MATH 216T TOPICS IN NUMBER THEORY

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1 California State University, Fresno MATH 6T TOPICS IN NUMBER THEORY Spring 00 Instructor : Stefaan Delcroix

2 Chapter Diophantine Equations Definition. Let f(x,..., x n ) be a polynomial with integral coefficients in x,..., x n. We call the equation f(x,..., x n ) = 0 a diophantine equation if we are looking for all solutions (x,..., x n ) Z n or (x,..., x n ) Q n.. The Diophantine Equation ax + by = c Theorem. Let a, b Z 0 and c Z. Then the following holds about the diophantine equation ax + by = c : (a) The equation has a solution if and only if gcd(a, b) c. (b) Suppose that (x 0, y 0 ) is a solution. Then all the solutions are given by where d = gcd(a, b). x = x 0 t b d y = y 0 + t a d, t Z Proof : Put d = gcd(a, b). (a) Suppose first that (x 0, y 0 ) is a solution. Then ax 0 + by 0 = c. Since d a and d b, we get that d (ax 0 + by 0 ) and so d c. Suppose next that d c. Then c = dk for some k Z. We ve seen in MATH 6 that the greatest common divisor of a and b can be written as a linear combination of a and b. So there exist u, v Z with au + bv = d. Hence a(uk) + b(vk) = dk = c. So the equation ax + by = c has a solution (namely (x, y) = (uk, vk)). (b) Suppose that (x 0, y 0 ) is a solution. Then d c by (a). Pick t Z. Put x = x 0 t b d and y = y 0 + t a d. Then ax + by = a ( x 0 t b d ) + b ( y 0 + t a d ) = ax 0 t ab d + by 0 + t ba d = ax 0 + by 0 = c

3 So (x, y) is a solution. Next, we show that every solution is of this form. Suppose that (x, y) is a solution. Then ax + by = c = ax 0 + by 0 Hence b(y y 0 ) = a(x 0 x). Put a = da and b = bd. Then db (y y 0 ) = da (x 0 x) and so b (y y 0 ) = a (x 0 x). But gcd(a, b ) =. Since b a (x 0 x), we get that b (x 0 x). Hence x 0 x = b t for some t Z. Then y y 0 = a t. So x = x 0 b t and y = y 0 + a t. Example : Find all the solutions of the diophantine equation 457x + 67y = Are there any solutions (x, y) N? We start by calculating (457, 67) and writing it as a linear combination of 457 and 67. Hence = ( ) so R 3 = R 7R and change signs 7 67 = 6 + ( 5) so R 4 = R 6R 3 and change signs = 5 + so R 5 = R 3 R = + so R 6 = R 4 R = + 0 so we stop Multiplying both sides by 6074, we get that gcd(457, 67) = = ( 8) ( 69999) = 60, 74 So ( 69999; ) is a particular solution of 457x + 67y = Hence all integral solutions of 457x + 67y = 6074 are { x = t y = t, t Z Are there any solutions (x, y) N? So Hence we get t 0 and t 0 t and t Since t is an integer, we get that t = So x = = 66 and y = = 456 the only natural solution of 457x + 67y = 6074 is (x, y) = (66, 456)

4 We have the following result for natural solutions of ax + by = c. Proposition.3 Let a, b, c N 0 with gcd(a, b) = and c > ab. Then the equation ax + by = c has a solution (x, y) N 0. Proof : Let (x 0, y 0 ) be a solution of ax + by = c. Then all the solutions are given by { xt = x 0 tb y t = y 0 + ta, t Z Hence the solutions are equally spaced along the line L ax + by = c : for all t Z, the distance between (x t+, y t+ ) and (x t, y t ) is ([(x0 (t + )b] [x 0 tb]) + ([y 0 + (t + )a] [y 0 + ta]) = a + b Note that the X-intercept (resp. Y -intercept) of L are ( c ) ( a, 0 resp. 0, c ) b The distance between these two intercepts (which is the length of the part of L that is in the first quadrant) is given by c a + c b = c a + b ab > a + b since c > ab. Hence there exists t Z with (x t, y t ) N 0 (so (x t, y t ) is in the first quadrant).. Unique Prime Factorization over N The fact that we can write every n N with n uniquely (up to order) as a product of primes can be a very powerful tool. We illustrate this with a question from the International Math Olympiad. The original question was : (IMO 997, B) : Find all (a, b) N 0 with a b = b a. We start with an easy lemma : Lemma.4 Let n N with n. Then we have the following : (a) n k > k for all k N with k 5. (b) n k > k for all k N 0. 3

5 Proof : (a) We use induction on k. For k = 5, we have n 5 = n 3 3 = 8 > 5. So suppose that n k > k for k = 5, 6,..., m for some m N with m 5. Then using induction, we get n (m+) = n m = n n m > n m m m + since m 5. (b) Similar to the proof in (a). Now we tackle the question. One easily checks that a = b =. So we may assume that a, b >. Hence we can consider the prime factorization of a and b. We only use the primes that show up in either factorization : a = n i= p α i i and b = where p < p < < p n are primes, α i, β i N and not both α i and β i are zero for i =,,..., n. Substituting this into a b = b a, we find n i= Using unique prime factorization, we get p α ib i = n i= p β ia i n i= p β i i α i b = β i a for i =,,..., n ( ) Suppose first that a b. From (*), we get that α i β i for i =,,..., n. Hence b = n i= p β i i divides n i= p α i i = a So a = kb for some k N with k. Hence by (*), α i = kβ i for i =,,..., n. So a = n i= p α i i = n i= p kβ i i = b k Putting everything together, we find kb = a = b k and so b k = k By Lemma.4(a), k {, 3, 4}. One easily checks that k = does no lead to a solution while k = 3, 4 lead to the solutions (a, b) {(7, 3), (6, )}. Suppose next that a b <. From (*), we get that α i < β i for i =,,..., n. Hence a = n i= p α i i divides but doesn t equal n i= p β i i = b 4

6 So b = ka for some k N with k. Hence by (*), β i = kα i for i =,,..., n. So b = n i= p β i i = n i= p kα i i = a k Putting everything together, we find ka = b = (a k ) = a k and so By Lemma.4(b), there are no solutions. a k = k All couples (a, b) N 0 with a b = b a are {(, ), (6, ), (7, 3)}..3 Pythagorean Triples In this section, we will find all the solutions (x, y, z) N 3 0 of the diophantine equation x + y = z Definition.5 () A triple (x, y, z) N 3 0 is a Pythagorean triple if x + y = z. () A Pythagorean triple (x, y, z) is primitive if gcd(x, y, z) =. Remarks : () If (x, y, z) is a Pythagorean triple, then so is (dx, dy, dz) for any d N 0. () Let (x, y, z) be a Pythagorean triple. Put d = gcd(x, y, z). Then x = dx, y = dy and z = dz for some x, y, z N 0. So (dx ) + (dy ) = (dz ). Hence x + y = z. Note that gcd(x, y, z ) =. So (x, y, z ) is a primitive Pythagorean triple and (x, y, z) = (dx, dy, dz ). Hence it is enough to find all primitive Pythagorean triples. (3) Let (x, y, z) be a Pythagorean triple. Then gcd(x, y, z) = gcd(x, y) = gcd(y, z) = gcd(x, z) =. Lemma.6 Let (x, y, z) be a primitive Pythagorean triple. Then either x is even and y is odd or x is odd and y is even. Proof : Since gcd(x, y) =, we get that x and y can not both be even. Suppose x and y are both odd. Then z x + y + mod 4 a contradiction since is not a quadratic residue modulo 4. So either x is even and y is odd or x is odd and y is even. If (x, y, z) is a primitive Pythagorean triple, then so is (y, x, z). By Lemma.6, it is enough to find all primitive Pythagorean triples with x even and y odd. 5

7 Theorem.7 Let (x, y, z) N 3 0 with x even. Then (x, y, z) is a primitive Pythagorean triple if and only if (x, y, z) = (mn, m n, m + n ) for some m, n N 0 with m > n, gcd(m, n) = and m and n not both odd. Proof : Suppose (x, y, z) is a primitive Pythagorean triple. By Lemma.6, we get that y is odd. Since x + y = z, we get that z is odd. Hence z + y and z y are even. We get that x = z y = (z + y)(z y) and so ( ) x = ( z + y Put d = gcd, z y ). Then ( z + y d + z y ) ( ) ( ) z + y z y ( z + y and d z y ) So d z and d y. Hence d gcd(z, y). But gcd(z, y) =. So d =. By HW #, there exist m, n N such that z + y = m z y and = n ( z + y Then gcd(m, n) = since gcd, z y ) =. We easily get that x = mn, y = m n and z = m + n Since y > 0, we have that m > n. Since gcd(m, n) =, m and n can not both be even. If m and n are both odd, then y is even, a contradiction. So m and n are not both odd. One easily checks that (mn, m n, m + n ) is a primitive Pythagorean triple if m, n N 0 such that gcd(m, n) =, m > n and not both m and n are odd. Remark : It follows now that all Pythagorean triples (x, y, z) are given by (x, y, z) {d(mn, m n, m + n ), d(m n, mn, m + n )} where d, m, n N 0, m > n, gcd(m, n) = and not both m and n are odd. Example : Find all Pythagorean triples (x, y, z) such that one of the variables has value 5. We may assume that (x, y, z) = d(mn, m n, m + n ) where d, m, n N 0, m > n, gcd(m, n) = and not both m and n are odd. So either d(m n ) = 5 or d(m + n ) = 5. So d 5. Hence d {, 3, 5, 5}. Suppose first that d(m + n ) = 5. We easily get (by going over all the cases for d) that the only solution is (d, m, n) = (3,, ) and so (x, y, z) = (, 9, 5). 6

8 Suppose next that d(m n ) = 5. Then d(m n)(m + n) = 5. Note that m n < m + n. This leads to the following possibilities : (d, m n, m + n) {(,, 5), (, 3, 5), (3,, 5), (5,, 3)} Hence (d, m, n) {(, 8, 7), (, 4, ), (3, 3, ), (5,, )}. So (x, y, z) {(, 5, 3), (8, 5, 7), (36, 5, 39), (0, 5, 5)} All Pythagorean triples (x, y, z) such that one of the variables has value 5 are given by (, 9, 5), (, 5, 3), (8, 5, 7), (36, 5, 39), (0, 5, 5) (9,, 5), (5,, 3), (5, 8, 7), (5, 36, 39), (5, 0, 5).4 The Chord-Tangent Method In this section, we illustrate with an example how to find all rational points on an irreducible quadratic curve if one rational point is known. This method is called the Chord-Tangent Method of Diophantus. We want to find all the rational points on the circle C with equation x + y =. Clearly, the point P = (0, ) is a rational point on C. Suppose Q is another rational point on C. Then the line through P and Q is either vertical or has a rational slope. Conversely, let l be a line through P that is either vertical or has a rational slope. Then this line l will intersect the circle C in a second point Q, which turns out to be rational. So all the rational points on C can be found by intersecting C with lines through P that are either vertical or have a rational slope P. We now find that second point of intersection between the circle C and such a line l. The vertical line through (0, ) intersects the circle in (0, ) and (0, ). So suppose l has a rational slope. Then an equation for l is y = mx where m Q. So the points of intersection between the circle C and the line l are the solutions of { y = mx We get that or x + y = x + (mx ) = (m + )x mx = 0 This is a quadratic equation in x and we know that x = 0 is a solution. The second solution is So x = y = mx = m m m + m m + = m m + 7

9 Hence the second point of intersection is ( ) m Q = m +, m m + Notice that we get the point (0, ) if we consider the limit as m + in the expression for Q. This is normal since we can view a vertical line as a line with infinite slope. We get that all the rational points on x + y = are {(0, )} {( m m +, m m +.5 The Method of Infinite Descent ) } : m Q In this section, we prove that the diophantine equation x 4 + y 4 = z 4 has no solutions with xyz 0. We begin with the following lemma. Lemma.8 Let x, y, z N 0 with x 4 + y 4 = z. Then there exist a, b, c N 0 such that a 4 + b 4 = c and c < z. Proof : Suppose first that gcd(x, y) >. Then there exists a prime p such that p x and p y. So x = pa and y = pb for some a, b N 0. Hence z = x 4 + y 4 = p 4 (a 4 + b 4 ). So p 4 z. Hence p z. So z = p c for some c N 0. Then a 4 + b 4 = c and c < z. Suppose next that gcd(x, y) =. Note that (x ) + (y ) = z and that gcd(x, y, z) =. So (x, y, z) is a primitive Pythagorean triple. We may assume that y is even. Then by Theorem.7, we get that there exist m, n N 0 such that gcd(m, n) =, m > n, not both m and n are odd and x = m n, y = mn and z = m + n So x + n = m. Since gcd(m, n) =, we get that (x, n, m) is a primitive Pythagorean triple. Since y is even, we get that x (and so also x) is odd. Hence n is even. By Theorem.7, there exist r, s N 0 such that gcd(r, s) =, r > s, not both r and s are odd and x = r s, n = rs and m = r + s Since x + n = m, we get that m (and so also m), is odd. Since gcd(m, n) = and m is odd, we get that gcd(m, n) =. But y = m(n). So there exist w, c N 0 such that n = w and m = c. Clearly, w is even, say w = v with v N 0. Then we have that 4v = w = n = 4rs and so rs = v. Since gcd(r, s) =, there exist a, b N 0 such that r = a and s = b. Since r + s = m, we have that a 4 + b 4 = c Note that z = m + n > m m = c c. 8

10 Corollary.9 Let x, y, z N such that x 4 + y 4 = z. Then xyz = 0. Proof : Suppose that xyz 0. Put x 0 = x, y 0 = y and z 0 = z. By Lemma.8, there exist x, y, z N 0 such that x 4 + y 4 = z and z < z 0. Continuing to apply Lemma.8, we get that for all n N, there exist x n, y n, z n N 0 such that x 4 n + yn 4 = zn and z 0 > z > z >. This is impossible since z n N 0 for all n N. Hence xyz = 0. Theorem.0 (Fermat) Let x, y, z N with x 4 + y 4 = z 4. Then xyz = 0. Proof : Note that x 4 + y 4 = (z ). By Corollary.9, xyz = 0. Hence xyz = 0..6 Sums of Two Squares In this section, we study the diophantine equation x + y = n For which n does this equation have a solution? Definition. Let n N. We say that n is the sum of two squares if there exist x, y N with x + y = n. We start with a little lemma that tells us that certain numbers are not the sum of two squares. Lemma. Let n N with n 3 mod 4. Then n is not the sum of two squares. Proof : Suppose that n = x + y for some x, y N. Note that a 0 mod 4 or a mod 4 for all a N. Hence x + y {0,, } mod 4, a contradiction since n 3 mod 4. Being the sum of two squares is a multiplicative property : Lemma.3 Let m, n N. If m and n are both the sum of two squares then mn is also the sum of two squares. Proof : that Suppose that m = a + b and n = c + d for some a, b, c, d N. One easily checks So mn is the sum of two squares. mn = (a + b )(c + d ) = (ac bd) + (ad + bc) Corollary.4 Let a i N for i =,,..., n. If a i is the sum of two squares for i =,,..., n then a a a n is the sum of two squares. Proof : This follows from Lemma.3 by induction on n. 9

11 Lemma.3 makes the following question quite natural : which primes are the sum of two squares? If p is an odd prime that is the sum of two squares then p mod 4 by Lemma.. It turns out that this condition is also sufficient. In order to prove this, we use a special type of complex numbers. Definition.5 () A Gaussian integer is a complex number of the form a+bi with a, b Z. () The set of all Gaussian integers is denoted by Z[i]. Note that Z Z[i]. (3) Let α, β Z[i] with β 0. We say that β divides α (notation : β α) if α = βγ for some γ Z[i]. (4) Let α Z[i]. Then α is a unit if αβ = for some β Z[i]. (5) Let α := a + bi Z[i]. We define the norm of α (notation : N(α)) as the natural number N(α) = a + b. The norm function N is quite a powerful tool to convert Gaussian integers into natural numbers. Lemma.6 The following holds about Gaussian integers and the norm N: (a) N(αβ) = N(α)N(β) for all α, β Z[i]. (b) Let α Z[i]. Then α is a unit if and only if N(α) =. (c) The units of Z[i] are {,, i, i}. Proof : (a) Note that N(α) = α (where stands for the norm of a complex number) for all α Z[i]. Since z z = z z for all z, z C, we get that N(αβ) = N(α)N(β) for all α, β Z[i]. (b) Suppose first that α is a unit. Hence αβ = for some β Z[i]. Applying the norm N to both sides and using (a), we get N(α)N(β) = N(αβ) = N() = + 0 = Since N(α), N(β) N, we have that N(α) = N(β) =. Suppose next that N(α) =. Put α = a + bi with a, b Z. Then a + b =. Put β = a bi. Then β Z[i] and αβ = (a + bi)(a bi) = a + b = So α is a unit. (c) Let α := a + bi Z. By (b), we get that α is a unit N(α) = a + b = Since a, b Z, we have (a, b) {(, 0), (, 0), (0, ), (0, )}. Hence α {,, i, i}. The Gaussian integers Z[i] behave pretty much like the integers Z. We need the concept of a Gaussian prime number in order to proceed. There are two ways of thinking of prime numbers over the integers. In general domains, this leads to two different concepts. 0

12 Definition.7 Let α Z[i] such that α 0 and α is not a unit. () We say that α is irreducible if β, γ Z[i] : α = βγ β is a unit or γ is a unit (b) α is reducible if α is not irreducible. Hence α is reducible if and only if α = βγ for some β, γ Z[i] \ {,, i, i}. (c) α is prime if β, γ Z[i] : α (βγ) α β or α γ To distinguish between p Z being prime and α Z[i] being prime, we call a prime p Z a rational prime while a prime α Z[i] is called a Gaussian prime. In Math 5, we proved the following relation between being irreducible and being a prime in Z[i]. Proposition.8 Let α Z[i]. Then α is irreducible if and only if α is a Gaussian prime. Examples (a) 5 is a rational prime but not a Gaussian prime Note that 5 = ( + i)( i). Since neither + i nor i is a unit, we get that 5 is reducible and hence not a Gaussian prime. (b) 3 is both a rational prime and a Gaussian prime. Suppose that 3 = αβ for some α, β Z[i]. Applying the norm N to both sides, we get that 9 = = N(3) = N(αβ) = N(α)N(β) Since N(α), N(β) N, we get that (N(α), N(β)) {(, 9), (9, ), (3, 3)}. Note that there are no Gaussian integers with norm 3 (indeed, let a, b Z with N(a + bi) = 3; then a + b = 3, a contradiction since a, b Z). Hence N(α) = or N(β) =. So α is a unit or β is a unit. Hence 3 is irreducible and so a Gaussian prime. (c) + i is a Gaussian prime Suppose that + i = αβ for some α, β Z[i]. Applying the norm N to both sides, we get that 5 = + = N( + i) = N(αβ) = N(α)N(β) Since N(α), N(β) N, we get that (N(α), N(β)) {(, 5), (5, )}. Hence N(α) = or N(β) =. So α is a unit or β is a unit. Hence + 5i is irreducible and so a Gaussian prime. Note that this example generalizes : if α Z[i] such that N(α) is a rational prime, then α is a Gaussian prime.

13 (d) 3i is not a Gaussian prime How can we come up with a factorization of 3i? The answer : use the norm! Suppose that 3i = αβ for some α, β Z[i]. Applying the norm N to both sides, we get that 0 = + ( 3) = N( 3i) = N(αβ) = N(α)N(β) Since N(α), N(β) N, we may assume that (N(α), N(β)) {(, 0), (5, )}. If we can exclude (5, ) then 3i will be irreducible. First, we ll find all α Z[i] with N(α) = 5. Putting α = a + bi, we get that N(α) = a + b = 5. Since a, b Z we get So (a, b) {(, ), (, ), (, ), (, ), (, ), (, ), (, ), (, )} α { + i, i, + i, i, + i, i, + i, i} We don t have to check all eight possibilities. If 3i is divisible by + i, then it is also divisible by any unit times + i. So we can divide the eight possibilities in groups of four : { ( + i), ( ) ( + i), i ( + i), ( i) ( + i)} = { + i, i, + i, i} { ( i), ( ) ( i), i ( i), ( i) ( i)} = { i, + i, + i, i} Now we try one element of these groups of four. If neither group works, 3i is irreducible; if one of the groups works then we found a factorization of 3i. First, we try + i : is 3i divisible by + i? We easily get 3i + i = ( 3i)( i) ( + i)( i) = 7i 5 = 5 7 i / Z[i] 5 So 3i is not divisible by + i. Next we try i : 3i i ( 3i)( + i) = ( i)( + i) = 5 5i 5 = i Z[i] Hence we came up with the following factorization : 3i = ( i)( i) Since neither i nor i is a unit (in fact, they are Gaussian primes since their norms are rational primes), we get that 3i is reducible and hence not a Gaussian prime. The following proposition shows the relation between rational primes being Gaussian primes and sums of squares.

14 Proposition.9 A rational prime p is a Gaussian prime if and only if p is not the sum of two squares. Proof : Let p be a rational prime. We will prove that p is reducible if and only if p is the sum of two squares. Suppose first that p is the sum of two squares, say p = a +b with a, b N. Then p = a +b = (a + bi)(a bi). Note that neither a + bi nor a bi is a unit (since p is a rational prime, we have that a 0 b). Hence p is reducible. Suppose next that p is reducible. So p = αβ for some α, β Z[i] \ {,, i, i}. Applying the norm to both sides, we find p = p + 0 = N(p) = N(αβ) = N(α)N(β) Since N(α), N(β) N \ {} and p is a rational prime, we get that N(α) = N(β) = p. Put α = a + bi with a, b Z. Then p = N(α) = a + b. So p is the sum of two squares. Corollary.0 Let p be a rational prime with p 3 mod 4. Then p is a Gaussian prime. Proof : By Lemma., p is not the sum of two squares. So by Proposition.9, p is a Gaussian prime. Next, we prove that a rational prime p with p mod 4, is not a Gaussian prime. We need Euler s Criterion (seen in Math 6). Proposition. (Euler s Criterion) Let p be an odd prime and a Z with gcd(a, p) =. Then a is a square modulo p if and only if a p mod p. Corollary. Let p be an odd prime. Then is a square modulo p if and only if p mod 4. Proof : Suppose first that p mod 4. Then p = 4k + for some k N. Hence ( ) p ( ) k mod p So by Euler s Criterion, is a square modulo p. Suppose next that p 3 mod 4. Then p = 4k + 3 for some k N. Hence ( ) p ( ) k+ mod p So by Euler s Criterion, is not a square modulo p. 3

15 Proposition.3 Let p be a rational prime with p mod 4. Then p is not a Gaussian prime. Proof : Suppose that p is a Gaussian prime. By Corollary., there exists n N with n mod p. Hence p (n + ). So p ((n + i)(n i)). Since p is a Gaussian prime, we get that p (n + i) or p (n i). Hence there exist a, b Z with n ± i = p(a + bi) = pa + pbi Hence n = pa and pb =, a contradiction since b Z and p is a rational prime. So p is not a Gaussian prime. Theorem.4 Let p be a rational prime. Then the following are equivalent : (a) p is a Gaussian prime. (b) p 3 mod 4. (c) p is not the sum of two squares. Proof : This follows from Proposition.9, Corollary.0, Proposition.3 and the fact that is not a Gaussian prime (since = ( + i)( i)). We are now able to describe which natural numbers are the sum of two squares. Note that 0 and are the sum of two squares. Theorem.5 Let n N with n and let n = k i= pα i i be the prime factorization of n. Then n is the sum of two squares if and only if for i =,,..., k, we have that α i is even if p i 3 mod 4. Proof : Assume first that for i =,,..., k, we have that α i is even if p i 3 mod 4. Note that is the sum of two squares. If p is an odd rational prime, then p is the sum of two squares if p mod 4 by Theorem.4 and p is the sum of two squares if p 3 mod 4. So n is the sum of two squares by Corollary.4. Assume next that n is the sum of two squares. Suppose that there exists i {,,..., k} such that p i 3 mod 4 and α i is odd, say i =. Then n = a + b for some a, b N. Let f = gcd(a, b). Put a = fc and b = fd. Then So f n. Put n = f m. Then we have that Let f = k i= pδ i i n = a + b = f c + f d = f (c + d ) m = c + d be the prime factorization of f. Then m = n f = k i= pα i i ( k i= pδ i i ) = k i= p α i δ i i 4

16 Since α is odd, we get that α i δ. So p m. If p c, then p d since d = m c and so p d since p is a prime, a contradiction since gcd(c, d) =. Hence gcd(p, c) =. As seen in Math 6, there exists t N with ct d mod p (indeed, the equation cx d mod p has a solution). So we get 0 m c + d c + (ct) c ( + t ) mod p Since gcd(p, c) =, we get that + t 0 mod p So is a square modulo p, a contradiction to Corollary.. Hence for i =,,..., k, we have that α i is even if p i 3 mod 4..7 Sums of k-th Powers In the previous section, we found all the natural numbers that can be written as a sum of two squares. We now could ask the question : which natural numbers can be written as the sum of three squares? The answer is a theorem similar to Theorem.5. Note that not every natural number can be written as the sum of three squares (7, 5, 3,... are all examples; in fact, if n 7 mod 8 then n can not be written as the sum of three squares). Lagrange however proved the following amazing result : Theorem.6 (Lagrange,770) Every natural number can be written as the sum of four squares. In 770, Edward Waring suggested the following generalization : Given k N, does there exist g(k) N such that every n N can be written as the sum of g(k) k-th powers of natural numbers? He also conjectured that g(3) = 9 and g(4) = 9. In 906, Hilbert proved that the function g(k) indeed exists. g(3) = 9. Only in 986 was it shown that g(4) = 9. We finish this section with a result from Euler (that we can write now as) : [ ( Theorem.7 g(k) k + 3 ) ] k for all k. In 909, it was proven that Notice that in Euler s time, it was only conjectured that g(k) existed. Euler s proof is constructive but his lower bound is truly remarkable : [ (3 ) ] k Mahler proved in 957 that g(k) k + for at most a finite number of values for k. [ (3 ) ] k Stemmler, Kubina and Wunderlich showed in 990 that g(k) = k + for all k 47, 600,

17 .8 The Pell Equation x ny = Let n N that is not a perfect square. Put Q( n) = {a + b n a, b Q} and Z[ n] = {a + b n a, b Z} Since n is irrational and every element of Q( n) can be written uniquely as a+b n for some a, b Q. Recall that the norm of a + b n is N(a + b n) = a nb. We proved that α Z[ n] is a unit if and only N(α) = ±. Putting α = x + y n where x, y Z, this leads to solving the Diophantine equations In this section, we concentrate on the latter : x ny = or x ny = x ny = This Diophantine equation is called the Pell equation or Pell-Fermat equation. We want to find all (x, y) N with x ny =. Clearly (x, y) = (, 0) is a natural solution. This solution is called the trivial solution. We also put (x 0, y 0 ) = (, 0). Are there any other solutions? We will prove that there are infinitely many solutions that can be described in terms of a smallest or fundamental solution. Suppose that (x, y ), (x, y ) N are solutions of x ny =. Then we have that x < x x < x + ny < + ny y < y y < y So we can order the natural solutions of x ny = : (x, y ) < (x, y ) x < x y < y We start by proving that the equation x ny = always has a non-trivial solution. We give a non-constructive existence proof that relies heavily on the Pigeonhole Principle : If we distribute more than k pigeons into k holes then at least one hole contains more than one pigeon. If we distribute infinitely many pigeons into finitely many holes then at least one hole contains infinitely many pigeons. The main step is Dirichlet s Approximation Theorem. 6

18 Theorem.8 [Dirichlet] Let n, B N 0 with n not a perfect square. Then there exist a, b Z such that b B and a b n < B. Proof : For k =,,..., B +, put α k = k n [k n] (where for x R, [x] is the integral part of x : it is the largest integer smaller than or equal to x). Note that α k is irrational and 0 < α k < for k =,,..., B +. Moreover, {α, α,..., α B+ } are B + different numbers because n is irrational. By the Pigeonhole Principle, we have that α k α l B for some k < l B + Put a = [l n] [k n] and b = l k. Then a, b Z, b = l k B and a b n = ([l n] [k n]) (l k) n = (k n [k n]) (l n [l n]) = α k α l B Since b 0, we have that a b n is irrational and so a b n < B. Corollary.9 Let n N with n not a perfect square. Then there are infinitely many couples (a, b) Z N 0 with a b n < b. Proof : Suppose that there are only finitely many couples (a, b) Z N 0 with a b n < b, say {(a, b ),..., (a k, b k )}. Pick B N with B min{, a b n,..., ak b k n }. By Dirichlet s Approximation Theorem, there exists (a, b) Z N 0 such that b B and a b n < B. Since b B, we have that a b n < B b. Hence (a, b) = (a i, b i ) for some i k. But then a contradiction. a i b i n = a b n < B a i b i n Hence there are infinitely many couples (a, b) Z N 0 with a b n < b. Corollary.30 Let n N with n not a perfect square. Then there are infinitely many numbers a b n Z[ n] with the same norm. Proof : By Corollary.9 there are infinitely many couples (a, b) Z N 0 with a b n < b. If (a, b) is such a couple then a + b n = (a b n) + b n a b n + b n < b + b n < 3b n and so N(a b n) = a nb = a b n a + b n < b 3b n = 3 n Hence there are infinitely many numbers a b n Z[ n] with N(a b n) < 3 n. Since the norm is an integer, it follows from the Pigeonhole principle that there are infinitely many numbers a b n Z[ n] with the same norm. 7

19 Corollary.3 There exists an integer N and two different numbers a b n, a b n Z[ n] such that N(a b n) = N(a b n) = N, a a mod N, b b mod N and (a b n)(a b n) > 0. Proof : By Corollary.30, there exists N Z and infinitely many numbers a b n Z[ n] with N(a b n) = N. Since there are only finitely many congruence classes modulo N, it follows from the Pigeonhole Principle that there exist p, q {0,,..., N } such that infinitely many of these numbers a b n satisfy a p mod N and b q mod N. Since every number is either positive or negative, we get (using the Pigeonhole Principle one more time) that infinitely many of these numbers a b n have the same sign. So there are two different numbers a b n, a b n Z[ n] such that N(a b n) = N(a b n) = N, a a mod N, b b mod N and (a b n)(a b n) > 0. We are now ready to prove the existence of a non-trivial solution to the Pell-Fermat equation. Theorem.3 Let n N with n not a perfect square. Then the equation x ny = has a non-trivial solution. So there exists (x, y) N such that x ny = and (x, y) (, 0). Proof : Let N and a b n, a b n Z[ n] be as in Corollary.3. Since a nb = N(a b n) = N, we get that a b n = (a b n)(a + b n) = a a nb b a b n (a b n)(a + b n) N + a b b a n N Recall that a nb = N(a b n) = N, a a mod N and b b mod N. Hence a a nb b a a nb b a nb 0 mod N So a := a a nb b N Z. Similarly, we get that b := a b b a N a + b n = a b n a b n Z[ n] Z. Hence Since N(a b n) = N(a b n), we get that a nb = N(a + b ( ) a b n n) = N = N(a b n) = a b n N(a b n) Since (a b n)(a b n) > 0, we get that a + b n > 0. Finally, since a b n and a b n are different numbers, we have that a + b n. So ( a, b ) is a non-trivial solution of x ny =. So the equation x ny = has a non-trivial solution. Recall that we can order the natural solutions of x ny =. We call the smallest non-trivial natural solution of x ny = the fundamental solution of x ny = and denote it by (x, y ). 8

20 By trial and error, we compiled a list of the fundamental solution for some small square-free values of n : n (x, y ) (3, ) 3 (, ) 5 (9, 4) 6 (5, ) 7 (8, 3) 0 (9, 6) Later, we will use continued fractions to find the fundamental solution which can be enormous, even for relatively small values of n. For example, the fundamental solution of x 6y = is ( , ) Let (x, y ) be the fundamental solution of x ny =. For k N, we have that (x +y n) k Z[ n]. Hence we can define x k, y k N by x k + y k n = (x + y n) k for k = 0,,,... Note that it follows from Newton s Binomium that x k, y n are indeed natural numbers for all k 0. We now prove that all the natural solutions of x ny = are {(x k, y k ) k = 0,,,...}. One final remark : let x, y Q with x ny = (so N(x + y n) = ). Then x + y n = x y n (x + y n)(x y n) = x y n x ny = x y n Theorem.33 Let n N with n not a perfect square and let (x, y ) be the fundamental solution of x ny =. For k 0, define x k, y k N by x k + y k n = (x + y n) k. Then the following holds : (a) All the natural solutions of x ny = are given by {(x k, y k ) k = 0,,,...} x k = (x + y n) k + (x y n) k (b) For all k 0, we have that y k = (x + y n) k (x y n) k n (c) For all k, we have that { xk = x x k + ny y k y k = y x k + x y k 9

21 Proof : (a) Using the norm, we get that for all k 0 x k ny k = N(x k + y k n) = N ( (x + y n) k ) = ( N(x + y n) ) k = (x ny ) k = k = So (x k, y k ) is a natural solution of x ny = for all k 0. Suppose that x, y N with x ny =. Since (x +y n) 0 = and lim k + (x + y n) k = +, it follows that there exists a unique k N such that (x + y n) k x + y n < (x + y n) k+ () Note that x + y n (x + y = x + y n = (x+y n)(x n) k k y k n) = xxk nyy k +(yx k xy k ) n := a+b n x k + y k n where a, b Z. Applying the norm, we find It follows from () that a nb = N(a + b n) = N((x + y n)(x k y k n)) = N(x + y n)n(x k y k n) = (x ny )(x k ny k ) = = (x + y n) k (x + y x + y n n) k (x + y = a + b n < (x + y n) k+ n) k (x + y n) k = x + y n () Thus Hence and so x + y n < a + b n x y n < a b n Adding () and (3), we get that So a + b n < x + y n < 0 (3) 0 b n < y n 0 b < y Since a nb = and (x, y ) is the fundamental solution of X ny =, we get that ( a, b) = (, 0). It follows from () that a =. So x + y n = (a + b n)(x + y n) k = (x + y n) k = x k + y k n 0

22 Hence (x, y) = (x k, y k ), which proves (a). (b) Pick k 0. Recall that x k + y k n = (x + y n) k (4) Hence x k y k n = x k + y k n = ( (x + y = n) k x + y n ) k = (x y n) k (5) Solving equations (4) and (5) for x k and y k, we get x k = (x + y n) k + (x y n) k which proves (b). (c) Pick k. Then and y k = (x + y n) k (x y n) k n x k + y k n = (x + y n) k = (x + y n)(x + y n) k = (x + y n)(xk + y k n) = x x k + ny y k + n(x y k + y x k ) Hence which proves (c). x k = x x k + ny y k and y k = x y k + y x k Example : Consider the equation x 3y = We see that (, ) is the fundamental solution. Hence all the natural solutions of x 3y = are given by {( ( + 3) k + ( 3) k, ( + 3) k ( ) } 3) k k N 3 We can also get these solutions by considering x k + y k 3 = ( + 3) k for k = 0,,,...

23 Chapter Analytic Number Theory. The Riemann-Zeta Function.. The Riemann-Zeta Function and the Euler Product Riemann saw that there was a connection between the distribution of primes and the zeros of a function, now called the zeta function. The zeta function is a function of a complex variable traditionally denoted by s with real part σ and imaginary part t (so s = σ + it). Proposition. The series of the half plane Re(s) >. Re(s) >. + n= converges absolutely and uniformly on any compact subset ns In particular, the series + n= n s is analytic on the half-plane Proof : This proposition is a special case of Proposition.9 with a n = for all n. Definition. We define the Riemann-zeta function ζ(s) by ζ(s) = + n= n s for all s C with Re(s) > The following property of the zeta function will be used to prove that certain sets of prime numbers are infinite. Lemma.3 For s R with s >, we have lim ζ(s) = + s + Proof : For all s (, + ) and all integers N >, we have that N n= N n < s dx x = N s s s < N n= n s

24 Taking the limit as N goes to +, we find Hence So Hence lim ζ(s) = +. s + + n= n + s s n s n= ζ(s) s ζ(s) s ζ(s) s + for all s (, + ) for all s (, + ) for all s (, + ) The following theorem gives a relation between the zeta function and prime numbers. Theorem.4 (Euler Product for ζ(s)) For all s C with Re(s) >, we have that ζ(s) = p prime p s Proof : This theorem is a special case of Theorem.7 with m = and χ = χ 0. This Euler Product leads to an easy analytic proof of Euclid s Theorem. Theorem.5 (Euclid) There are infinitely many prime numbers. Proof : Suppose that there are only finitely many prime numbers, say p < p < < p k. Then it follows from the Euler Product that lim ζ(s) = lim s + s + p prime = lim p s s + k i= p s i = k i= p i < + This is a contradiction to Lemma.3. Hence there are infinitely many prime numbers... Analytic Continuation of ζ(s) and the Riemann Hypothesis The zeta function ζ(s) is only defined for s C with Re(s) >. One can prove that the series η(s) = + n= ( ) n is an analytic function on the half-plane Re(s) > 0 and that ζ(s) = n s η(s) s for all s C with Re(s) > 3

25 η(s) So is an analytic continuation of ζ(s). It follows from the theory of analytic continuation s that this is the only way to extend the definition of ζ(s) in an analytic way to Re(s) > 0. Riemann showed that there exists an analytic continuation ζ(s) of ζ(s) that is defined over C \ {} (again, this continuation is unique). For s C with Re(s) > 0, put Γ(s) = + Then Γ(s) is analytic on C \ {0,,,...}. Riemann proved that ( πs ) ζ(s) = s π s sin Γ( s) ζ( s) for all s C with Re(s) < 0 ( ) 0 x s e x dx. It follows from this formula that ζ(s) = 0 for s =, 4, 6,.... These are called the trivial zeros of ζ(s). Using the Euler Product, one can show that ζ(s) 0 for all s C with Re(s) >. Since Γ(s) 0 for all s C with Re(s) > 0, it follows from (*) that ζ(s) has no non-trivial zeros with Re(s) < 0. So if s is a non-trivial zero of ζ(s) then 0 Re(s). It turns out that the non-trivial zeros of ζ(s) are symmetric around the line Re(s) = : if s is a non-trivial zero of ζ(s) then ζ(s) = 0 ζ( s) = 0. This leads to Riemann s famous conjecture : Riemann Hypothesis : If s is a non-trivial zero of ζ(s) then Re(s) =. In 896, Hadamard and de la Vallée-Poussin were able to prove the following : If s is a non-trivial zero of ζ(s) then Re(s).. The Prime Number Theorem For x 0, let π(x) be the number of primes less than or equal to x. In 859, Riemann found a connection between π(x) and the non-trivial zeros of ζ(s). He also x showed how a proof of his conjecture would result in a proof that π(x) is asymptotic to ln(x). In 896, this major result in number theory was finally proven by Hadamard and de la Vallée- Poussin (although they were unable to prove Riemann s conjecture, they did prove a related result). Theorem.6 (Prime Number Theorem) lim x + π(x) x ln(x) = lim x + x π(x) = dt ln(t) In 949, Selberg and Erdős found elementary proofs of the Prime Number Theorem (in number theory, an elementary proof is a proof that does not use complex analysis or abstract algebra; it can still be extremely complicated). 4

26 .3 Dirichlet s Theorem This section is devoted to the proof of the following theorem (due to Dirichlet) : Let a, m N with gcd(a, m) =. Then there exist infinitely many primes p with p a mod m. For certain values of a and m, one can prove quite easily that there are infinitely many primes p with p a mod m. For example, if a = and m =, we need to prove that there are infinitely many odd primes. This easily follows from Euclid s Theorem (there are infinitely many primes) and the fact that is the only even prime. The following theorem gives an elementary proof of Dirichlet s Theorem in the special case that a = 3 and m = 4. Theorem.7 There are infinitely many primes p with p 3 mod 4. Proof : Suppose there are only finitely many primes p with p 3 mod 4, say p < p < < p k. Put N = p p p k +. Modulo 4, we find N p p p k mod 4 In particular, N is odd. So does not divide N. If p i divides N for some i =,,..., k then p i (N p p p k ) and so p i divides, a contradiction. Hence if a prime p divides N then p mod 4. So the prime factorization of N is of the form N = q q... q n where q i is a prime and q i mod 4 for i =,,..., n. Again considering modulo 4, we find that N q q q n mod 4 a contradiction since N 3 mod 4. Hence there are infinitely many primes p with p 3 mod Dirichlet Series Definition.8 Let a n C for all n. Then the series + n= a n n s is called a Dirichlet series with coefficients a n. Remark : The Riemann-Zeta function is an example of a Dirichlet series. We prove some sufficient conditions for a Dirichlet series to be analytic. 5

27 Proposition.9 Suppose the sequence a n n is bounded. Then the series + a n n s n= converges absolutely and uniformly on any compact subset of the half plane Re(s) >. In particular, the series + n= a n is analytic on the half-plane Re(s) >. ns Proof : Let T be a compact subset of the half plane Re(s) >. Then there exists p > such that Re(s) p for all s T. Pick M > 0 with a n M for all n. Note that Since the series a n a n = n s n s M e ln(n)s = M e ln(n)re(s) = M n Re(s) M n p + M n p n= for all s T converges (it s a p-series with p > ), we get that the series + converges absolutely and uniformly on T by the Weierstrass M-test. Since a n n is analytic on T for all n, we get that + s n= Proposition.0 Suppose the partial sums of the series + n= a n n s the series a n is analytic on T. n s + n= n= a n n s a n are bounded. Then the series converges uniformly on any compact subset of the half-plane Re(s) > 0. In particular, + n= a n is analytic on the half-plane Re(s) > 0. ns Proof : For n, we put s n = n a k. k= It is given that the sequence s n n is bounded, say by M. So s n M for all n. First, we prove the following claim: For all s with Re(s) > 0 and all u, v N 0 with u < v, we have that v a ( n n s M (u + ) Re(s) + s ) Re(s) n=u+ Consider the function S : [, + ) C : t n t a n So S(t) = s [t] for all t [, + ). Pick s with σ := Re(s) > 0. Put f(t) = t s for all t [, + ). 6

28 Pick u, v N 0 with u < v. Note that a n = s n s n for all n. Hence v n=u+ a n n s = = v n=u+ v n=u+ s n f(n) v n=u+ v s n f(n) s n f(n) s n f(n + ) n=u = s v f(v) s u f(u + ) v n=u+ But if n N 0 then S(t) = s n for all t [n, n + ) and so Hence v n=u+ Recall that So s n (f(n + ) f(n)) = a n n s = s vf(v) s u f(u+) v n=u+ n+ n n+ n s n f (t) dt = s n (f(n + ) f(n)) n+ n S(t)f (t) dt v S(t)f (t) dt = s v f(v) s u f(u+) S(t)f (t) dt u+ r c = r Re(c) for all r > 0 and all c C s v f(v) = s v v σ M v σ and s u f(u + ) = s u (u + ) σ M (u + ) σ Note that S(t) = s [t] M for all t and f (t) = st s = s t σ. Hence v v S(t)f (t) dt S(t) f (t) dt u+ Putting everything together, we find that v a n n s M v + M σ (u + ) + M s ( σ σ n=u+ u+ v M s = M s σ (u + ) σ v σ u+ t σ dt ( (u + ) ) σ v σ ) = Mv ( s ) + σ σ ( M + s ) (u + ) σ σ Note that s Re(s) = σ. So v a n n s n=u+ ( M + s ) (u + ) σ σ which proves the claim. 7

29 We can now finish the proof of the theorem. Let T be a compact subset of the half-plane Re(s) > 0. Then there exist δ, D > 0 such that Re(s) δ and s D for all s T. Pick ɛ > 0. Pick N N with ( ( M N > + D )) δ ɛ δ Pick u, v N with u, v > N and s T. We may assume that v > u. By the claim, we have that v a ( n n s M (u + ) Re(s) + s ) M ( + D ) < MN ( + D ) < ɛ Re(s) (u + ) δ δ δ δ n=u+ + a n Hence the series n is uniformly Cauchy on T. So the series + a n converges uniformly on s ns n= n= T. Since a n n is analytic on T for all n, we get that + a n is analytic on T. s n s.3. Group Characters Definition. Let (G, ) be a finite abelian group. n= () A character of G is a homomorphism φ : G (C 0, ) () Ĝ is the set of all characters of G. (3) For ϕ, ψ Ĝ, we define the map ϕ ψ : G C 0 : g ϕ(g)ψ(g) One easily checks that (Ĝ, ) is an abelian group with identity element ϕ 0 : G C 0 : g (also called the trivial character) the inverse of ϕ Ĝ is the character ϕ : G C 0 : g ϕ(g) The following proposition gives us some basic properties about characters. Proposition. Let G be a finite abelian group. Then the following holds : (a) Let ϕ Ĝ, g G and n N with gn = G. Then ϕ(g) is a n-th root of unity. In particular, ϕ(g) = ϕ(g). (b) Let G g G. Then there exists ϕ Ĝ with ϕ(g). 8

30 Proof : (a) Since ϕ is a homomorphism, we get (ϕ(g)) n = ϕ(g n ) = ϕ( G ) = Hence ϕ(g) =. Put ϕ(g) = a + bi with a, b R. Then a + b = ϕ(g) =. So ϕ(g) = a + bi = a bi (a + bi)(a bi) = a bi = a bi = ϕ(g) a + b (b) It follows from the Fundamental Theorem for Finite Abelian Groups that G is the direct sum of a finite number of cyclic subgroups (say t). So there exist g,..., g t G and n,..., n t N \ {0, } such that g i is of order n i for i =,,..., t and every x G can be written uniquely as x = g m g mt t with 0 m i < n i for i =,,..., t (hence G = g g g t ). So we can write g = g k gt kt with 0 k i < n i for i =,,..., t. Since g G, there exists j {,,..., t} with k j 0. Define the map ϕ : G C 0 : g m gt mt e πm j i n j where 0 m i < n i for i =,,..., t. One easily checks that ϕ Ĝ and ϕ(g) = e πk j n j i. If G is a finite abelian group and χ is a character of G then χ(g) is a G -th root of unity for all g G. Hence G has only finitely many characters. We will prove that Ĝ = G. The characters of a finite abelian group satisfy some nice relations. Proposition.3 (Orthogonality Relations) Let G be a finite abelian group and ϕ 0 the trivial character. Then the following relations hold : (a) For all ϕ Ĝ, we have that { G if ϕ = ϕ0 ϕ(g) = 0 if ϕ ϕ 0 g G (b) For all g G, we have that { G if g = G ϕ(g) = 0 if g G ϕ Ĝ Proof : (a) Pick ϕ Ĝ. If ϕ = ϕ 0 then ϕ(g) = = G g G g G So we may assume that ϕ ϕ 0. Then there exists h G with ϕ(h). Note that G = {hg g G}. Since ϕ is a homomorphism, we get ( ) ϕ(h)ϕ(g) = ϕ(h) ϕ(g) g G ϕ(g) = g G ϕ(hg) = g G g G 9

31 ( ) Hence (ϕ(h) ) ϕ(g) = 0. Since ϕ(h), we get that ϕ(g) = 0. g G g G (b) Pick g G. Assume first that g G. By Proposition.(b), there exists ψ Ĝ with ψ(g). Note that Ĝ = {ψϕ ϕ Ĝ}. Then we get ϕ(g) = = ϕ Ĝ ϕ Ĝ(ψϕ)(g) ψ(g)ϕ(g) = ψ(g) ϕ(g) ϕ Ĝ ϕ Ĝ Hence (ψ(g) ) ϕ(g) = 0. Since ψ(g), we get that ϕ(g) = 0. ϕ Ĝ ϕ Ĝ If g = G then ϕ(g) = = Ĝ ϕ Ĝ ϕ Ĝ Combining (a) and (b), we get that ϕ(g) = ϕ(g) = Ĝ = Ĝ g G,ϕ Ĝ g G ϕ Ĝ = ( ) ϕ(g) = G = G ϕ Ĝ g G Hence Ĝ = G..3.3 Dirichlet Characters and L-Series Throughout this section, m N 0. Most results are trivial when m =. Definition.4 (a) For n Z, we put n = n + mz Z/mZ. (b) Put Z m = {n n Z, gcd(m, n) = }. Note that (Z m, ) is an abelian group of order ϕ(m) (where ϕ is the Euler-Phi function). (c) A Dirichlet character (mod m) is a character of Z m. We denote the trivial character of Z m by χ 0. (d) Let χ be a Dirichlet character. We extend the definition of χ to N as follows : { χ(n) if gcd(m, n) = χ : N C : n 0 if gcd(m, n) One easily checks that χ(n n ) = χ(n )χ(n ) for all n, n N. 30

32 (e) The L-series associated to the Dirichlet character χ is the series L(s, χ) = + n= χ(n) n s Note that an L-series is a Dirichlet series. Since L-series are Dirichlet series, we can easily prove the following propositions. Proposition.5 Let χ be a Dirichlet character. Then the series L(s, χ) = converges absolutely and uniformly on any compact subset of the half plane Re(s) >. In particular, L(s, χ) is analytic on the half-plane Re(s) >. + n= χ(n) n s Proof : Note that χ(n) for all n N. Hence the result follows from Proposition.9. χ(n) Proposition.6 Let χ be a non-trivial Dirichlet character. Then the series L(s, χ) = n s n= converges uniformly on any compact subset of the half-plane Re(s) > 0. In particular, L(s, χ) is analytic on the half-plane Re(s) > 0 if χ χ 0. Proof : For n, we put s n = n χ(k). k= Pick n N 0. Put n = qm + r with q, r N and 0 r < m. Then ( n qm ) ( r ) ( m ) ( r ) χ(k) = χ(k) + χ(qm + k) = q χ(k) + χ(k) k= By Proposition.3(a), k= m χ(k) = k= k= m k= gcd(k,m)= n s n = χ(k) = k= χ(k) = 0. So r χ(k) k= k= r χ(k) = r < m k= k= + by Proposition.(a). The result now follows from Proposition.0. Although not obvious, there is a relation between L(s, χ) and the prime numbers! 3

33 Theorem.7 (Euler Product for L(s, χ)) Let χ be a Dirichlet character. s >, we have that L(s, χ) = χ(p)p s p prime Then for all Proof : have Pick s >. Let p < p < p 3 < be the list off all primes. Then by definition, we Note that for any prime p, i= p prime = lim χ(p)p s k + + j=0 χ(p) j p js = k i= χ(p)p s χ(p)p s i where the series converges absolutely since χ(p)p s <. Using the Cauchy Product for absolutely convergent series and the fact that χ is completely multiplicative, we get ( k k + ) + χ(p)p s = χ(p) j p js i = (χ(p ) α χ(p k ) α k )(p α... p α k k ) s = χ(n) i n s n N k i= j=0 α,...,α k =0 where N k is the set of natural numbers that have no prime divisors bigger than p k. Note that + N N N 3 and + k= N χ(n) k = N. Since the series converges absolutely, we get n s p prime = lim χ(p)p s k + k i= χ(p)p s i = lim k + n= χ(n) = n s n N k + n= χ(n) n s = L(s, χ) From now on, we assume that m. The following property of L(s, χ) is not so easy to prove. Theorem.8 Let χ be a non-trivial Dirichlet character. Then L(, χ) 0. Proof : L(s, χ 0 ) behaves quite differently. Theorem.9 Let χ 0 be the trivial Dirichlet character. Then for s R with s >, we have lim L(s, χ 0) = + s + Proof : Pick s R with s >. Using the Euler Product for L(s, χ 0 ) and ζ(s), we get 3

34 Note that L(s, χ 0 ) = lim s + p prime = = p prime p m ( p prime = ζ(s) p prime p m χ 0 (p)p s p s p prime p m ) p s ( p s ) ( p s ) = p prime p m p prime p m ( p s ) ( p ) is a finite strictly positive real number. Hence it follows from Lemma.3 that lim s + L(s, χ 0) = +. Proposition.0 Let χ be a Dirichlet character. Then the following holds : (a) The series ( + ) χ(p) k converges to an analytic function M(s, χ) on the half plane kp ks Re(s) > p prime k= (b) e M(s,χ) = L(s, χ) for all s C with Re(s) > (c) There exists a function Φ(s, χ) defined on the half plane Re(s) > such that Φ(s, χ) is bounded on the half plane Re(s) > and for all s C with Re(s) >. M(s, χ) = Φ(s, χ) + p prime χ(p) p s (d) M(s, χ) is bounded on (, + ) if χ is not the trivial character. Proof : (a) Let T be a compact subset of the half plane Re(s) >. Then there exists σ > such that Re(s) σ for all s T. Note that χ(p) k kp ks for all k N, all primes p and all s T kp kσ 33

35 and p prime ( + k= ) kp kσ p prime ( + k= ) = p kσ p σ < p prime since the latter series is a p-series with p = σ >. p prime Hence the series p < + σ n < + σ n= p prime ( + k= ) χ(p) k kp ks converges absolutely and uniformly on T. Since( T was an arbitrary compact subset of the half + ) χ(p) k plane Re(s) >, we get that the series converges to an analytic function kp ks M(s, χ) on the half plane Re(s) >. p prime (b) Let p ( < p < p 3 < be the list of all primes. The above also shows that for i =,,..., + ) χ(p i ) k the series converges to an analytic function f i (s) on the half plane Re(s) >. k= kp ks i Recall the following from complex analysis : The series + k= z k k k= converges to an analytic function g(z) on {z C z < } with e g(z) = for all z C with z <. z Note that χ(p i ) p s = < for i =,,... and all s C with Re(s) >. Hence i pre(s) e fi(s) = χ(p i )p s i n Since the sequence f i (s) i= n analytic on C, we get that the sequence Re(s) >. But e n i= f i(s) = n e fi(s) = i= Hence by Theorem.7, we get n e M(s,χ) = lim n + i= for i =,,... and all s C with Re(s) > converges to M(s, χ) on the half plane Re(s) > and e z is n i= e n i= f i(s) χ(p i )p s i χ(p i )p s i n = converges to e M(s,χ) on the half plane for all s C with Re(s) > p prime = L(s, χ) χ(p)p s (c) For s C with Re(s) >, put Φ(s, χ) = + k= ( p prime ) χ(p) k kp ks 34

36 Since the series p prime ( + k= ) χ(p) k converges absolutely, we get that kp ks M(S, χ) = = p prime + k= = p prime = p prime ( + ( k= p prime χ(p) p s + ) χ(p) k kp ks ) χ(p) k kp ks + k= ( χ(p) p s + Φ(s, χ) p prime for all s C with Re(s) >. Again because of absolute convergence, we get that ( ) + χ(p) k Φ(s, χ) = kp ks k= p prime ( + ) χ(p) k = kp ks p prime k= ( + ) χ(p) k kp ks p prime k= ( + ) p k p prime = p prime p prime + n= = ζ() k= p p n p ) χ(p) k kp ks for all s C with Re(s) >. So Φ(s, χ) is bounded on the half plane Re(s) >. (d) Suppose that χ is not the trivial character. Then L(, χ) 0 by Theorem.8. Hence there exists 0 < ɛ < such that L (s, χ) is analytic on B(, ɛ) := {s C s < ɛ}. L(s, χ) By the Antiderivative Theorem, there exists a function N(s, χ) such that N(s, χ) is analytic on B(, ɛ) and N (s, χ) = L (s, χ) L(s, χ) for all s B(, ɛ). Since em(s,χ) = L(s, χ), we get that 35