is divisible by x Thus x a + dy, so there exist an integer c such that a + dy = cx Therefore and so Next we see that cxy = (a + dy)y = xb dx = x(b dx) cy + dx = b N = a + b = (cx dy) + (cy + dx) = (x + y )(c + d ) = q(c + d ) Since (a, b) = 1 it follows that (c, d) = 1 and the roof is comlete And now for the actual descent ste, assume that we have an odd rime (and thus > ) that divides a number M of the form N = a + b with (a, b) = 1 We want to show that 1 (mod 4) First, note that we can add or subtract any multile of from a or b without changing the roblem That is, we can find integers a 1, b 1 with a 1, b 1 < / such that N 1 = a 1+b 1 In articular, N 1 < / Denote d = (a 1, b 1 ) Then d < /, so d We also know that a 1 = da, b 1 = db and (a, b ) = 1 Note that a a 1 < / and likewise b < / Therefore N = a + b < / We have a 1 + b 1 = d (a + b ) Since is a rime that does not divide d, it follows that N = a + b So we showed that our rime has to divide a number M = u +v < / with (u, v) = 1 and u, v < / The ositive integer m = M/ will have to be m < / Let q be a rime divisor of m Clearly q since q m < / In articular q < and M q Assume that q can be written as the sum of two squares By Lemma 3, we have M/q = x + y for some integers (x, y) = 1 But then x + y < u + v = M So if all the rime factors of M different from can be written as sums of two squares, then so can Since we assumed that this is not the case, it follows that M has some rime divisor 1 < that cannot be written as the sum of two squares By reeating the argument for 1 it follows that there must exist another rime < 1 that cannot be written as the sum of two squares This argument cannot continue indefinitely, so at some oint we are bound to hit the rime number 5 = + 1 which can obviously be written as the sum of two squares The descent ste is now roven and this comletes the roof of Theorem 1 6

7 Note that we imlicitly used the fact that if (x, y) = 1 then 3 x + y To see this, recall that for any integer x we have x 0, 1 or 1 (mod 3), so x 0 or 1 (mod 3) Since (x, y) = 1 we cannot have x y 0 (mod 3), so x + y 0 (mod 3) 3 Pythagorean triles We want to find all right triangles with all three sides of integral length In other words, we want to solve the diohantine equation x + y = z (31) Note that any solution generates a ositive solution by changing the sign, hence solving the equation in Z is equivalent to solving it in Z >0, which is the same as finding all right triangles with integral sides We can further reduce the roblem to finding solutions with (x, y, z) = 1, that is we exclude similar triangles Each such solution will generate infinitely many solutions (dx, dy, dz) with gcd = d and vice versa It is worth noticing that if a rime divides two of the number x, y, z then it would have to divide the third one as well Hence we must have (x, y) = (y, z) = (x, z) = 1 There is one more observation we can make to simly our roblem Claim x y (mod ) Proof We know that we cannot have x y 0 (mod ) because that force x and y to not be relatively rime We are going to argue by contradiction for the other case as well Assume that x y 1 (mod ) Then x y 1 (mod 4), and this would mean that z (mod 4), which is imossible Since x and y are interchangeable in our roblem, we can assume without loss of generality that x is odd and y is even This also imlies that z is odd We can rewrite our equation as y = z x = (z x)(z + x) and further as ( y ) z x = z + x All the fractions above are really ositive integers since y is even and x, z are both odd with z > x Next we want to use the following observation Fact If a, b, c Z with (a, b) = 1 and ab = c, then there exist integers a 1, b 1 such that a = a 1 and b = b 1 Clearly a 1 and b 1 have to be relatively ( rime as well z x In order to do that, we need to show that gcd, z + x ) = 1 Assume that is a rime that divides both of them Then divides both their sum and their difference, that is it has to divide both x and z That would imly that divides y as well, and this contradicts the fact that (x, y, z) = 1 7

8 Hence the gcd of the two fractions is indeed 1 and there must exist ositive integers u and v with (u, v) = 1 such that This leads to z x = v and x = u v y = uv z = u + v z + x = u Note that since x and z are odd, we must also have u v (mod ) Also, x > 0 imlies u > v In short, we roved that all ositive Pythagorean triles are of the form x = d(u v ) y = duv z = d(u + v ) where u, v Z, u > v > 0 and u v (mod ) 4 More descent We want to study the Fermat equation for n = 4, x 4 + y 4 = z 4 (41) Fermat himself roved that it has no non-trivial solutions (ie no integer solutions with xyz 0) His roof uses again the method of descent Assume that x, y, z are ositive integers satisfying (41) Set d = (x, y, z) Then x = dx 1, y = dy 1 and z = dz 1 where (x 1, y 1, z 1 ) = 1 and x 1, y 1, z 1 are also ositive integers satisfying the same equation (41) In articular, x 1, y 1, t 1 = z 1 is a relatively rime Pythagorean trile In articular, x 1, y 1, t 1 are relatively rime ositive integers that form a solution to the equation X 4 + Y 4 = T (4) Note that x 1 and y 1 are interchangeable, so we can assume without loss of generality that x 1 is odd and y 1 is even It follows from our study of Pythagorean triles (Section 3) there exist integers u > v > 0 such that (u, v) = 1 and u v (mod ) such that x 1 = u v y1 = uv t 1 = u + v 8

9 Since x 1 is odd, we have x 1 1 (mod 4) and therefore u is odd and v is even Note that this imlies further that (u, v) = 1 Since u(v) = y 1 and v is even, we have u = t and v = 4d for some ositive relatively rime integers t and d, with t odd We can rewrite the formula for x 1 as x 1 + v = u Since (u, v) = 1 it follows that x 1, v, u is a relatively rime Pythagorean trile with x 1 odd and v even Alying again the results from Section 3, there exist integers a > b > 0 such that (a, b) = 1, a b (mod ) and x 1 = a b v = ab u = a + b Since v = ab and v = 4d it follows that ab = d But (a, b) = 1 and therefore a = x, b = y for some integers x > y > 0 with (x, y ) = 1 and x y (mod ) To reca, we have u = a + b a = x b = y u = t Therefore x, y, t are relatively rime ositive integers that satisfy But we also have x 4 + y 4 = t t t 4 = u < u + v = t 1 We roved that if we start with a relatively rime ositive solution (x 1, y 1, t 1 ) to (4) we can roduce another relatively rime solution (x, y, t ) with 0 < t < t 1 Alying this fact over and over again we obtain infinitely many ositive solutions (x n, y n, t n ) to (4) with 0 < < t n < t n 1 < < t 1 This is imossible because there are only finitely many integers between 0 and t 1 (In fact, there are t 1 1 of them!) In short, the assumtion that we can find a ositive solution to (41) led to a contradiction, and that roves that no such solution can exist 9

10 5 Diohantine equations and congruences As we have already seen in some isolated examles, one can try to show that a diohantine equation does not have solutions by showing that it has no solution modulo some integer n Examle 1 x 3y = 1 Looking at this equation modulo 3, we see that which we know it is imossible since 3 1 Examle x 7y = 1 x 1 (mod 3), This imlies that x (mod 7) and that is imossible since 7 is a rime and 7 3 (mod 4) Examle 3 x 15y = This imlies that x (mod 5) But the only squares modulo 5 are 0, 1, 4 Examle 4 x 5y = 3z Assume that we have a ositive solution with (x, y, z) = d Then x = dx 1, y = dy 1, z = dz 1 with (x 1, y 1, z 1 ) = 1 and x 1 5y 1 = 3z 1 In articular, 3 x 1 5y 1 and, since obviously 3 6y 1, we get 3 x 1 + y 1 We know that this is only ossible if 3 x 1 and 3 y 1 But then 9 3z 1 and so 3 z 1 This cannot haen since (x 1, y 1, z 1 ) = 1 6 Fermat-Pell equations We will consider equations of the form and x dy = 1 (61) x dy = 1 (6) We want to figure out for which d they have (non-trivial) integer solutions If the answer is affirmative, we want to find a way to write down all solutions First a few examles Examle 1 x 3y = 1 Looking at this equation modulo 3, we see that which we know it is imossible since 3 1 x 1 (mod 3), 10

11 Examle x 3y = 1 For instance (, 1) is a solution In fact, it has infinitely many solutions as we shall see shortly Examle 3 x 7y = 1 This imlies that x (mod 7) and that is imossible since 7 is a rime and 7 3 (mod 4) Examle 4 x y = 1 has no solutions when is a rime 3 (mod 4) The argument is the same as in the revious examle We start our systematic study by roving the following result Theorem 61 Let d be a ositive integer 1 If the equation x dy = 1 (61) has one ositive solution, then it has infinitely many ositive solutions If the equation x dy = 1 (6) has one ositive solution, then both (61) and (6) have infinitely many ositive solutions The theorem follows immediately from the following lemma Lemma 6 Assume that a, b, d Z >0 and let c = a db Then for any n 1 there exist ositive integers x n, y n such that x n dy n = c n Moreover, we can choose these integers such that {x n } n and {y n } n are two strictly increasing sequences Proof By induction on n First, we have to check for n = 1 This is resolved by taking x 1 = a and y 1 = b Now assume that we found x n, y n Then c n+1 = c n c = (x n dy n)(a db ) = a x n+d b y n d(a y n+b x n) = (ax n +dby n ) d(ay n +bx n ) 11

12 Then { x n+1 = ax n + dby n y n+1 = ay n + bx n have the roerty that x n+1 dy n+1 = c n+1 It remains to verify that x n+1 > x n and y n+1 > y n This is so because x n+1 = ax n + dby n > ax n x n and y n+1 = ay n + bx n > ay n y n They are of course ositive because x n > x 1 = a > 0 and y n > y 1 = b > 0 for all n > 1 Even though we roved that if a solution exists, then infinitely many solutions exits, we are far from done We still have to figure out exactly when the Pell equations have solutions and how to generate all solutions For the rest of this section we assume that d > 0 is not a square 61 Diohantine aroximation We want to figure out when these equations have solutions equation (61), namely x dy = 1 We will deal first with the Examle 1 The smallest ositive solution to x 61y = 1 is ( , ) Examle The smallest ositive solution to x 1y = 1 is (55, 1) Examle 3 The smallest ositive solution to x 58y = 1 is (19603, 574) If (x, y) is a ositive integer solution to Pell s equation (61), then we must have (x + y d )(x y d ) = 1 x y d = 1 x + y d (63) If x, y are ositive integers, the quantity x + y d will be relatively large Hence x y d must be a small ositive real number So, if we can find x, y ositive integers such that x y d is as small as ossible, then the roduct (x + y d )(x y d ) has to be a relatively small ositive integer And a small ositive integer has a good chance of being equal to 1, or at least close to it That chance increases if we can kee the size of our large factor x + y d under control 1

13 We start by investigating the question of how small we can make x y d and we will follow Dirichlet s answer to this question For any integer y > 0 we can choose x to be the closest integer to y d and this ensures that x y d 1 But we can do much better than that Examle Take d = 13 y = 1 x = 4 = x y y = 5 x = 18 = x y Dirichlet emloyed a simle, yet very owerful, idea to answer the question of how small x y d can become: the igeonhole rincile This rincile simly says that if there more igeons than igeonholes, then at least one of igeonholes contains more than one igeon Theorem 63 (Dirichlet s diohantine aroximation theorem) Let α R >0 \ Q be a ositive irrational number Then there are infinitely many airs of ositive integers (x, y) such that x αy < 1 y x y α < 1 y Proof Pick a large integer T > 0 For 0 t T write each of the numbers tα = N t +F j t where N t = jα is the floor of jα and F t [0, 1) is the fractional art of jα Our igeons are the T + 1 numbers F 0, F 1,, F T Our igeonholes are the T subintervals of [0, 1) given by [ t 1, ) t T T, 1 t T The igeonhole rincile ensures that there exist 0 j < k T such that F k F j < 1 T But F j = jα N j and F k = kα N k Substituting above, we get (N k N j ) (k j)α < 1 T Set x = N k Nj and y = k j Then x, y Z >0 and x yα < 1 On the other hand, T since 0 j < k T it follows that 0 < y = k j T To reca we have that and 0 < y T (64) x yα < 1 T (65) 13

14 Note that by taking T larger and larger we automatically get new airs of ositive integers (x, y) that satisfy (64) and (65) This haens because x yα > 0 (recall that α / Q!) For any fixed air of ositive integers (x 0, y 0 ) we can make T large enough so that (65) does not hold anymore Therefore we obtain another air of ositive integers (x 1, y 1 ) for this new T Continuing the rocedure and making T larger and larger yields infinitely many airs of ositive integers that satisfy (64) and (65) Note also that (64) imlies that and our roof is comlete 1 T 1 y, Going back to Pell s equation (61), we will aly Dirichlet s theorem to α = d The fact that d > 0 is not a square ensures that d / Q Hence there are infinitely many airs of ositive integers (x, y) such that x y d < 1/y We would like the left hand side here to be equal to x + y d since that would yield a solution to Pell s equation Idea: take two airs (x, y) for which x dy gives the same value and divide them Let s see on an examle what we mean Examle d = 13 (x 1, y 1 ) = (11, 3) = x 1 13y 1 = 4 (x 1, y 1 ) = (119, 33) = x 13y = 4 We divide by taking the quotient x y d x 1 y 1 d = = = We still don t have what we want because 11/ and 3/ are not integers Let s try again with another air (x 3, y 3 ) Namely, Then (x 3, y 3 ) = (14159, 397) = x 3 13y 3 = 4 x 3 y 3 d x 1 y 1 d = = The key oint here is that (x 1, y 1 ) (x 3, y 3 ) (mod 4), but (x 1, y 1 ) (x, y ) (mod 4) Lemma 64 Let d be an integer that is not a erfect square Set F d = {x + y d; x, y Q} C and R d = {m + n d; m, n Z} F d 14

15 (i) If x, y Q, then x + y d = 0 x = y = 0 (ii) Z R d, Q F d (iii) z 1, z R d = z 1 ± z, z 1 z R d (iv) z 1, z F d = z 1 ± z, z 1 z F d (v) z F d, z 0 = 1 z F d Proof Exercise Lemma 65 With the same notation as in Lemma 64, if (x + y d) n = A + B d with x, y, A, B Z and n N, then (x y d) n = A B d Same statement for x, y, A, B Q Proof Induction on n Theorem 66 Let d be a ositive integer that is not a erfect square (i) Then Pell s equation always has ositive integer solutions x y d = 1 (ii) Moreover, let (a 1, b 1 ) be the ositive integer solution with the smallest a 1 and set z = a 1 + b 1 d (Alternatively, we could choose the air that gives the smallest z > 1) Define (a k, b k ) to be the ositive integers given by the formula a k + b k d = z k = (a 1 + b 1 d) k k 1 (66) Then (a k, b k ), k 1, are all the ositive integer solutions to Pell s equation Proof To rove the first art we will emloy again the igeonhole rincile Dirichlet s Theorem 63 imlies that there are infinitely many airs (x, y) of ositive integers such that x y d < 1 y For any such (x, y) we have 0 < x < y d + 1, and so y x + y d < d + 1 y < 3y d Therefore x y d = x y d x + y d < 1 y 3y d = 3 d (67) Let T = 3 d We will aly the igeonhole rincile to 15

16 igeons: airs of ositive integers (x, y) such that x y d < 1 There are infinitely many y igeons (cf Theorem 63) igeonholes: integers T m T There are T + 1 igeonholes Since we have infinitely many igeons and only finitely many igeonholes, it follows that at least one igeonhole must contain infinitely many igeons Therefore there exist an integer M such that M < 3 d and an infinite sequence (x k, y k ) such that x k < x k+1, y k < y k+1, x k y kd = M (68) We would like to divide two such airs to get a solution to Pell s equation But as we ve seen in our examle, we need to choose our airs carefully We do that by emloying the igeonhole rincile again This time the rotagonists are: igeons the infinitely many airs of ositive integers (x k, y k ); igeonholes airs of integers (A, B) with 0 A, B M 1 There are M igeonholes Once again, we have infinitely many igeons and finitely many igeonholes It follows that there are two different airs (x k, y k ) and (x j, y j ) with k < j and x k x j (mod M), y k y j (mod M) Then, by (68), x j y j d = x j y j d xk + y k d = (x jx k dy j y k ) + (x j y k x k y j ) d x k y k d x k y k d x k + y k d M Set First we see that x = x jx k dy j y k M y = x jy k x k y j M x y d = (x j y j d)(x k y k d) M = 1 Then we use the congruence relations to show that x, y are integers Indeed, and x j x k dy j y k x k dy k (mod M) 0 (mod M) x j y k x k y j x k y k x k y k (mod M) 0 (mod M) Changing signs if necessary, we found a nonnegative integers x, y such that x y d = 1 Clearly this imlies that x 1 We want to show that y 0 We argue by contradiction Assume y = 0 Then x j y k = x k y j 16

17 which imlies that y km = y k(x j dy j ) = x ky j dy ky j = My j We obtain y k = y j which is a contradiction Therefore y > 0 and we found our solution Now for the second statement of the theorem First we should note that z > 1 since a 1, b 1 1 and that Lemma 64 ensures that a k, b k are well defined (ie z k can be indeed written as integer + and integer multile of d; ositivity I leave to you) The same lemma ensures that we do not have any reeat airs among the (a k, b k ) s We should also note that so By Lemma 65, Thus z(a 1 b 1 d) = (a1 + b 1 d)(a1 b 1 d) = a 1 db 1 = 1, 1 z = a 1 b 1 d z k = a k b k d k 1 a k b kd = (a k + b k d)(ak b k d) = zk z k = 1 k 1, which is to say (a k, b k ) is a solution of the Pell equation (61) And now we have to show that there are no other ositive integer solutions Assume (u, v) is a ositive integer solution We want to show that there exist a ositive integer k such that (u, v) = (a k, b k ) Let r = u + v d Then u a 1 > 0 and this imlies in turn that v = u 1 a 1 1 = b d d 1 Since we are dealing with ositive numbers, this imlies v b 1, so r z > 1 It follows that there exist an integer k 1 such that z k r < z k+1 Then 1 z k r < z, which can be rewritten as 1 (a k b k d)(u + v d) < z Hence where Moreover, 1 s + t d < z (69) s = a k u b k vd and t = a k v b k u s t d = (a k u b k vd) d(a k v b k u) = (a k db k)(u dv ) = 1 (610) We need to show that s+t d = z k r? = 1 We do that by excluding all other ossibilities Assume s < 0, t < 0 Then s + t d < 0 which contradicts (69) 17

18 Assume s < 0, t 0 Then s + d > s + t d 1 and thus 1 = s + t d = ( s + t d)(s + t d) 1 1 (contradiction) Assume s 0, t < 0 Then s t d > s + t d 1 and thus 1 = s t d = (s t d)(s + t d) s t d > 1 (contradiction) Assume s > 0, t > 0 Then s a 1 > 0 since (a 1, b 1 ) is the ositive integer solution with the smallest a 1 By (610) we have t = s 1 d a 1 1 d = b 1 Again we are dealing with ositive integers, so t b 1 and s + t d a 1 + b 1 d = z which contradicts (69) Assume s = 0 Then t d = 1 which is imossible since d > 0 The only ossibility left is that t = 0 This imlies s = 1, hence z k = r 6 Continued fractions We still want a method for finding that smallest solution to Pell s equation (61) The answer will be rovided in terms of continued fractions (and dates back to the dawn of time, or at least to VI century India) Given a real number A one comutes its continued fraction exansion as follows x 0 = A, a 0 = x 0 x i+1 = 1 x i a i, a i+1 = x i+1 i 0 (611) This formula ensures that x i, a i 1 for all i 1 Definition We say that [a 0, a 1, ] = a a a + is the continued fraction of the real number A, where the a i s are comuted according to the rocedure given in (611) Examle 67 The continued fraction exansion of 5 3 A = 5 3 = = is [1, 1, ] =

19 Examle 68 A finite continued fraction [a 0, a 1,, a n ] with a i 1 for all 1 i n, is a rational number Vice versa, if a is a rational number, the rocedure (611) outlines the b Euclidean algorithm for a and b Thus the rational fraction of a rational number is finite, unique, and a is equal to to its continued fraction b Examle 69 The continued fraction of is [1,,,,, ] Definition Let [a 0, a 1, ] be a continued fraction The rational number is called the nth convergent of [a 0, a 1, ] Examle 610 n q n = [a 0,, a n ] (n 0) n = 0 : n = 1 : 0 q 0 = [a 0 ] = a 0 = 0 = a 0, q 0 = 1 1 = [a 0, a 1 ] = a = a 0a = 1 = a 0 a 1 + 1, q 1 = a 1 q 1 a 1 a 1 We can continue this rocedure and see that in general n+1 = a n+1 n + n 1 for all n 1, 0 = a 0, 1 = a 0 a 1 + 1; q n+1 = a n+1 q n + q n 1 for all n 1, q 0 = 1, q 1 = a 1 (61) Note that Furthermore, (61) imlies that 1 q 0 0 q 1 = a 0 a a 0 a 1 = 1 n+1 q n n q n+1 = (a n+1 n + n 1 ) q n n (a n+1 q n + q n 1 ) = n 1 q n n q n 1 It follows by induction that n+1 q n n q n+1 = ( 1) n n 0 (613) In articular, n, q n are relatively rime for all n and therefore the convergents n indeed written in lowest terms with n, q n comuted using the recurrence relations (61) Moreover, if [a 0, a 1, ] is an infinite continued fraction (61) imlies that the denominators q n kee growing, while (613) tells us that n n+1 = ( 1)n Therefore the convergents q n q n+1 q n q n+1 n, n 0, form a Cauchy sequence We can now make the following definition q n 19 q n are

20 n Definition When we write α = [a 0, a 1, ], we mean that α = lim n q n Theorem 611 If A R \ Q, then the continued fraction exansion of A is infinite and A is indeed equal to its continued fraction obtained according to (611), ie Proof Exercise n A = lim = [a 0, a 1, ] n q n Definition We say that a continued fraction is urely eriodic if it is of the form [b 0, b 1,, b m ] = [b 0, b 1,, b m, b 0, b 1,, b m, b 0, b 1, ] We say that a continued fraction is eriodic if it is of the form Examles [a 0,, a k, b 0, b 1,, b m ] = [a 0,, a k, b 0, b 1,, b m, b 0, b 1,, b m, b 0, b 1, ] = [1, ] 3 = [1, 3, 1, 5, 1, 1, 4, 1, 1, 8, 1, 14, 1, 10,, 1, ] π = [3, 7, 15, 1, 9, 1, 1, 1,, 1, 3, 1, 14,, 1, ] e = [, 1,, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, ] Examle 61 Let A = [a, b] Note that our rocedure (611) ensures that b 1 Then On the other hand, so A = a + 1 B, where B = [ b ] > 1 B = b + 1 B, B = bb + 1 Since B > 0, the quadratic formula tells us that B = b + b + 4, 0

21 and therefore A = a + b + b + 4 b + b + 4 b + b + 4 = a b + b + 4 In articular, if b = a we get that [a, a, a,, a, ] = a + 1 We have already seen this for a = 1, namely we have seen that = [1, ] Lemma 613 If B = [b 1,, b m ] has a urely eriodic continued fraction with, then there are ositive integers x, y, u, v such that B = xb + y ub + v Proof The key observation here is, as above, that B = b 1 + The rest is just algebraic maniulation Theorem 614 ie b m+ 1 B (i) If the continued fraction exansion of a real number A is eriodic, A = [a 0,, a k, b 0, b 1,, b m ], then there exist integers r, s, t, d with d > 0 not a square, s, t 0 such that A = r + s d t (ii) Let d be a ositive integer that is not a erfect square and r, s, t Z, s, t 0 Then the continued fraction of A = r + s d t is eriodic Proof For the first art, denote B = [b 0, b 1,, b m ] Lemma 613 imlies that B satisfies a quadratic equation ab + bb + c = 0 Note that the discriminant = b 4ac > 0 since B R \ Q (This is because B is the limit of a sequence of ositive rational numbers, but it cannot be rational itself since its continued fraction is infinite) In other words, B is of the form B = r + s d t 1

22 for some integers r, s, t In articular B F d and since F d is closed under addition, division and multilication, it follows that A F d, so A is of the desired form The argument for the second art is more comlicated First observe that A = 0 + D q 0 for some integers D > 0, 0, q 0 and that we can choose them so that q 0 0 D (In order to see this, you might want to take advantage of the fact that A is the root of some quadratic equation with integer coefficients) The rocedure (611) for comuting the continued fraction of A is the following Let x 0 = A, a 0 = x 0 Next, we set Then at each ste we set x i+1 = x 1 = 1 x 0 x 0 1 and a 1 = x 1 1 x i x i > 1 and a i+1 = x i+1 1 The statement of the second art of the theorem is equivalent to showing that there exist ositive integers m, k such that x k = x k+m (In this case, the continued fraction rocedure ensures that a k = a m+k and x k+1 = x m+k+1, etc ) We have x 1 = 1 q = D q 0 a 0 0 a 0 q 0 + D = q 0 ( 0 a 0 q 0 + D) 0 D + q 0 (a 0q 0 a 0 0 ) Since q 0 0 D it follows that 1 = 0 a 0 q 0 and q 1 = 0 D q 0 and x 1 = 1 + D q 1 Note that q 1 = 1 D q 0, hence q 1 1 D An induction argument shows that, for i 1 we have 1 < x i = i + D q i i+1 = i a i q i Z + a 0q 0 a 0 0 are integers q i+1 = i+1 D q i Z

23 q i i D q i 0 (this comes down to D / Q) q i > 0 0 < i D < q i < i + D < D It follows that ( i, q i ) must be ositive integers smaller than D There are only finitely many ossibilities for such airs, and thus, the igeonhole rincile ensures that there are ositive integers m, k such that ( k, q k ) = ( m+k, q m+k ) and so x k = x m+k Going back to our initial goal, the study of Pell s equations, we can now formulate the following results We state them without roofs, which you can find in any standard textbook One of my favorites is Davenort s Higher Arithmetic Theorem 615 Let d > 0 be a ositive integer that is not a square (i) Then the eriodic fraction of d is of the form d = [a, b1,, b m 1, a] with b i = b m i for 1 i m 1 (ii) Let written in lowest terms equation q = [a, b 1,, b m 1 ] Then (, q) is the smallest ositive integer solution to the x dy = ( 1) m (iii) The smallest ositive integer solution to Pell s equation is given by { (, q) x dy = 1 (61) if m is even; ( + dq, q) if m is odd Theorem 616 Let d > 0 not a erfect square Pell s equation x dy = 1 (6) has ositive integer solution if and only if the continued fraction of d has odd eriod 3

24 7 Primes of the form = x + ny In Section we roved that a rime can be written as the sum of two squares if and only if = or 1 (mod 4) One direction was easy, but the other one was comletely non-trivial The roof consisted of two stes Recirocity ste: A rime 1 (mod 4), then it divides N = a + b with a and b relatively rime integers The roof was a bit ad-hoc We used the fact that 4 φ() to find an integer a for which a (mod ) Descent ste: If a rime divides a number N of the form N = a + b, where (a, b) = 1, then itself can be written as = x + y for some (x, y) = 1 This ste was based on Lemma 3 which said that if a rime q = x +y divides a sum of squares a + b = N with (a, b) = 1, then N/q can be written as a sum of relatively rime squares Furthermore, we used in an essential way the fact that if a number N is the sum of two squares, then all its rime divisors can be written as sums of two squares One can look at other questions of this tye For instance, Fermat himself stated (and Euler roved) the following two results Theorem 71 A rime is of the form = x + y if and only if = or 1 or 3 (mod 8) Theorem 7 A rime is of the form = x + 3y if and only if = 3 or 1 (mod 3) Again, it is easy to show that if the rime has the given form in terms of squares, then it lands in the desired congruence class For the other direction, let us try to imitate the rocedure from Section Descent ste We start by considering the generalization of the Lemma 3 This is the first comonent of our descent ste Lemma 73 Fix n Z >0 Suose M is an integer of the form M = a +nb with (a, b) = 1 and that q = x + ny is a rime divisor of M Then there exist integers (c, d) = 1 such that M/q = c + nd Proof The general case is one of the homework roblems Here we discuss only the roof in the case n = We know that M = a + b, (a, b) = 1, q = x + y is rime and q M Since q is rime x and y are forced to be relatively rime Just as in the roof of Lemma 3 we look at x M b q = x (a + b ) b (x + y ) = (ax by)(ax + by) 4

25 Since q (x M b q) it follows that q (ax by) or q (ax + by) Without loss of generality (we can always change the sign of b), we can assume that q ax by Thus, there exist an integer d such that ax by = dq We can rewrite this as by = ax dq = ax dx dy, which imlies that x y(b + dy) Not only is x relatively rime to y, but it is also odd (if x is even then q cannot be rime) Therefore x (b + dy), so b + dy = cx for some integer c On the other hand, cxy = y(b + dy) = x(a dx), so But then a dx = cy M = a + b = (dx + cy) + (cx dy) = (x + y )(c + d ) = q(c + d ) Note that since (a, b) = 1 we must also have (c, d) = 1 And now we try to reroduce the second comonent of the descent ste That is we would like to say that rime, a + nb with (a, b) = 1 = = x + ny (71) As in the Section we can assume that Then, if is odd a, b a + nb < n If n 3, this imlies that a + nb < and therefore any rime divisor q of a + nb has to be q < Now we can comlete the descent ste using the same argument as in Section For n = 1 : done in Section For n = : assume that cannot be written as x + y (7) 5

26 If all the other rime divisors of a + b could be written in the form (7), then Lemma 73 would imly that can also be written as in (7) and we assumed that this is not the case (Here we used that a + b because a + b < ) Hence there must exist a rime divisor 1 of a + b that cannot be exressed as (7) But we have seen that any other rime divisor 1 of a +b has to be 1 < By the same argument now there must exist yet another rime < 1 < that cannot be written in the given form (7) And then another, and another There is nothing to revent us from reeating this rocess indefinitely (note that is of the desired form) and thus we get an infinite decreasing sequence of ositive (and rime) numbers This contradiction finishes the descent ste For n = 3 : see the homework roblems Note that (71) cannot hold in general For instance, in the case n = 5 we see that 3 1 = 1 + 5, but 3 cannot be written as x + 5y So we need to figure out how the rime divisors of a + nb can be reresented The answer will come from Legendre s theory of reduced quadratic forms (See Section 9) Recirocity ste We need to find congruence conditions which will guarantee that x + ny for some (x, y) = 1 The roblem is that we cannot adat directly our roof from the n = 1 case (Section ) This is because our roof was done in an ad-hoc manner Namely, to reca, we said that if 1 (mod 4), then φ() = 4k for some integer k Therefore the olynomial X 4k 1 has 4k roots (mod ) But X 4k 1 = (X k 1)(X k+1 + 1) Since X k 1 can have at most k roots (mod ), it follows that there must exist an integer (a, ) = 1 such that a k +1 0 (mod ) Thus (a k ) +1 and since a k and 1 are relatively rime, we are done But this cannot be relicated directly for n = for instance One more thing that is worth noticing We have the following conjectures (due to Fermat) n = 1 : 1 (mod 4) = a + b for some (a, b) = 1 n = : 1, 3 (mod 8) = a + b for some (a, b) = 1 n = 3 : 1 (mod 3) = a + 3b for some (a, b) = 1 The key observation is that these are all congruences modulo 4n (The last one can be restated as 1, 7 (mod 1)) And indeed, we are going to find conditions (mod 4n) that would ensure that a rime is of the form x +ny A systematic aroach will be formulated in terms of the Legendre symbol (see Section 8) 6

27 8 Quadratic recirocity 81 Legendre symbol In this section will be an odd rime Definition An integer a 0 (mod ) is called a quadratic residue modulo if there exist x Z such that x a (mod ); otherwise the integer a 0 (mod ) is called a quadratic nonresidue modulo Note that the definition deends only on the residue class of a (mod ) Examle = 3 = 5 = 7 quadratic residues 1 1, 4 1,, 4 quadratic nonresidues, 3 3, 5, 6 Lemma 81 In any reduced residue system modulo, there are exactly 1 quadratic residue and 1 quadratic nonresidues Proof Exercise Note: We could try to make a similar definition modulo an odd ositive integer n But Lemma 81 would not hold For instance, if we take n = 15 we have 8 modulo classes relatively rime to 15 : 1,, 3, 7, 811, 13, 14 But only 1 and 4 are quadratic residues Definition The Legendre symbol modulo is the function Z C given by ( ) a 0 if a = 1 if a and a is a quadratic residue (mod ) 1 if a and a is a quadratic nonresidue (mod ) Examle ( ) ( ) ( ) 1 43 = 1 = 1 = ( ) 1 In general = 1 for any odd rime ( ) = 1 3 ( ) = 1 7 ( ) 14 = 0 7 The connection to the recirocity ste in Section 7 is rovided by the following fact Proosition 8 Let n be an integer relatively rime to Then ( ) n a + nb for some integers (a, b) = 1 = 1 7

28 Proof First assume that there exist integers (a, b) = 1 such that a +nb 0 (mod ) Since a and b are relatively rime, it follows that b 0 (mod ) Therefore there exist c Z such that bc 1 (mod ) But then ( ) n a c + n 0 (mod ) = = 1 The other direction is even simler Since n is a quadratic residue (mod ), there exists an integer a such that a n (mod ) Hence a + n 1 and (a, 1) = 1 Corollary 83 ( ) 1 = ( 1) 1 = { 1 1 (mod 4) 1 3 (mod 4) Proof Follows immediately from Proosition 8 and Theorem 1 Lemma 84 (Euler s criterion) ( ) a a 1 (mod ) Proof If a we get 0 on both sides and the equality holds If a, then a 1 1 (mod ), so a 1 ±1 (mod ) We have two cases If If ( ) a = 1, there exists x 0 (mod ) such that a x (mod ), so a 1 x 1 (mod ) 1 (mod ) ( ) a (mod ) ( ) a = 1, it is enough to show that a 1 1 (mod ) Consider the olynomial f(x) = X 1 1 It has at most 1 roots modulo On the other hand, we have seen from the revious case that all the quadratic residues are roots of f(x) By Lemma 81, there are exactly 1 quadratic residues (mod ) Hence no quadratic residue can be a root of f(x), and we are done Proosition 85 The Legendre symbol modulo is a comletely multilicative function 8

29 Proof We aly Euler s criterion twice ( )( ) a b a 1 b 1 (mod ) (ab) 1 (mod ) ( ) ab (mod ) The result follows since 1 1 (mod ) and 0 ±1 (mod ) Proosition 86 ( ) { = ( 1) 1 1 ±1 (mod 8) 8 = 1 ±3 (mod 8) Proof By Euler s criterion we know that ( ) 1 (mod ) There are exactly 1 even integers between 1 and We have the following congruences for them 1 1( 1) 1 (mod ) ( 1) (mod ) 3 3( 1) 3 (mod ) 4 4( 1) 4 (mod ) One of the columns will end with either 1 the roduct of all these relations we obtain ( ) 1 4 ( 1)!( 1) which can be rewritten as 1 ( 1 Since ( ) 1! this simlifies to and the desired result follows Corollary 87 )! ( 1 or (whichever one is even) Taking (mod ), )!( 1) 1 8 (mod ) 1 ( 1) 1 8 (mod ) a + b for some integers (a, b) = 1 1, 3 (mod 8) 9

30 Proof By Proosition 8 we know that a + b for some integers (a, b) = 1 ( ) = 1, so all ) we need to do is figure out for which residue classes (mod 8) the Legendre symbol is equal to 1 Since the Legendre symbol is comletely multilicative we have ( ( ) = 1 ( )( ) 1 = 1 In other words, we need to see when ( ) 1 = ( ) ( 1) 1 = ( 1) 1 8 If = 8k + 1, then 1 1 = 4k, (mod ) = 8k + k both even If = 8k + 3, then 1 = 4k + 1, 1 8 = 8k + 6k + 1 both odd If = 8k + 5, then 1 = 4k +, 1 8 = 8k + 10k + 3 one even, one odd If = 8k + 7, then 1 = 4k + 3, 1 8 = 8k + 14k + 6 one odd, one even The above Corollary, together with the descent ste for n = that we roved in Section 7, roves Theorem 71 Lemma 88 (Gauss s Lemma) Assume n 0 (mod ) For 1 t 1 denote by x t the remainder of the division of tn by Let Then Proof Denote r = 1 m = #{x t ; x t >, 1 t 1 } ( ) n = ( 1) m First note that x 1,, x r are distinct integers between 1 and 1 Indeed, since they are remainders to divisions by, then have to be 0 x t 1 On the other hand, since n, cannot divide any of the integers n, n, 3n,, 1 n So x t 1 for all 1 t 1 30

31 On the other hand, if x t = x s for some 1 s, t 1, we must have tn sn (mod ) This means t s and given the range of ossible values for s and t, the only way this could haen is for s = t Denote by A the set of x t s that are < / and by B the set of x t s that are > / Note that by definition m = #B Denote k = #A Since A B = {x 1,, x r } and A B = and not two x t s are the same, it follows that k + m = r = 1 Denote by a 1,, a k the elements of A and by b 1,, b m the elements of B Let Then #C = m and both C = {c 1,, c m } where c j = b j, 1 j m A, C { 1,,, 1 } (81) Claim A C = If we had a i = c j for some 1 i k and some 1 j m, then a i + b j = By the very definition of the sets A and B, there exist integers 1 s, t 1 such that a i = x s sn (mod ) and b j = x t tn (mod ) Therefore sn + tn 0 (mod ) Since (n, ) = 1, this means that s + t But this is imossible given that 0 < s + t 1 The claim imlies that #A C = m + k = 1 (8) Taken together, (81) and (8) imly that { A C = 1,,, 1 } Therefore the roduct of all elements of A C is ( ) 1 a 1 a k c 1 c m =! Therefore ( ) 1! a 1 a k ( b 1 ) ( b m ) (mod ) ( 1) m a 1 a k b 1 b m (mod ) 31

32 Going back to the definition of the sets A and B, this can be rewritten as ( ) 1! ( 1) m x t (mod ) ( 1) m tn (mod ) 1 t r 1 t r Hence ( ) ( ) 1! ( 1) m n 1 1! (mod ) Since ( ) 1! 0 (mod ), multilying both sides by ( 1) m gives us ( 1) m n 1 (mod ), and the result follows from Euler s criterion (Lemma 84) Note that we are interested only in the arity of m The following result deals with said arity Proosition 89 Let n be an integer not divisible by With the same notation as in Gauss s Lemma 88, we have m (n 1) 1 + tn (mod ) 8 In articular, if n is odd, then Proof Denote m 1 t 1 γ = 1 t 1 tn (mod ) 1 t 1 tn We will use the same notation from the roof of Gauss s Lemma 88 For each 1 t 1 we have { } tn tn tn = + and the fractional art is strictly between 0 and 1 It follows that { } tn x t = = tn tn (83), 3

33 Recall that we defined sets A = {a 1,, a k }, B = {b 1,, b m } and C = {c 1,, c m } with { c j = } b j, 1 j m By definition, A and B are disjoint and their union is xt ; 1 t 1, so k m a i + b j = x t Let α = k a i and β = i=1 i=1 j=1 1 t 1 m b j Substituting (83) above we get that j=1 α + β = 1 t 1 i=1 tn j=1 1 t 1 1 t 1 tn = n 1 γ (84) 8 We have also seen that the sets A and C are disjoint and their union is { } 1,,, 1 Therefore k m a i + c j = t = 1 8 We can rewrite this as which imlies that Adding u (84) and (85) yields α + m j=1 ( b j ) = 1 8, α β + m = 1 (85) 8 When we reduce this α + m = (n + 1) 1 8 γ mod, taking into account that is odd, we obtain m m (mod ) (n + 1) 1 8 γ (mod ) (n 1) γ (mod ) Theorem 810 (Quadratic recirocity law) If and q are odd rimes, then ( ) ( ) = ( 1) ( 1)(q 1) q 4 q 33

34 Proof If the two rimes are equal, the relation obviously holds If they are different, then the Legendre symbols are nonzero, and so the relation is equivalent to ( )( ) q = ( 1) ( 1)(q 1)/4 (86) q By Gauss s Lemma 88 and Proosition 89, the two Legendre symbols are ( ) q = ( 1) m 1 where m 1 ( ) = ( 1) m where m q Hence (86) would follow if we roved that tq + 1 t 1 1 s q 1 s = 1 q 1 t 1 1 s q 1 tq s q (mod ); (mod ) q 1 (87) To this end, consider the function f(x, y) = qx y on the domain x <, y < q A coule of observations about f(x, y) are in order x, y Z = f(x, y) Z (x 1, y 1 ) (x, y ) airs of integers in our domain = f(x 1, y 1 ) f(x, y ) The first observation is immediate For the second, note that, if f(x 1, y 1 ) = f(x, y ) then q(x 1 x ) = (y 1 y ) Thus x 1 x and q y 1 y Given the range in which these integers live, this is ossible only if x 1 x = 0 and y 1 y = 0 Therefore f(x, y) takes 1 q 1 nonzero values as the integer x ranges from 1 to 1 and the integer y ranges from 1 to q 1 Now we count the number of ositive and negative values of f(x, y) in this range Fix the integer 1 x 1 Then f(x, y) > 0 qx > y y < qx 1 y tx and so, the number of ositive values that f(x, y) takes is recisely tx 1 x 1 Similarly, fix an integer 1 y q 1 Then f(x, y) < 0 qx < y x < y q 1 x y, q 34

35 and the number of negative values that f(x, y) takes is y q 1 y q 1 Therefore (87) holds and the theorem is roved Note: This roof has a nice interretation in terms of lattice oints in the lane Find it! Examle: Determine whether 583 is a quadratic residue or nonresidue (mod 907) ( ) ( )( ) ( ) = = ( 1) ( ) ( )( ) 3 3 = = ( 1) (53 1)(53+1) ( 1) ( 1) 53 1 ( ) ( ( 9 = 11 ) = ( 3 )( ) 6 53 ) = 1 Therefore 583 is a quadratic nonresidue (mod 907) Now we are ready to rove the recirocity ste for rimes of the form x + 3y For that, we need to figure out for which rimes 3 is a quadratic residue, and for which it is not For > 3 we have ( ) 3 = ( 1) ( ) = ( 1) 1 3 ( ) 3 The first factor yields a condition modulo 4, while the second yields a condition modulo 3 Thus we need to look at congruence classes modulo 1 There are four cases (mod 1) (mod 4) (mod 3) ( ) ( ) ( ) ( 1) Here we used the fact that ( ) ( )( ) = = ( 1) 1 ( ) 3 Therefore, Proosition 8 imlies the recirocity ste for n = 3 below Proosition 811 Let > 3 be a rime Then a + 3b for some integers (a, b) = 1 1, 7 (mod 1) 35

36 The above result, together with the descent ste outlined in Problem 4 of Homework 4, rove Theorem 7 The general roblem of which rimes can be written as x + ny with n a fixed ositive integer is more comlicated though However, quadratic recirocity allows us to get closer to our goal of understanding the recirocity ste Proosition 81 If and q are distinct odd rimes, then ( ) q = 1 ±a (mod 4q) for some odd integer a Proof Let = ( 1) 1 Then But we know that and therefore ( ) ( ) ( ( 1) ( 1)/ 1 = = q q q ( ) q ( ) 1 q = ( 1) q 1, = ( 1) 1 By quadratic recirocity (Theorem 810), ( ) = q q 1 ( ) q ( ) q ) 1 ( ) q Therefore it remains to rove that ( ) = 1 ±a (mod 4q) for some odd integer a q The roof of this last equivalence is left as an exercise 8 Jacobi symbol In order to move forward toward our goal of comleting the recirocity ste in Euler s strategy we need to extend the Legendre symbol beyond rimes This extension is due to Jacobi Definition Let m be an odd ositive integer ( ) If m = 1, the Jacobi symbol : Z C is the constant function

37 If m > 1, it has a decomosition ( as ) a roduct of (not necessarily distinct) rimes m = 1 r The Jacobi symbol : Z C is given by m ( ) ( ) ( ) a a a = m Note: The Jacobi symbol does not necessarily distinguish between quadratic residues and nonresidues That is, we could have ( a m) = 1 just because two of the factors haen to be 1 For instance, ( ) ( )( ) = = ( 1)( 1) = 1, but is not a square modulo 15 The following roerties of the Jacobi symbol are direct consequences of its definition Proosition 813 Let m, n be ositive odd integers and a, b Z Then ( ) 1 (i) = 1; m ( ) a (ii) = 0 (a, m) > 1; m ( ) ( ) a b (iii) a b (mod m) = = ; m m ( ) ( )( ) ab a b (iv) = ; m m m ( ) ( )( ) a a a (v) = ; mn m n ( ) ( ) a b b (vi) (a, m) = 1 = = m m Proof Exercise Theorem 814 Let m, n be ositive odd integers Then ( ) 1 (i) = ( 1) m 1 ; m ( ) (ii) = ( 1) m 1 8 ; m ( ) ( ) n (iii) = ( 1) m 1 n 1 m m n 37 1 r

38 Proof The first two formulas are trivially true when m = 1 and so is the third if m = 1 or n = 1 or if (m, n) > 1 We assume that m, n > 1 and (m, n) = 1 Thus m = 1 r and n = q 1 q s for some rimes i and q j and i q j for all 1 i r, 1 j s Then m = r i = i=1 r r (1 + ( i 1)) = 1 + ( i 1) + ( i1 1)( i 1)+ i=1 i=1 1 i 1 <i r roducts of 3, 4 and so on factors Since m is odd, so are the rimes i Therefore i 1 0 (mod ) and ( i1 1)( i 1) 0 (mod 4) Therefore all the terms in the above sum that are imlicit are also divisible by 4 Hence r m 1 + ( i 1) (mod 4), which is to say m 1 i=1 r ( i 1) (mod 4) i=1 Since m and the i s are odd, it follows that m 1 0 (mod ) and (mod ), 1 i r Thus we can divide each term above by and still get integers It follows that so Similarly, m 1 r i=1 ( 1) m 1 = ( 1) P r i 1 i=1 = i 1 r i=1 ( 1) i 1 = (mod ), (88) r ( ) 1 = i=1 i ( ) 1 m m = r i = i=1 r ( 1 + ( i 1) ) r = 1 + ( i 1) + ( i 1 1)( i 1)+ i=1 i=1 1 i 1 <i r roducts of 3, 4 and so on factors We use again the fact that both m and the i are odd That means that m 1 = (m 1)(m + 1) is the roduct of two consecutive even integers, so one of them is divisible by 4 Thus m 1 0 (mod 8) and likewise i 1 0 (mod 8), 1 i r It follows that the roduct of two or more factors in the above summation is divisible by 64, hence m 1 r ( i 1) (mod 64) i=1 38