Chapter 2 Bézout and Gauss

Chapter 2 Bézout and Gauss 2 Bachet Bézout s Theorem Let a be a positive integer In this section, we denote by D(a) (resp M(a)) theset of positive divisors (resp multiples) of a Theorem 2 Let a, b be any positive integers (i) The set D(a, b) = D(a) D(b) has a greatest element d called the greatest common divisor of a and b denoted by d = (a, b) (ii) The set M(a, b) = M(a) M(b) has a smallest element m called the least common multiple of a and b denoted by m =[a,b] Proof We only prove (i) The set D(a, b) is a non-empty subset of Z 0 since it contains and is upper bounded by min(a, b) We conclude the proof by appealing to Axiom Definition 22 Two positive integers a and b are said to be coprime, orrelatively prime, if and only if (a, b) = Examples 23 (6, 5) = 3 and [6, 5]=30 The integers 25 and 4 are coprime and [25, 4]=350 Remark 24 Theorem 2 shows at once that (b, a) = (a, b) and [b,a]=[a,b] We obviously have (a, b) [a,b] We also extend Theorem 2 to all integers by setting (a, b) = ( a, b ) and [a,b]= [ a, b ] so that, for any integers a,b Z \{0}, we will always have (a, b) and [a,b] Also, we set (a, 0) = a and [a,0]=0 so that (a, ) =, [a,]= a and, if a b then (a, b) = a and [a,b]= b for all a,b Z O Bordellès, Arithmetic Tales, Universitext, DOI 0007/978--447-4096-2_2, Springer-Verlag London 202 27

28 2 Bézout and Gauss The following result gives essential information about the gcd Theorem 25 Let a, b be any integers and d be a positive integer Then d = (a, b) if and only if d a, d b and there exist integers u, v such that d = au + bv Proof Suppose first that d = (a, b) so that obviously d a and d b Now define the set E by E = { ax + by : (x, y) Z 2 and ax + by } Clearly E Z 0 and E since a or b is an element of E By Axiom, we infer that E contains a smallest element δ and we set (u, v) Z 2 such that δ = au + bv Let us prove that δ = d If we had δ a, the Euclidean division of δ into a would give and then we would have a = δq + r and 0 <r<δ r = a δq = a q(au+ bv) = a( qu)+ b( qv) which implies that r E, and then r δ, which contradicts the inequality r<d Hence we have δ a, and we prove δ b in a similar way Now let c be another positive integer dividing a and b We then have c (au + bv) = δ, and thus c δ, which proves that d = δ = au + bv Conversely, since d a and d b, we infer that d (a, b) On the other hand, since (a, b) a and (a, b) b, wehave(a, b) (au + bv) = d which implies that (a, b) d, so that d = (a, b), and the proof is complete As a corollary we get at once the important Bachet Bézout theorem Corollary 26 (Bachet Bézout) Let a, b be any integers Then a and b are coprime integers if and only if there exist integers u, v such that au + bv = Remark 27 The integers u and v, called Bézout s coefficients of a and b, are not unique since we can write au + bv = a(u + kb) + b(v ka) (k Z) Theorem 25 provides a useful criterion for the gcd Corollary 28 Let a, b be integers and d be a positive integer Then { (i) d a, d b d = (a, b) (ii) c a, c b = c d Proof The sufficient condition is obvious Conversely, if d = (a, b), then d satisfies (i) and Theorem 25 provides integers u, v such that d = au+ bv, so that if c a and c b, then c (au + bv) = d as required

22 The Euclidean Algorithm 29 22 The Euclidean Algorithm The computation of the gcd can be achieved by using the following observation discovered by Euclid Theorem 29 Given a,b,k Z \{0} we have (a, b) = (a, b + ka) Proof If d = (a, b) and d 2 = (a, b + ka), then it is easy to see that d a and d b imply that d a and d (b + ka) so that d d 2 Conversely, since d 2 a and d 2 (b + ka), we infer that d 2 a and d 2 (b + ka ka) = b so that d 2 d Thisis the desired result since d,d 2 Now let a b be positive integers for which we want to compute their gcd By making use of the former result, we could proceed in the following way (i) If b a then (a, b) = b Otherwise, the Euclidean division of r 0 = a by r = b provides two unique integers q and r 2 such that a = bq + r 2 and 0 <r 2 <b, and Theorem 29 with k = q gives (a, b) = (b, a bq ) = (b, r 2 ) (ii) If r 2 b then (a, b) = (b, r 2 ) = r 2 Otherwise, the proof proceeds as before so that we get two unique integers q 2 and r 3 such that (b, r 2 ) = (r 2,r 3 ) (iii) And so on This algorithm, derived by Euclid, gradually constructs a strictly decreasing sequence of natural integers (r k ) Hence there exists a non-negative integer n such that r n 0 and r n+ = 0 and we get (a, b) = (b, r 2 ) = (r 2,r 3 ) = =(r n,r n ) = (r n, 0) = r n With a little more work, one can even estimate the number of divisions required in this algorithm to obtain the gcd The following result is a version of a theorem proved by Lamé Proposition 20 (Lamé) Let b a be integers Then the number of divisions necessary to compute (a, b) in the Euclidean algorithm does not exceed [ ] log b + log where = + 5 2 is the golden ratio

30 2 Bézout and Gauss Proof The proof uses the Fibonacci numbers F n and their connection with the golden ratio Recall that is the positive solution of the equation x 2 = x + and that Fibonacci numbers are the terms of the Fibonacci sequence (F n ) defined by { F 0 = 0, F = F n+2 = F n+ + F n (n 0) An easy induction shows that the following lower bound F n n 2 (2) holds for any positive integer n Now consider the Euclidean algorithm with n 2 steps written as r 0 = r q + r 2 0 <r 2 <r r = r 2 q 2 + r 3 0 <r 3 <r 2 r n 2 = r n q n + r n 0 <r n <r n r n = r n q n (with a = r 0 and b = r ) Let us prove by induction that r n k F k+2 (k =,,n) (22) Note first that q n 2 since r n = q n r n and r n <r n and therefore r n = r n q n q n 2 = F 3 so (22) istruefork = It is also true for k = 2 since r n 2 = r n q n + r n r n + r n 2 + = 3 = F 4 Now suppose that (22)istrueforsomekWehave r n k = r n k q n k + r n k+ r n k + r n k+ and by induction hypothesis we get r n k F k+2 + F k+ = F k+3 which proves (22) We now use (22) with k = n to get b = r F n+ and (2) implies then that which completes the proof n log b log +

23 Gauss s Theorem 3 Example 2 Let us show how the Euclidean algorithm could be used to get a gcd and Bézout s coefficients associated to it Let d = (8 459, 3 809)Wehave 4 5 2 8 459 3 809 3 223 586 293 3 223 586 293 0 where the quotients are written in the first line and the remainders are in the second and third lines We then obtain d = 293Wealsohave 8 549 = 3 809 (4) + 3 223 3 809 = 3 223 () + 586 3 223 = 586 (5) + 293 which we can rewrite as 8 459 + 3 809 ( 4) + 3 223 ( ) = 0 6 3 809 () + 3 223 ( ) + 586 ( ) = 0 ( 5) 3 223 () + 586 ( 5) = 293 which immediately gives 8 459 (6) + 3 809 ( 29) = 293 23 Gauss s Theorem The following result is fundamental Theorem 22 (Gauss) Let a, b, c be any integers Then } a bc = a c (a, b) = Proof Since (a, b) =, using Bachet Bézout s theorem we have integers u, v such that au + bv = Multiplying both sides of this identity by c and using the fact that there exists k Z such that bc = ka, we get c = acu + bcv = acu + kav = a(cu + kv) which is the desired result since cu + kv Z This result has, along with Bachet Bézout s theorem, a lot of consequences The following proposition summarizes some of them Proposition 23 Let n, r be positive integers and a,b,c,a,a 2,,a r,k be any integers

32 2 Bézout and Gauss (i) (ka, kb) = k (a, b) and [ka, kb]= k [a, b] (ii) If d = (a, b) and a = da and b = db, then (a,b ) = (iii) (a, b) [a, b]= ab (iv) (a, b) = (a,bc) = (a, c) (v) (a n,b n ) = (a, b) n and [a n,b n ]=[a,b] n (vi) a c b c = ab c (a, b) = More generally, we have a b,,a r b i j = (a i,a j ) = } = a a r b (vii) Suppose (a, b) = and k If ab = c k for some positive integer c, then there exist positive integers r, s such that a = r k and b = s k (viii) Suppose k<n Then n (n, k) ( ) n k Proof Without loss of generality, we prove the theorem by considering that a, b, c and a,a 2,,a r,k are positive integers (i) We set d = (a, b) Clearly we have kd (ka, kb) since kd ka and kd kb By Theorem 25, there exist integers u, v such that d = au + bv so that kd = u(ka) + v(kb) and therefore, using Theorem 25 again, we infer that kd = (ka, kb) Next we set m =[a,b] and M =[ka,kb] km is a common multiple of ka and kb so that km M Conversely, since ka M and kb M, we deduce that the integer M/k is a common multiple of a and b, and thus M/k m, which gives M km, and therefore M = km (ii) We have d = (a, b) = (da,db ) = d(a,b ) so that (a,b ) = (iii) Suppose first that (a, b) = We obviously have [a,b] ab since these two numbers are common multiples of a and b On the other hand, since a [a,b], there exists a positive integer q such that [a,b] =qa Since b [a,b] =qa and (a, b) =, Gauss s theorem implies that b q, so that b q and thus ab qa =[a,b] Hence we proved that [a,b] =ab as soon as (a, b) = Now if (a, b) = d>, we set a = da and b = db with (a,b ) = We then have [a,b]= [ da,db ] = d [ a,b ] = da b = ab d = ab (a, b) giving the asserted result (iv) We set d = (a, c) and D = (a,bc) Clearly, d D Conversely, since D a and D bc, wehaved ac and D bc so that D (ac,bc) = c (a, b) = c, and thus D d

23 Gauss s Theorem 33 (v) We first note that (iii) implies by induction that (a, b) = = (a, b n ) = and similarly (b n,a)= = (b n,a n ) =, so that we have (a, b) = = ( a n,b n) = Now set d = (a, b) and a = da, b = db with (a,b ) = We get ( a n,b n) = ( d n a n,d n b n) = d n( a n,b n) = d n = (a, b) n Next we use (iii) [ a n,b n] = an b n ( ) ab n (a n,b n ) = =[a,b] n (a, b) (vi) There exists a positive integer q such that c = qa and then we get } b qa = b q (a, b) = by Gauss s theorem Thus, q = hb for some positive integer h and then c = qa = hba so that ab c The second assertion follows by an easy induction (vii) Let d = (a, c) First,wehaved k c k = ab and since (d, b) = ((a, c), b) = (c, (a, b)) = (c, ) =, we have (d k,b)= and Gauss s theorem implies that d k a On the other hand, if we set a = da and c = dc so that (a,c ) =, then the equality ab = c k is equivalent to a b = d k c k, so that a d k c k and since (a,c k ) =, we have a d k by Gauss s theorem again, so that a d k Hence we get a = d k and b = (c/d) k (viii) We set d = (n, k) and n = dn, k = dk with (n,k ) = We have k ( n k ) ( ) n = n k and the binomial coefficient of the right-hand side is an integer since n, k, therefore we have ( ) n n k k Since (n,k ) =, Gauss s theorem implies that which is the desired result n ( ) n k The proof is complete

34 2 Bézout and Gauss 24 Linear Diophantine Equations In this section, a, b, n are any integers satisfying ab 0 and we set d = (a, b)any equation of the form ax + by = n with (x, y) Z 2 is called a linear Diophantine equation The aim of this section is the resolution of such equations We begin with an important observation Lemma 24 Let a,b,n Z such that ab 0 and set d = (a, b) Then the Diophantine equation ax + by = n has a solution in Z 2 if and only if d n Proof If n = 0 then d n and (x, y) = (0, 0) is a solution Now suppose n 0 We also define (u, v) Z 2 such that d = au + bv Suppose first that the equation has at least one solution (x, y) Z 2 Since d a and d b, we get d (ax + by) = n Conversely, suppose d n Set x 0 = nu/d and y 0 = nv/d Note that (x 0,y 0 ) Z 2 and we have ax 0 + by 0 = Thus, the pair (x 0,y 0 ) = ( nu d, nv d anu + bnv d = n (au + bv) = n d ) is solution of the equation The solutions are completely determined by the following result Proposition 25 Let a,b,n Z such that ab 0 and set d = (a, b) Suppose that d n and let (x 0,y 0 ) be a particular solution of the Diophantine equation ax +by = n Then the solutions are given by the following formulae x = x 0 + kb d y = y 0 ka (k Z) d Proof Define a = da, b = db and n = dn so that (a,b ) = The equation is equivalent to a x + b y = n Let(x, y) be a solution distinct from (x 0,y 0 )From we get a x + b y = a x 0 + b y 0 a (x x 0 ) = b (y 0 y) (23) with y y 0 Z, so that b a (x x 0 ) Since (a,b ) =, Gauss s theorem implies that b (x x 0 ) so that there exists k Z such that x = x 0 + kb Replacing x x 0 by kb in (23) gives y = y 0 ka Conversely, we check that the pairs (x, y) = (x 0 +kb,y 0 ka ) with k Z are solutions of the equation

25 Congruences 35 Example 26 Solve in Z 2 the equation 8 459x + 3 809y = 879 Answer Set d = (8 459, 3 809) Using Example 2 we know that d = 293 and since 879 = 3 293, Lemma 24 implies that this equation has at least one solution in Z 2 It is equivalent to 63x + 3y = 3 and we have seen in Example 2 that 63(6) + 3( 29) =, so that the pair (x 0,y 0 ) = (8, 87) is a particular solution of the equation Let (x, y) Z 2 be a solution We have 63x + 3y = 63(8) + 3( 87) 63(x 8) = 3( y 87) so that 3 63(x 8) Since (63, 3) =, Gauss s theorem implies that 3 (x 8), and hence there exists k Z such that x = 8+3k This gives y = 87 63k Conversely, we check that the pairs (8 + 3k, 87 63k) are solutions of the equation Thus, the set of solutions is S ={(8 + 3k, 87 63k) : k Z} Example 27 (Money changing problem) In how many ways can we obtain a sum of$34withonly$2coinsand$5bills? Answer We have to count the number of solutions in (Z 0 ) 2 of the equation 2x + 5y = 34 Using Proposition 25 we get x = 2 + 5k and y = 6 2k (k Z) as solutions in Z 2 The condition (x, y) (Z 0 ) 2 holds if and only if 2/5 k 3, which gives k {0,, 2, 3}, and hence the equation has four solutions in (Z 0 ) 2, and thus there are four ways to get $34 with only $2 coins and $5 bills 25 Congruences Introduced by Gauss, congruences are a very efficient tool in number theory In what follows, n is a positive integer Definition 28 Let a, b be any integers and let n be a positive integer We say that a is congruent to b modulo n, and we write a b(mod n), if and only if n (a b)if n (a b), we say that a and b are incongruent modulo n and write a b(mod n) Examples 29 42 0 (mod 2), 42 0 (mod 2), 42 5 (mod 2), 42 20 (mod 2) 2 0 (mod 9) and 0 (mod ) 3 n is even (resp odd) if and only if n 0 (mod 2) (resp n (mod 2))

36 2 Bézout and Gauss 4 If the decimal representation of a positive integer n is n = a r a r a 0 with 0 a i 9 (for i = 0,,r) and a r 0, then, for any integer k {,,r + } we have n a k a 0 ( mod 0 k ) 5 For any integers a, b,wehavea b(mod ) Remark 220 Givenafixedintegera, there is no unicity of the integer b such that a b(mod n) since this relation means nothing but that there exists k Z such that a = b + kn Thus, we can choose b depending on the result aimed at In particular, it is of great interest to take b as the remainder of the Euclidean division of a by n Indeed, we have in this case 0 b<n,ieareduction phenomenon Let us also mention that the relation a 0 (mod n) means n a, and it is not hard to see that, if a b(mod n), then for any positive divisor d of n, wealsohavea b(mod d) The following theorem summarizes the main properties of congruences Proposition 22 Let a,b,c,d,x,y Z and n, d be positive integers (i) a a(mod n) (reflexivity) a b(mod n) b a(mod n) (symmetry) } a b(mod n) = a c(mod n) (transitivity) b c(mod n) In other words, congruence is an equivalence relation (ii) If a b(mod n) and c d(mod n), then we have ax + cy bx + dy (mod n) and ac bd (mod n) (iii) If m is a positive integer, then we have a b(mod n) = a m b m (mod n) (iv) (Gauss s theorem) Here is a congruence version of Theorem 22 } ac bc (mod n) = a b(mod n) (c, n) = (v) (Generalization of Gauss s theorem) We set d = (c, n) Then we have ac bc (mod n) a b(mod n/d) Proof We leave the proofs of (i) and (ii) to the reader as an exercise They directly derive from the definition

25 Congruences 37 (iii) Using Lemma 6 we know that (a b) divides (a m b m ) so if n (a b), then n (a m b m ) (iv) The system can be written as { n (a b)c (c, n) = so that Gauss s theorem immediately yields n (a b) (v) We set d = (c, n) and c = dc, n = dn so that (c,n ) = Suppose first that ac bc (mod n) Then there exists k Z such that ac bc = kn and so ac bc = kn Thus we have { n (a b)c (c,n ) = and Gauss s theorem then implies that n (a b) Conversely, if (n/d) (a b), then (nc/d) = nc c(a b) and since n nc, we obtain n c(a b) The proof is complete Example 222 Find the remainder of the Euclidean division by 7 of A = 0 k= 0 0k Answer 0 2 = 2 + 7 4 and hence 0 2 2 (mod 7), and thus 0 6 = ( 0 2) 3 2 3 8 (mod 7) On the other hand, for every integer k 2, we have 3 ( 4 k ) by using Lemma 6, and hence 6 divides 4(4 k ), and so 4 k 4 (mod 6) Since 0 4 (mod 6), we deduce that 0 k 4 (mod 6) for k 2 The congruence being We could also use Fermat s little theorem See Theorem 35

38 2 Bézout and Gauss also true for k =, we have 0 A = 0 0k = k= 0 0 4 k= 0 k= 0 k= 4 0 4+6h 0 4 3 4 5 (mod 7) Example 223 Let a,b Z and n be a positive integer such that a b(mod n) Prove that Answer Using Lemma 6 we get and we know that a b 0 (mod n) and n k=0 a n b n (mod n 2 ) n a n b n = (a b) a k b n k k=0 n a k b n k a n na n 0 (mod n) k=0 Example 224 Prove that 3 (2 70 + 3 70 ) Answer We have 2 6 = 5 3 and 3 3 = 2 3 +, hence 2 6 (mod 3) and 3 3 (mod 3) Writing 70 = 6 + 4 and 70 = 3 23 + we obtain 2 70 + 3 70 = 2 6+4 + 3 3 23+ 2 4 + 3 3 0 (mod 3) Example 225 Solve the Diophantine equation 2 m 3 n = n where m, n are nonnegative integers such that n 5 (mod 8) Answer (0, 0) is the only solution when m = 0 and there is no solution when m = Furthermore, if m, then n = 2 m 3 n (mod 2) so that one may suppose that m 2 and n is an odd integer satisfying n 5 (mod 8) by assumption Writing n = 2k + forsomek Z 0, we infer that k ±2 (mod 8) and therefore 3 n + n = 3 9 k + 2k + 4 + 2k 0 (mod 8) Now let m and n be two solutions with m 2 and n is odd such that n 5 (mod 8) By the argument above we get 8 2 m, so that m 2, and then m = 2, giving n = Therefore, under the above hypotheses, the equation has two solutions, namely the pairs (0, 0) and (2, )

25 Congruences 39 Our next goal is to count the number of solutions of the equation ax b(mod n) Here we mean the number of integer solutions x such that ax b(mod n) and 0 x n Lemma 24 and Proposition 25 then immediately give Proposition 226 Let a,b Z and n be a positive integer We set d = (a, n) Then the congruence ax b(mod n) has a solution if and only if d b Furthermore, if x 0 is a particular solution, then there are exactly d solutions given by x 0 + kn d with 0 k d In particular, if (a, n) =, then the congruence ax b(mod n) has exactly one solution An important application of this result is the so-called Chinese remainder theorem which treats several congruences simultaneously The name of this result comes from the fact that a 2-congruence version was discovered by the Chinese mathematician Sun Tse ( st century AD) Suppose we have a system of congruences x a (mod n ) x a k (mod n k ) where a,,a k Z and n,,n k are positive integers The following result is very useful Theorem 227 (Chinese remainder theorem) Suppose that the positive integers n,,n k are pairwise coprime, ie for every pair (i, j) {,,k} 2 such that i j, we have (n i,n j ) = Then the system x a (mod n ) x a k (mod n k ) has a unique solution modulo n n k

40 2 Bézout and Gauss Proof The proof is constructive, meaning that we construct a solution and then show that it is unique We set n = n n k and, for every integer i =,,k, we define n i = n/n i Obviously we have n i N and (n i,n i ) = Proposition 226 implies that there exist integers n i such that n i n i (mod n i) We then set x = a n n + +a kn k n k We notice that if i j, then n j 0 (mod n i) and thus x a i n i n i a i (mod n i ) (i =,,k) Suppose now that the system has two solutions x and y Then x y 0 (mod n i ) for every i =,,k Since the integers n i are pairwise coprime, Proposition 23 (vi) implies that x y 0 (mod n), which completes the proof Example 228 Solve the system x (mod 2) x 2 (mod 3) x 3 (mod 5) x 4 (mod 7) x 5 (mod ) Answer Note that the modules are pairwise distinct prime numbers, and then are trivially pairwise coprime We have n = 230 and the computations are summarized in Table 2 so that x = 55 + 3080 + 458 + 320 + 050 523 (mod 230) Remark 229 Let a, n be coprime integers It will be shown in Theorem 42 that a ϕ(n) (mod n) where ϕ is Euler s totient function It follows that x = a ϕ(n) is a solution of the equation ax (mod n) and thus one may take n i = (n/n i) ϕ(n i) Therefore Table 2 Calculations of the n i n n 2 n 3 n 4 n 5 55 770 462 330 20 n n 2 n 3 n 4 n 5 2 3

26 Further Developments 4 an explicit formula for the system of Theorem 227 in terms of n i and a i can be derived as follows x k ( n i= n i ) ϕ(ni ) a i (mod n n k ) 2 An important particular case, often used in some recreational mathematics, is when we have a = =a k = a, where the whole computational machinery is not needed The trivial solution x a(mod n n k ) is guaranteed to be the only one by Theorem 227 3 When the moduli are not pairwise coprime, some potential sets of remainders may be ruled out For instance, the system { x 2 (mod 6) x 3 (mod 4) has no solution in Z, otherwise there would exist (h, k) Z 2 such that 6h = + 4k, and hence 2h (mod 4), contradicting Proposition 226 A necessary and sufficient condition for a solution to exist is ( a i a j mod(ni,n j ) ) for all i, j {,,k} 2 In this case, all solutions are congruent modulo [n,,n k ] 26 Further Developments 26 The Ring (Z/nZ, +, ) A ring (R, +, ) is a non-empty set endowed with two binary operations, usually denoted by + and, such that (R, +) is an abelian group with identity element denoted by 0, the operation being left- and right-distributive over + If the operation also has an identity element, often denoted by, then the ring is called a unitary ring An element a R, where R is unitary, is a unit of R if there exists b R such that ab = ba = Such a b is then unique and is called the inverse of a Finally, an element a R is called a zero divisor if there exists b R \{0} not necessarily unique such that ab = 0, and if R is unitary, neither a nor b are units Now let n be a positive integer The set of residue classes modulo n, often denoted by 0,,,n, is a unitary abelian ring, denoted by Z/nZ, endowed with the binary operations + and defined by x + y = x + y and x y = x y (one can check that they are indeed well-defined) The basic result is then the following theorem

42 2 Bézout and Gauss Theorem 230 The ring Z/nZ is a disjoint union Z/nZ = E F where E, resp F, is the set of units, resp zero divisors, of Z/nZ Furthermore, x Z/nZ is a unit if and only if (x, n) = Proof Let d = (x, n) Suppose first that x is a unit and define y = (x) We then have xy (mod n) and then there exists k Z such that xy = kn Thus we have d (xy kn) = and then d = Conversely, suppose that d = By Bachet Bézout s theorem there exist u, v Z such that xu + vn = Therefore xu (mod n) and then x is a unit in Z/nZIfd>, we set x = dx, n = dn with (x,n ) = Then we have n x = dn x = nx 0 (mod n) so that x is a zero divisor in Z/nZ Conversely, a zero divisor cannot be a unit, and then d> Euler s totient function is the function ϕ which counts the number of units in Z/nZ, with the convention that ϕ() = Theorem 230 has then the following immediate application Theorem 23 Let n be a positive integer Then ϕ() = and, for every integer n 2, ϕ(n) is the number of positive integers m n such that (m, n) = In other words, we have ϕ(n) = m n (m,n)= We shall find this important function again in Chap 4 Finally, note that Lemma 34 tells us that if n = p is prime, then Z/pZ does not have any zero divisor, and then the set of units of Z/pZ is (Z/pZ) = {,,p } Thus, for every prime number p,wehaveϕ(p) = p More generally, group theory tells us that, if n = p α pα r r is the factorization of n into prime powers, then we have the following isomorphism of cyclic groups Z/nZ Z/p α Z Z/pα 2 2 Z Z/pα r r Z which induces the group isomorphism (Z/nZ) ( Z/p α Z) ( Z/p α 2 2 Z) ( Z/p α r r Z) whence we deduce that ϕ(n) = ϕ ( p α ) ϕ ( p α 2 2 ) ( ) ϕ p α r r characterizing the multiplicativity of Euler s totient function We shall prove this again in Chap 4 by using purely arithmetic arguments

26 Further Developments 43 262 Denumerants Let k, n be positive integers and A k ={a,,a k } be a finite set of pairwise relatively prime positive integers 2 The aim of this section is to study the number D k (n) of solutions of the Diophantine equation a x + +a k x k = n where the unknowns are (x,,x k ) (Z 0 ) k The number D k (n) is called the denumerant of n with respect to k and to the set A k When a = =a k =, this denumerant will be denoted by D (,,) (n) where the vector (,,) is supposed to have k components Exercise 0 deals with the case k = 2 where it is shown that D 2 (n) = [ n ab ] + r where r = 0orr = and for which it is convenient to rename the coefficients a, b instead of a, a 2 Thus,ifn<ab, there is at most one solution The Frobenius problem deals with the largest number n for which this equation is not soluble, and it can be proved that this number is given by n = ab a b By using the theory of generating functions, several authors derived some formulae for the exact value of D 2 (n) The following lines borrow the ideas essentially from [Tri00] Expanding the generating function into partial fractions and comparing the coefficients of z n in each side gives 3 D 2 (n) = n ab + ( 2 a + ) + a b a k= e a (nk) e a ( bk) + b b k= e b (nk) e b ( ak) (24) where e a (x) = e 2πix/a Note that the sums are periodic in n, the first one with period a and the second one with period b, and since (a, b) =, the sum of these two sums is periodic of period ab, and then the expression of D 2 (n) is essentially determined modulo ab Now define a and b to be integers satisfying a b b a aa (25) n(mod b) bb n(mod a) 2 When k =, this obviously means A ={} 3 See Exercise for another proof of this identity

44 2 Bézout and Gauss Note that hence we get { ea ( bk) } b ( e a ( bj k) = e a b bk ) = e a (nk) and using the formulae j=0 e a (nk) e a ( bk) = b e a ( bk) e a ( bj k) a k= valid for positive integers a, m, and j=0 { a, if a m e a ( mk) =, otherwise a k= e a ( mk) = a 2 valid for positive integers a, m such that (a, m) =, we then obtain a k= a e a (nk) e a ( bk) = k= = a 2 b e a ( bk) = b a + 2 a e a ( bj k) j=0 k= b (a ) + j= and the same is true for the second sum, so that we finally get D 2 (n) = n ab + a b + b a Note that by (25) both a and b divide n+aa +bb, so that n+aa +bb is divisible by ab by Proposition 23 (vi) Furthermore, since n + aa + bb n + a + b, we infer that the number on the right-hand side above is indeed a non-negative integer We can then state the following result Theorem 232 Let a,b be coprime integers and define positive integers a, b as in (25) Then the number D 2 (n) of solutions in non-negative integer pairs (x, y)

26 Further Developments 45 of the equation ax + by = n is given by D 2 (n) = n ab + a b + b a Note that the following four properties characterize uniquely D 2 (n) D 2 (n + kab) = D 2 (n) + k (k 0) D 2 (n) = if ab a b<n<ab D 2 (n) + D 2 (m) = if m + n = ab a b (m,n 0) D 2 (n) = if n = ax 0 + by 0 <ab a b (x 0,y 0 0) See [BR04, BZ04, Shi06, Wil90] for more details Another formula, discovered by Popoviciu [Pop53] in 953, is given in Exercise In the general case k 3, a similar method involving the generating function can be used, leading to the asymptotic formula D k (n) = n k (k )! a a k + O ( n k 2) as n, but computations become harder as soon as the number of unknowns increases On the other hand, a strong upper bound can be derived in a more elementary way by taking into account the following observation Since a x + +a k x k = n a x + +a k x k = n a k x k we obtain x k [n/a k ] and thus n/a k D k (n) = D k (n ja k ) (26) j=0 The aim is then to prove the following result Theorem 233 For every positive integer k we set { 0, if k = s k = a 2 + a 3 + +a k, 2 if k 2 and r k = {, if k = a 2 a k, if k 2

46 2 Bézout and Gauss with s 2 = a 2 Then D k (n) (n + s k) k (k )! r k Furthermore, if a = =a k =, then ( ) k + n D (,,) (n) = n The proof uses induction and the following lemma Lemma 234 For every real number x 0 and every integer k 2 we have 0 j x (x j) k k ( x + 2) k Proof We shall in fact prove the slightly stronger following inequality 0 j x (x j) k xk k + xk + 2 (k )xk 2 8 since we easily see using Newton s formula that ( x + ) k = xk k 2 k + xk (k )xk 2 + + positive terms 2 8 xk k + xk + 2 (k )xk 2 8 By partial summation with a = 0, f(t)= and g(t) = (x t) k we get x (x j) k = (k ) (x t) k 2( [t]+ ) dt 0 j x and since [t]=t ψ(t) /2, we obtain ( x (x j) k = (k ) 0 j x and integrating by parts gives 0 j x (x j) k = xk k + xk 2 0 + (k 2) 0 (x t) k 2 ( t + 2 ) dt x ( [ψ2 (k ) (t)(x t) k 2] x 0 x 0 ) (x t) k 3 ψ 2 (t) dt 0 ) (x t) k 2 ψ(t)dt

26 Further Developments 47 = xk k + xk (k )(k 2) 2 xk k + xk (k )(k 2) + 2 8 x 0 x 0 (x t) k 3 ψ 2 (t) dt (x t) k 3 dt = xk k + xk (k )xk 2 + 2 8 which completes the proof of the lemma Proof of Theorem 233 By induction on k, the case k = being immediate since D (n) = n ( )! = (n + s ) ( )!r Now suppose k 2 We let the case k = 2inExercise0, and suppose the inequality is true for some k 2 Using (26), induction hypothesis and Lemma 234 we get n/a k+ D k+ (n) = D k (n ja k+ ) j=0 (k )! r k ak k+ (k )! r k ak k+ k! r k ( n + sk n/a k+ j=0 (n+s k )/a k+ j=0 a k+ + 2 (n ja k+ + s k ) k ( n + sk a k+ ) k j ) k = (n + s k+) k k! r k+ which is the desired result The last part of the theorem may be treated similarly, using the well-known identity (see [Gou72, identity 49]) n ( ) ( ) k + j k + n = j n j=0 in the induction argument 263 Generating Functions We saw in the last section that it was essential to study the generating function of a sequence to get some information about the sequence itself Let (u n ) be a sequence

48 2 Bézout and Gauss The generating function of (u n ) is the function F(z)formally defined by F(z)= n 0 u n z n If F(z) can be written as an elementary function of z, then the comparison of the coefficients of z n can give a closed formula for u n Nevertheless, we must notice that such a series could either be studied formally or must be taken as a function of the complex variable z In the latter case, convergence problems must be studied Generating functions are the main tools in combinatorial theory, where results from complex analysis are frequently used (see [Odl95]or[Wil90]) For example, if F(z) is analytic in the open disc z <R, then, for any r with 0 <r<rand any nonnegative integer n we have u n r n max z =r F(z) by Cauchy s theorem For a wide class of generating functions, the saddle point method enables us to obtain more accurate estimates To this end, we first pick up the concept of H- function from [Odl95] A generating function F(z)= n 0 u nz n is said to be an H- function,orhayman s function,iff(z)is analytic in z <Rfor some 0 <R, real when z ] R,R[ and satisfies max F(z) = F(r) z =r for some R 0 <r<r Furthermore, let a(z) = zf (z) F(z) and b(z) = za (z) and let r δ(r) be a function defined in the range ]R 0,R[ and satisfying 0 < δ(r) < π in this range Suppose that Uniformly for θ <δ(r),wehave F ( re iθ) F(r)e iθa(r) 2 θ 2 b(r) and F ( re iθ) ( ) F(r) = o b(r) as r R lim r R b(r) = It can be shown [Odl95] that if P(z) = a n z n + +a 0 R[z] and if F(z) and G(z) are H-admissible in z <R, then e F(z), F (z)g(z) and F(z)+ P(z) are H- admissible in z <R Furthermore, if a n > 0, then P(F(z)) is H-admissible in z <R Now we may state one of the main results of the saddle point theory

26 Further Developments 49 Theorem 235 Let F(z)= n 0 u nz n be an H-function and set a(z) = zf (z) F(z) and b(z) = za (z) For all n Z 0, let z n be the solution of a(z n ) = n such that R 0 <z n <R Then, for n, we have u n z n n F(z n ) 2πb(zn ) For example, with u n = /n! we have F(z)= e z, R = and a(z) = b(z) = z so that z n = n, and we obtain the following version of Stirling s formula n! ( ) n n 2πn e In addition to () which is the simplest example of a generating function with u n =, it must be interesting to take the following series into account, which generalizes Newton s formula For any positive integer k and any real number x < we have ( ) k + j x j = j ( x) k (27) j=0 As an example, let us try to find a closed formula for the generating function of the denumerant D k (n) seen above Using () we have formally ( z a )( z a 2) ( z a k ) = z a x z a 2x2 x 0 x 2 0 x k 0 = z a x + +a k x k x,, x k 0 = n 0 ( x,, x k 0 a x + +a k x k =n = D k (n)z n n 0 ) z n z a kx k so that the generating function of D k (n) is given by F(z)= k i= z a i (28)

50 2 Bézout and Gauss Now let us show with an example how this series could be used We take k = 3 and A 3 ={, 2, 3} so that D 3 (n) counts the number of non-negative integer solutions of the equation x + 2y + 3z = n The generating function of D 3 (n) is then F(z)= and partial fractions expansion theory gives ( z)( z 2 )( z 3 ) ( z)( z 2 )( z 3 ) = 6( z) 3 + 4( z) 2 + 7 72( z) ( + 8( + z) + 9 z/ρ + z/ρ 2 where ρ = e 3 () = e 2πi/3 Using(27) we obtain D 3 (n)z n = { ( ) n + 2 z n + ( ) n + + 7 6 n 4 n 72 n 0 n 0 + ( )n + ( ρ n + ρ 2n)} 8 9 and comparing the coefficients of z n in each side gives D 3 (n) = ( ) n + 2 + 6 n 4 = 6n2 + 36n + 47 72 ( n + n + ( )n 8 ) + 7 72 + ( )n + 8 9 + 2 9 cos ( 2πn 3 ) ) ( ρ n + ρ 2n) Thus we can check the values of D 3 (n) according to the residue class modulo 6 of n which we summarize in Table 22, and then we have D 3 (n) = (n+3)2 2 + ε n where ε n 3 so that (n + 3) 2 D 3 (n) = 2 where x is the nearest integer to x Table 22 Values of D 3 (n) n mod 6 0 ± ±2 3 D 3 (n) n 2 + 6n + 2 2 (n + )(n + 5) 2 (n + 2)(n + 4) 2 (n + 3) 2 2

27 Exercises 5 27 Exercises Let a, b, n, x, y, z, t be any positive integers (a) Prove that if (a, b) =, then (a + b,a b) = or2 (b) Show that if (a, b) =, then (a + b,ab) = (c) Prove that a b a n b n (d) Suppose that xt yz = Prove that (a, b) = (ax + by, az + bt) 2 (Thue s lemma) The aim of this exercise is to provide a proof of the following result Lemma (Thue) Let a> be an integer and p 3 be a prime number such that p a Then the equation au v(mod p) has a solution (u, v) Z 2 such that { u < p, v < p (a) Let S ={0,,[ p]} and define a map f : S 2 {0,,p } by f(u,v) au v(mod p) Show that f is not injective (b) We denote by (u,v ) and (u 2,v 2 ) two pairs such that (u,v ) (u 2,v 2 ) and f(u,v ) = f(u 2,v 2 ), and set u = u u 2 and v = v v 2 Check that au v(mod p) and that u < p and v < p Show that u 0, and then v 0 3 (Euclidean algorithm revisited) Let a>bbe positive integers and let (r k ) be the sequence of remainders of the divisions in the Euclidean algorithm with r 0 = a and r = b r 0 = r q + r 2 0 <r 2 <r r = r 2 q 2 + r 3 0 <r 3 <r 2 r k = r k q k + r k+ 0 <r k+ <r k r n = r n q n r n = (a, b) We define the sequences (s k ) and (t k ) so that, for k =,,nwe have { s0 = s = 0 s k+ = q k s k + s k t 0 = 0 t = t k+ = q k t k + t k

52 2 Bézout and Gauss (a) Prove that r k = s k a + t k b (k = 0,,n+ ) (b) Prove that s k = ( ) k s k and t k = ( ) k+ t k (k = 0,,n+ ) Deduce that for k =,,n (c) Deduce that s k+ =q k s k + s k t k+ =q k t k + t k for k =,,n+ a = t k r k + t k r k 4 (Infinite sequence of coprime integers (Edwards)) Let u be an odd integer and (u n ) be the sequence of integers defined by u n = u 2 n 2 for any n 2 (a) Prove that u n is odd for every integer n (b) Deduce that (u n,u n ) = for every integer n 2 (c) Show that, for any integer n 3, we have u n 2 = u 2 n 2 u2 n 3 u2 (u 2 2) (d) Deduce that for any integers n 3 and r = 2,,n, we have (u n,u n r ) = 5 Solve the following equations and system: 2 825x + 9 450y = 2 025 (x, y Z 0 ) 5(x + y) 2 = 47[x,y] (x, y Z 0 ) { 3x (mod 5) (x Z) 5x 2 (mod 7) 6 Let a, b be positive integers and set d = (a, b) We consider a Cartesian coordinate system with origin O and x- and y-axis The rectangular coordinates x, y of a point M will be denoted by M x,y Aninteger point is a point with integer coordinates (a) Define A a,b Prove that the number N of integer points on the segment ]OA] is d

27 Exercises 53 (b) Define B a,0 Prove that the number N 2 of integer points inside the rightangled triangle OAB is 4 N 2 = (ab a b d)+ 2 7 (Oloa) Fermat s last theorem states that the Diophantine equation x n + y n = z n has no non-trivial solutions (x,y,z) N 3 as soon as n 3 This was finally proved by Wiles and Taylor in 995 Does the equation x n + y n = z n+ possess any nontrivial positive integer solutions for every positive integer n? 8 Let n be a non-negative integer (a) Prove that n 2 0, or4(mod 8) Deduce that an integer congruent to 7 modulo 8 cannot be written as the sum of three squares (b) Prove that n 3 ±or0(mod 9) Deduce that an integer congruent to 4 modulo 9 cannot be written as the sum of three cubes (c) Solve the Diophantine equation x 3 + y 3 + z 3 = 2005 where x, y, z are positive integers 9 Show that 64 (2 32 + ) 0 Let a<bbe positive integers such that b a = m 2 for some positive integer m The aim of this exercise is to study the difference x y of the solutions (x, y) (Z 0 ) 2 of the Diophantine equation (a) Prove that x>yand that (b) Prove that ax 2 + x = by 2 + y (x y) { + b(x + y) } = (mx) 2 (x y) { + a(x + y) } = (my) 2 x y = (b a,x y) (x, y) 2 (c) Provide some sufficient conditions so that x y may be a square 4 One can make use of Pick s formula which states that the number N int inside a convex integer polygon P is given by N int = area(p) N P + 2 where N P is the number of integer points on the edges of the polygon

54 2 Bézout and Gauss Let a, b be positive integers such that (a, b) = and let n be a positive integer Define D 2 (n) to be the number of non-negative integer solutions x, y of the Diophantine equation ax + by = n (a) Use Bachet Bézout s theorem to prove that D 2 (n) =[ ab n ]+r where r = 0or r = (b) Use Theorem 232 with its notation to prove that r = 0 aa + bb ab r = aa + bb >ab 2 (Beck & Robins) Let a, b be positive integers such that (a, b) = and let n be a positive integer Define D 2 (n) to be the number of non-negative integer solutions x, y of the Diophantine equation ax + by = n and set a,b to be positive integers such that aa (mod b) and bb (mod a) In 953, Popoviciu [Pop53] proved the following very elegant formula D 2 (n) = n ab + { } nb a { } na b where {x} is the fractional part of x The aim of this exercise is to provide an analytic proof of this formula Recall that e a (k) = e 2πik/a (k Z 0 ) and define the complexvalued function f by (a) Recall that f(z)= a j=0 z n+ ( z a )( z b ) a x j = (x e a (j)) (29) j= Prove that, for positive integers a, k,wehave a j= j k a j= e a (j) = a 2 e a (k) e a (j) = e a(k) a (b) Use the generating function F(z)of D 2 (n) to prove that Res f(z)= D 2(n) z=0 and compute the residues at all non-zero poles of f

References 55 (c) Prove that lim R and deduce the identity (24) (d) Check that if b = then we have f(z)dz = 0 2πi z =R D 2 (n) = n a { n a } + Deduce that a a k= e a (kn)( e a (k)) = ( ) 2 a { } n a (e) By noticing that a k= a e a (kn)( e a (kb)) = finalize the proof of Popoviciu s identity k= e a (knb)( e a (k)) References [BR04] Beck M, Robins S (2004) A formula related to the Frobenius problem in 2 dimensions In: Chudnovsky D, Chudnovsky G, Nathanson M (eds) Number theory New York seminar 2003 Springer, Berlin, pp 7 23 [BZ04] Beck M, Zacks S (2004) Refined upper bounds for the linear Diophantine problem of Frobenius Adv Appl Math 32:454 467 [Gou72] Gould HW (972) Combinatorial identities A standardized set of tables listing 500 binomial coefficients summations Morgantown, West Virginia [Odl95] Odlyzko AM (995) Asymptotic enumeration methods Elsevier, Amsterdam [Pop53] Popoviciu T (953) Asupra unei probleme de patitie a numerelor Stud Cercet Stiint [Shi06] Acad RP Rom, Fil Cluj 4:7 58 Shiu P (2006) Moment sums associated with linear binary forms Am Math Mon 3:545 550 [Tri00] Tripathi A (2000) The number of solutions to ax + by = n Fibonacci Q 38:290 293 [Wil90] Wilf HS (990) Generatingfunctionology Academic Press, New York

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