Number Theory and Cryptography

Transcription

1 Number Theory and Crytograhy Paul Yiu Deartment of Mathematics Florida Atlantic University Fall 017 Chaters 9 14 October 3, 017

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3 Contents 5 Quadratic Residues Quadratic residues The Legendre symbol ( 5.3 The Legendre symbol 1 ) The square roots of 1 (mod ) Gauss lemma....(. ) The Legendre symbol The law of quadratic recirocity Smallest quadratic nonresidue modulo Square roots modulo Square roots modulo rime 1 (mod 8) Square roots modulo for a generic rime The Jacobi symbol Primality Tests and Factorization of Integers Primality of Mersenne numbers Germain rimes Probabilistic rimality tests Pseudorimes Eulerb-seudorimes Strongb-seudorimes Carmichael numbers Quadratic sieve and factor base Fermat s factorization Factor base Pollard s methods Theρ-method The( 1)-method Pythagorean Triangles Construction of Pythagorean triangles Fermat s construction of rimitive Pythagorean triangles with consecutive legs

4 iv CONTENTS 7.3 Fermat Last Theorem for n = Two ternary trees of rational numbers Genealogy of Pythagorean triangles Homogeneous Quadratic Equations in 3 Variables Pythagorean triangles revisited Rational oints on a conic Integer triangles with a 60 angle Integer triangles with a 10 angle Heron triangles The Heron formula Construction of Heron triangles Heron triangles with sides in arithmetic rogression Heron triangles with integer inradii The equation P k,a +P k,b = P k,c for olynomial numbers Double ruling of S Primitive Pythagorean trile associated with ak-gonal trile Triangular triles Pentagonal triles

5 38 CONTENTS

6 Chater 5 Quadratic Residues 5.1 Quadratic residues Let n > 1 be a given ositive integer, and gcd(a,n) = 1. We say that a Z n is a quadratic residue modnif the congruencex a (mod n) is solvable. Otherwise, a is called a quadratic nonresidue mod n. 1. If a andbare quadratic residues mod n, so is their roduct ab.. If a is a quadratic residue, and b a quadratic nonresidue mod n, then ab is a quadratic nonresidue mod n. 3. The roduct of two quadratic nonresidues mod n is not necessarily a quadratic residue mod n. For examle, in Z 1 = {1,5,7,11}, only 1 is a quadratic residue; 5, 7, and are all quadratic nonresidues. Proosition 5.1. Let be an odd rime, and a. The quadratic congruence ax +bx+c 0 (mod ) is solvable if and only if(ax+b) b 4ac (mod ) is solvable. Theorem 5.. Let be an odd rime. Exactly one half of the elements of Z are quadratic residues. Proof. Each quadratic residue modulois congruent to one of the following 1( 1) residues. ( ) 1 1,,...,k,...,. We show that these residue classes are all distinct. For 1 h < k 1, h k (mod ) if and only if(k h)(h+k) is divisible by, this is imossible since each of k h andh+k is smaller than. Corollary 5.3. If is an odd rime, the roduct of two quadratic nonresidues is a quadratic residue.

7 40 Quadratic Residues In the table below we list, for rimes < 50, the quadratic residues and their square roots. It is understood that the square roots come in airs. For examle, the entry (,7) for the rime 47 should be interreted as saying that the two solutions of the congruence x (mod 47) are x ±7 (mod 47). Also, for rimes of the form = 4n+1, since 1 is a quadratic residue modulo, we only list quadratic residues smaller than. Those greater than can be found with the hel of the square roots of 1. Quadratic residues mod and their square roots 3 (1, 1) 5 ( 1, ) (1, 1) 7 (1, 1) (, 3) (4, ) 11 (1, 1) (3, 5) (4, ) (5, 4) (9, 3) 13 ( 1, 5) (1, 1) (3, 4) (4, ) 17 ( 1, 4) (1, 1) (, 6) (4, ) (8, 5) 19 (1, 1) (4, ) (5, 9) (6, 5) (7, 8) (9, 3) (11, 7) (16, 4) (17, 6) 3 (1, 1) (, 5) (3, 7) (4, ) (6, 11) (8, 10) (9, 3) (1, 9) (13, 6) (16, 4) (18, 8) 9 ( 1, 1) (1, 1) (4, ) (5, 11) (6, 8) (7, 6) (9, 3) (13, 10) 31 (1, 1) (, 8) (4, ) (5, 6) (7, 10) (8, 15) (9, 3) (10, 14) (14, 13) (16, 4) (18, 7) (19, 9) (0, 1) (5, 5) (8, 11) 37 ( 1, 6) (1, 1) (3, 15) (4, ) (7, 9) (9, 3) (10, 11) (11, 14) (1, 7) (16, 4) 41 ( 1, 9) (1, 1) (, 17) (4, ) (5, 13) (8, 7) (9, 3) (10, 16) (16, 4) (18, 10) (0, 15) 43 (1, 1) (4, ) (6, 7) (9, 3) (10, 15) (11, 1) (13, 0) (14, 10) (15, 1) (16, 4) (17, 19) (1, 8) (3, 18) (4, 14) (5, 5) (31, 17) (35, 11) (36, 6) (38, 9) (40, 13) (41, 16) 47 (1, 1) (, 7) (3, 1) (4, ) (6, 10) (7, 17) (8, 14) (9, 3) (1, 3) (14, ) (16, 4) (17, 8) (18, 1) (1, 16) (4, 0) (5, 5) (7, 11) (8, 13) (3, 19) (34, 9) (36, 6) (37, 15) (4, 18) 5. The Legendre symbol Letbe an odd rime. For an integer a, we define the Legendre symbol ( ) { a +1, ifais a quadratic residue mod, := 1, otherwise. Lemma 5.4. ( ) ab = ( a )( b ). Proof. This is equivalent to saying that modulo, the roduct of two quadratic residues (resectively nonresidues) is a quadratic residue, and the roduct of a quadratic residue and a quadratic nonresidue is a quadratic nonresidue.

8 ) 5.3 The Legendre symbol( 1 41 Theorem 5.5 (Euler). Letbe an odd rime. For each integeranot divisible by, ( ) a a 1 ( 1) mod. Proof. Suoseais a quadratic nonresidue mod. The modresidues1,,..., 1 are artitioned into airs satisfying xy = a. In this case, ( 1)! a 1 ( 1) (mod ). On the other hand, if a is a quadratic residue, with a k ( k) (mod ), aart from 0,±k, the remaining 3 elements of Z can be artitioned into airs satisfying xy = a. ( 1)! k( k)a 1 ( 3) a 1 ( 1) (mod ). Summarizing, we obtain ( 1)! ( ) a a 1 ( 1) (mod ). Note that by utting a = 1, we obtain Wilson s ( ) theorem: ( 1)! 1 (mod ). By comarison, we obtain a formula for : a ( ) a a 1 ( 1) (mod ). ( 5.3 The Legendre symbol ) 1 Theorem 5.6. Let be an odd rime. 1 is a quadratic residue modulo if and only if 1 (mod 4). Proof. ( ) If x 1 (mod ), then ( 1) 1 x 1 1 (mod ) by Fermat s little theorem. This means that 1 is even, and 1 (mod 4). ( ) If 1 (mod 4), the integer 1 is even. By Wilson s theorem, (( 1 )!) = j = 1 i=1 j ( j) 1 i=1 j ( j) = ( 1)! 1 (mod ). The solutions ofx 1 (mod ) are therefore x ±( 1 )!. Theorem 5.7. There are infinitely many rimes of the form4n+1. 1 i=1

9 4 Quadratic Residues Proof. Suose there are only finitely many rimes 1,,..., r of the form4n+1. Consider the roduct P = ( 1 r ) +1. Note that P 1 (mod 4). Since P is greater than each of 1,,..., r, it cannot be rime, and so must have a rime factor different from 1,,..., r. But then modulo, 1 is a square. By Theorem 5.6, must be of the form 4n + 1, a contradiction The square roots of 1 (mod ) Here are the square roots of 1 for the first0 rimes of the form 4k +1: ± 13 ±5 17 ±4 9 ±1 37 ±6 41 ±9 53 ±3 61 ±11 73 ±7 89 ±34 97 ± 101 ± ± ± ± ± ±8 173 ± ± ±81 In general, the square roots of 1 (mod ) can be found as n k for a quadratic nonresidue n modulo. Examle 5.1. Let = 7933 = Since 5 (mod 8), n = is a quadratic nonresidue. We comute 1983 by successive squaring and multilication making use of the binary exansion 1983 = ( ). Since this binary exansion has only one digit0at osition 6, = 11. From these, modulo 7933, k a k a k (mod 7933) 1983 = ( 1047) ( 3868) 950. The square roots of 1 (mod 7933) are ±950.

10 5.4 Gauss lemma Gauss lemma Theorem 5.8 ( (Gauss ) Lemma). Letbe an odd rime, andaan integer not divisible by. Then = ( 1) µ whereµis the number of residues among a falling in the range < x <. a, a, 3a,..., 1 a Proof. Every residue modulo has a unique reresentative with least absolute value, namely, the one in the range 1 x 1. The residues described above are recisely those whose reresentatives are negative. Now, among the reresentatives of the residues of a,a,..., 1 a, say, there are λ ositive ones, andµnegative ones r 1,r,...,r λ, s 1, s,..., s µ. Here,λ+µ = 1, and0 < r i,s j <. Note that no two of the r s are equal; similarly for the s s. Suose that r i = s j for some indices i and j. This means ha r i mod ; ka s j mod for some h, k in the range 0 < h, k < 1 ( 1). Note that (h + k)a 0 mod. But this is a contradiction since h+k < 1 and does not divide a. It follows that r 1,r,...,r λ,s 1,s,...,s µ are a ermutation of 1,,..., 1 ( 1). From this a a 1 a = ( 1)µ 1 1, ( ) anda 1 ( 1) = ( 1) µ. By Theorem 5.5, = ( 1) µ. ( 5.5 The Legendre symbol Theorem 5.9. Letbe an odd rime. ( ) = ( 1) 1 4 (+1) = ( 1) 1 8 ( 1). Equivalently, ) a ( ) = { +1 if ±1 mod 8, 1 if ±3 mod 8.

11 44 Quadratic Residues Proof. We need to see how many terms in the sequence 1,, 3,..., 1 are in the range < x <. If = 4k +1, these are the numbers k +,...,4k, and there are k of them. On the other hand, if = 4k + 3, these are the numbers k +,...,4k +, and there arek+1 of them. In each case, the number of terms is 1(+1). 4 Here are the square roots offor the first0 rimes = 8k ±1: The law of quadratic recirocity Theorem 5.10 (Law of quadratic recirocity). Letand q be distinct odd rimes. ( )( ) q = ( 1) 1 q 1. q Equivalently, when at least one of, q 1 mod 4, is a quadratic residue mod q if and only if q is a quadratic residue mod. 1 Proof. (1) Letabe an integer not divisible by. Suose, as in the roof of Gauss Lemma above, of the residues a, a,... 1 a, the ositive least absolute value reresentatives are r 1, r,..., r λ, and the negative ones are s 1, s,..., s µ. The numbers a,a,..., 1 a are a ermutation of hi a +r i, i = 1,,...,λ, and kj a +( s j ), j = 1,,...,µ, where h 1,..., h λ, k 1,..., k µ are a ermutation of 1,,..., 1. Considering the sum of these numbers, we have a 1 ( 1) m=1 m = 1 ( 1) m=1 ma + λ r i + i=1 µ ( s j ) 1 For q 3 mod 4,is a quadratic residue mod q if and only if q is a quadratic nonresidue mod. j=1

12 5.6 The law of quadratic recirocity 45 = = 1 ( 1) m=1 1 ( 1) m=1 In articular, if a is odd, then ma + ma + λ r i + i=1 1 ( 1) m=1 µ s j + j=1 µ ( s j ) j=1 m+µ µ s j. j=1 µ 1 ( 1) m=1 ma mod, and by Gauss lemma, ( ) a = ( 1) 1 ( 1) m=1 ma. () Therefore, for distinct odd rimes andq, we have ( ) q = ( 1) 1 ( 1) m=1 mq, and ( ) = ( 1) 1 (q 1) n=1 n q. q q n 1 1 m (3) In the diagram above, we consider the lattice oints (m,n) with 1 m 1 and 1 n q 1 1. There are altogether q 1 such oints forming a rectangle. These oints are searated by the line L of sloe q through the oint (0,0). For each m = 1,,..., 1, the number of oints in the vertical line through (m,0) underlis mq. Therefore, the total number of oints underl is 1 ( 1) m=1 mq.

13 46 Quadratic Residues Similarly, the total number of oints on the left side of L is 1 (q 1) n n=1. From q these, we have It follows that 1 ( 1) m=1 mq + 1 (q 1) n=1 n = 1 q 1 q. ( )( ) q = ( 1) 1 q 1. q The law of quadratic recirocity can be recast into the following form: ( ) ( ) q, if q 3 mod 4, = ( ) q +, otherwise. q Examle 5.. (a) 59 is a quadratic residue modulo 131: ( ) ( ) ( ) ( ) ( ) = = = = ( ) ( ) 13 1 = = = ( 1) = The square roots are ±37. ( (b)34 is a quadratic nonresidue modulo 97: 34 ( 97) = )( 17 ( 97 97). Now, 97) = +1 by Theorem 5.9, and ( ) ( ) ( ) ( )( ) ( ) = = = = = Therefore, ( 34 97) = (+1)( 1) = 1. ( ) 17 = 3 ( ) = 1. 3 Examle 5.3. For which rimes > 3 is 3 a quadratic residue? A rime > 3 is of the form6k+ε forε = ±1. For such a rime, ( ( 3) = ε 3) = ε. By the law of quadratic recirocity, ( ) 3 ( = ( 1) = ( 1) 3) k ( 1) ε 1 ε. ( ) Since ( 1) ε 1 3 ε = 1 for ε = ±1, we have = ( 1) k. This means that 3 is a quadratic residue mod if and only if k is even, i.e., = 1m±1.

14 5.7 Smallest quadratic nonresidue modulo 47 Exercise Comlete the following table for the rimes admitting a as a quadratic residue. ( ) a = +1 if and only if a 1 1 (mod 4) ±1 (mod 8) Smallest quadratic nonresidue modulo Lemma Let be an odd rime. The smallest quadratic nonresidue modulo is a rime. Proof. Let b be the smallest quadratic nonresidue. If b is not rime, we write b = cc, where 1 < c,c < b. Exactly one of c, c must be a quadratic nonresidue. This contradicts the minimality ofb. Letbe a rime. We want to determine the smallest quadratic nonresidue modulo, denoted by b(). Clearly, if 3 or 5 mod 8, then b() =. Recall that 3 is a quadratic residue mod if and only if is of the form 1n ± 1. It follows that b() = 3 for 5,7 (mod 1). Aart from the rime 3, modulo 4, every odd rime has residue 1, 5, 7, 11, 13, 17, 19, 3. The smallest quadratic nonresidue mod in these cases are given by (mod 4) b()? 3 3?

15 48 Quadratic Residues Smallest quadratic nonresidues for the first100 rimes of the form 4k ±1. b() b() b() b() b() b() Square roots modulo Square roots modulo rime 1 (mod 8) ( Proosition 5.1. Letbe a rime of the form4k+3. If roots of a (mod ) are ±a 1 4 (+1). Proof. ( a 1 4 (+1) ) a 1 (+1) = a 1 ( 1) a = ( Proosition Letbe a rime of the form8k+5. If roots of a (mod ) are (a)±a 1 8 (+3) if a 1 4 ( 1) 1 (mod ), (b) ± 1 4 ( 1) a 1 8 (+3) ifa 1 4 ( 1) 1 (mod ). Proof. Note that ( Since a a ) = 1, then the square ( ) a a = a (mod ). ( a 1 8 (+3) ) a 1 4 (+3) = a 1 4 ( 1) a (mod ). ) = a 1 ( 1) 1 (mod ), we have a 1 4 ( 1) ±1 (mod ). a ) = 1, then the square

16 5.8 Square roots modulo 49 If a 1 4 ( 1) 1 (mod ), then this gives a 1 8 (+3) as a square root ofa (mod ). If a 1 4 ( 1) 1 (mod ), then we have ( ) ( ) a a 1 y ( ) ( ) 8 (+3) a 1 8 (+3) y 1 4 ( 1) a 1 8 (+3) for any quadratic nonresidue y (mod ). Since 5 (mod 8), we may simly take y =. Examle 5.4. (a) Let = 3. Clearly is a quadratic residue mod 3. The square roots of mod 3 are ± 6 ±18 5 (mod 3). (b) Let = 9. Both 6 and 7 are quadratic residues mod 9. Since (mod 9), the square root of 7 mod 9 are ±7 4 ±3 6. On the other hand, Since (mod 9), the square roots of 6 mod 9 are ± ±1 0 ±8 (mod 9) Square roots modulofor a generic rime Let a be a quadratic residue mod. We find x Z such that x a (mod ). For this, we write 1 = s t with t odd, and begin with a quadratic nonresidue n mod. Comute the two residues (1) b := n t mod, which is a rimitive s -root of unity mod, and () r := a (t+1)/ mod. Lemma (a) b is a rimitive s -root of unity mod. (b) ρ 0 := a 1 r is a s 1 -root of unity mod. Proof. (a) b s = (n t ) s = n s t = n 1 1 (mod ). Therefore, b is a s -root of unity modulo. Note that b, being an odd ower of a quadratic nonresidue, is a quadratic nonresidue mod. If b is not a rimitive s -root of unity, there is an integer r, 0 r < s, such that b r 1 (mod ). Then b would be an even ower of a rimitive s -root of unity, and therefore a quadratic residue mod, a contradiction. (b) Modulo, ρ s 1 0 = (a 1 r ) s 1 (a 1 a t+1 ) s 1 = a s 1 t = a ( 1)/ = ( ) a = +1. Now we find a square root ofa mod in the formb j r, for somej,0 j < s 1. Let j = (j s j s 3 j 1 j 0 ) be the binary exression of j, each of j 0,..., j s 1 being 0 or 1. We find the binary digits of j inductively.

17 50 Quadratic Residues (i) From (b), ρ0 s 1 1 (mod ). Therefore,ρ0 s ±1 (mod ). Set { 0 ifρ0 s 1 mod, j 0 = 1 ifρ0 s 1 mod andρ 1 = a 1 (b (j 0) r) = a 1 (b j 0 r), which is a s -root of unity. (ii) Suose we have j 0,j 1,..., j k 1 such that ρ k := a 1 (b (j k 1 j 1 j 0 ) r) is a s k 1 -root of unity mod. Set { 0 if ρk s k 1 mod, j k = 1 if ρk s k 1 mod andρ k+1 := a 1 (b (j kj k 1 j 1 j 0 ) r), which is a s k -root of unity. (iii) With j 0,..., j s, we obtain ρ s 1 = a 1 (b (j s j 1 j 0 ) r) = 1. Therefore,b j r is a square root of a mod. Examle 5.5. Consider = 401 and a = 186; a is a quadratic residue mod, and a 1 = 35. Now, 1 = 400 = 4 5; (s,t) = (4,5). With the quadratic nonresidue n = 3, we take (i) b = n t = 3 5 = 68, (ii) r = a 1 (t+1) = = 103. With these, (iii) ρ 0 = a 1 r = = 98; ρ 0 = 1; j 0 = 1, andρ 1 = a 1 (br) = 1. (iv) ρ 1 1 = 1; j 1 = 0, andρ = 1. (v) ρ 0 = 1; j = 1. (vi) j = (101) = 5. The square roots of 186 mod 304 are ±b 5 r = ±304 = ±97. Examle 5.6. Consider = 7993 and a = 41; a is a quadratic residue mod, and a 1 = Now, 1 = 799 = ;(s,t) = (3,999). With the quadratic nonresidue n = 5, we take (i) b = n t = = 1654, (ii) r = a 1 (t+1) = = 487. With these, (iii) ρ 0 = a 1 r = ( 3899) 487 = 110; ρ 1 0 = 1; j 0 = 1, and ρ 1 = a 1 (br) = 799. (iv) ρ 0 1 = 799 = 1; j 1 = 1. (v) j = (11) = 3. The square roots of a mod are ±b 3 r = ±1975.

18 5.9 The Jacobi symbol The Jacobi symbol Definition. Let n = k i i be an odd integer. For an integer a with gcd(a, n) = 1, the Jacobi symbol is defined by ( a n) := ( ) ki a. i Proosition Letmandnbe odd numbers. (a) Ifgcd(a,n) = 1 anda b (mod n), then ( ( a n) = b n). (b) Ifgcd(a,n) = gcd(b,n) = 1, then ( ( ab n) = a b ) ) n)( n. In articular, (a n (c) Ifmand n are relatively rime, andgcd(a,m) = gcd(a,n) = 1, then ( a ) mn ( a = n )( a n). (d) ( ) 1 n = 1 if and only if n 1 (mod 4). (e) ( n) = 1 if and only ifn ±1 (mod 8). (f) (The law of recirocity) If gcd(m,n) = 1, then ( m n )( n m) = ( 1) m 1 n 1. = 1. Remark. If n is not a rime, ( a n) = +1 does not imly that a is a quadratic residue mod n. For examle, ( ( 15) = 3)( 5) = ( 1)( 1) = +1. But the quadratic residues modulo 15 are 1, 4, and 10.

19 5 Quadratic Residues

20 Chater 6 Primality Tests and Factorization of Integers 6.1 Primality of Mersenne numbers A Mersenne number of is one of the form M k := k 1. A Mersenne rime gives rise to an even erfect number. IfM k = k 1 is a rime, thenk must be a rime. The converse, however, is not true. The Mersenne numbers M = 3, M 3 = 7, M 5 = 31, M 7 = 17 are all rimes. ButM 11 = 047 is comosite. Theorem 6.1 (Fermat). If > is rime, then every rime divisor ofm := 1 is of the formk +1 for some integer k. Proof. Let q be a rime divisor of M, so that 1 (mod q). This means that order q () =. Since q 1 1 (mod q) by Fermat s little theorem, q 1 is an even multile of the odd rime. Therefore, q = k +1 for some integer k. Examle 6.1. (a)m 11 = 11 1 = 047. The divisors ofm 11 of the formk+1. For k = 1, it can be easily checked that 047 = (The other divisor 89 = 4+1). (b) M 13 = 13 1 = We need only check rime divisors of the form 6k +1 which are less than 90. These are 53 and 79. None of these divides We conclude that M 13 is rime. (c) M 9 = 9 1 = = There are 99 rimes of the form58k +1 smaller than the square root ofm 9. Of these,33,1103,089 divide M 9 (which are the second,6-th, and1-th rimes in increasing order). Exercise Factorize comletely each of the following Mersenne numbers, or rove that it is rime. (1) M 19 = 19 1 = () M 3 = 3 1 =

21 54 Primality Tests and Factorization of Integers (3)M 47 = 47 1 = Remark. In the beginning of the 0-th century, F. N. Cole, Professor of Mathematics at Columbia University, sent the Sundays of three consecutive years on the factorization ofm 67, and obtained M 67 = 67 1 = = Theorem 6.. Let 3 (mod 4) be a rime. The Mersenne numberm = 1 is divisible by +1 if and only if+1 is rime. Proof. Letq = +1. Note that q 7 (mod 8). ( ) If q divides 1, 1 (mod q), and order q () divides. Since is rime, order q () = = 1(q 1). On the other hand, order q() divides ϕ(q) q 1. Therefore, ϕ(q) = or. Since ϕ(q) (an odd number), we must have ϕ(q) = = q 1, andq is rime. ( q ( ) ( ) Since q 7 (mod 8), = 1. By Euler s theorem, = (q 1)/ q ) 1 (mod q). Therefore,q divides 1 = M. Table 1. The first100 Mersenne numbers M, 3 (mod 4), divisible by rime q = + 1 (,q) (,q) (,q) (,q) (,q) (11, 3) (3, 47) (83, 167) (131, 63) (179, 359) (191, 383) (39, 479) (51, 503) (359, 719) (419, 839) (431, 863) (443, 887) (491, 983) (659, 1319) (683, 1367) (719, 1439) (743, 1487) (911, 183) (1019, 039) (1031, 063) (1103, 07) (13, 447) (1439, 879) (1451, 903) (1499, 999) (1511, 303) (1559, 3119) (1583, 3167) (1811, 363) (1931, 3863) (003, 4007) (039, 4079) (063, 417) (339, 4679) (351, 4703) (399, 4799) (459, 4919) (543, 5087) (699, 5399) (819, 5639) (903, 5807) (939, 5879) (963, 597) (303, 6047) (399, 6599) (3359, 6719) (3491, 6983) (3539, 7079) (363, 747) (3779, 7559) (3803, 7607) (3851, 7703) (3863, 777) (3911, 783) (4019, 8039) (411, 843) (471, 8543) (4391, 8783) (4871, 9743) (4919, 9839) (4943, 9887) (5003, 10007) (5039, 10079) (5051, 10103) (5171, 10343) (531, 10463) (579, 10559) (5303, 10607) (5399, 10799) (5639, 1179) (5711, 1143) (5903, 11807) (6131, 163) (663, 157) (633, 1647) (6491, 1983) (6551, 13103) (6563, 1317) (6899, 13799) (6983, 13967) (7043, 14087) (7079, 14159) (7103, 1407) (7151, 14303) (711, 1443) (7643, 1587) (7691, 15383) (783, 15647) (7883, 15767) (8111, 163) (843, 16487) (8663, 1737) (8951, 17903) (9059, 18119) (9371, 18743) (9419, 18839) (9479, 18959) (9539, 19079) (9791, 19583) (10091, 0183)

22 6. Germain rimes Germain rimes An odd rimefor which+1 is also rime is called a Germain rime. Theorem 6. can be restated as saying that if 3 (mod 4) is a Germain rime, then the Mersenne number M has a rime divisor + 1. Table 1 gives the 100 Germain rimes of the form 4k +3. Here are the Germain rimes of the form 4k +1. Table. The first 100 Germain rimes = 4k + 1 (,q) (,q) (,q) (,q) (,q) (5, 11) (9, 59) (41, 83) (53, 107) (89, 179) (113, 7) (173, 347) (33, 467) (81, 563) (93, 587) (509, 1019) (593, 1187) (641, 183) (653, 1307) (761, 153) (809, 1619) (953, 1907) (1013, 07) (1049, 099) (19, 459) (189, 579) (1409, 819) (1481, 963) (1601, 303) (1733, 3467) (1889, 3779) (1901, 3803) (1973, 3947) (069, 4139) (19, 459) (141, 483) (73, 4547) (393, 4787) (549, 5099) (693, 5387) (741, 5483) (753, 5507) (969, 5939) (339, 6659) (3389, 6779) (3413, 687) (3449, 6899) (3593, 7187) (3761, 753) (381, 7643) (4073, 8147) (4349, 8699) (4373, 8747) (4409, 8819) (4481, 8963) (4733, 9467) (4793, 9587) (5081, 10163) (5333, 10667) (5441, 10883) (5501, 11003) (5741, 11483) (5849, 11699) (6053, 1107) (6101, 103) (6113, 17) (6173, 1347) (669, 1539) (639, 1659) (6449, 1899) (651, 13043) (6581, 13163) (6761, 1353) (711, 1443) (7193, 14387) (7349, 14699) (7433, 14867) (7541, 15083) (7649, 1599) (7841, 15683) (7901, 15803) (8069, 16139) (8093, 16187) (873, 16547) (8513, 1707) (8693, 17387) (8741, 17483) (8969, 17939) (909, 18059) (91, 18443) (993, 18587) (9473, 18947) (969, 1959) (9689, 19379) (10061, 013) (1053, 0507) (10313, 067) (1059, 1059) (10589, 1179) (10613, 17) (10709, 1419) (10733, 1467) (10781, 1563) (1131, 643) (11369, 739) Proosition 6.3. Letbe a Germain rime, so that q = +1 is also rime. (a) If 1 (mod 4), then+1 is rimitive root modulo q. (b) If 3 (mod 4), then is a rimitive root modulo q. Proof. (a) If 1 (mod 4),+ q +1 1 (mod q) and ( ) ( ) ( )( ) = = =. q q q q ( ) Note thatq 3 (mod 8), and = 1. From this q (+1) = (+1) q 1 ( ) +1 = 1 (mod q). q Therefore, the order of+1 mod q is, and +1 is a rimitive root.

23 56 Primality Tests and Factorization of Integers (b) If 3 (mod 4), then q 7 (mod 8) and 1 (mod q). It follows that ( ) ( ) ( )( ) ( ) 1 1 = = = =. q q q q q Again, andis a rimitive root for q. = q 1 ( ) = 1, q 6.3 Probabilistic rimality tests Pseudorimes The converse of Fermat s little theorem is not true. If 1 1 (mod ), one cannot conclude that is a rime. Here is an examle: = 341 = is comosite, but (mod 341). Definition. Given an integer b, an odd comosite n is called a b-seudorime ifgcd(n,b) = 1 andb n 1 1 (mod n). Examle is a 3-seudorime, since (mod 91). But91 is not a-seudorime, since (mod 91). To verify these, we make use of the binary exansion of90 = , and comute 90 and3 90 modulo91 by successive squaring and multilication: t k = t k (mod 91) 3 k (mod 91) ( 17)(16)(16) 4( 17)( 17) 4(16) 64 (mod 91), ( 9)( 10)( 10) 9 10 (90) 1 (mod 91). Proosition 6.4. Letb,b 1,b, andnbe an odd comosite number rime tob. (a)nis ab-seudorime if and only if order n (b) n 1. (b) If n is ab-seudorime, and ab 1 (mod n), thennis ana-seudorime. (c) Ifnis ab 1 - andb -seudorime, then it is also ab 1 b - andb 1 b 1 -seudorime.

24 6.3 Probabilistic rimality tests 57 Proof. (a) If n is a b-seudorime, thenb n 1 1 (mod n); order n (b) n 1. Conversely, if n 1 = k order n (b), then b n 1 (b ordern(b) ) k 1 (mod n) andnis a b-seudorime. (b) Let n be a b-seudorime. Since ab 1 (mod n), a n 1 a n 1 b n 1 (ab) n 1 1 (mod n). Therefore, n is ana-seudorime. (c) Fromb n (mod n) and b n 1 (mod n), we have Therefore, n is a b 1 b seudorime. Proosition 6.5. If n fails the test (b 1 b ) n 1 b n 1 1 b n 1 1 (mod n). b n 1 1 (mod n) for a single base b Z n, then it fails the test for at least half of the ossible bases b Z n. Proof. Suose b n 1 1 (mod n), i.e., n is not a b-seudorime. Let b 1, b,..., b s Z n be the bases for whichnis a seudorime. Ifnasses the test for any of the bases bb i, then by Proosition 6.4 (c), it would be a seudorime for (bb i )b 1 i b (mod n), a contradiction. Thus, the distinct residues bb 1, bb,..., bb s are bases for which the test fails. There are at least as many bases for which n fails the test as there are bases for whichnasses the test. Corollary 6.6. Letn > 1 be a given integer. If n asses the test b n 1 1 (mod n) for b = b 1, b,...,b k, then the robability that n is a rime is at least1 ( 1 ) k. Proof. For i = 1,,..., k, n is a b i -seudorime. By Proosition 6.5, the robability thatnis still comosite desite assing thek tests is at most1out of k. From this the result follows. Examle 6.3. n = is ab-seudorime for the9values ofb =,3,5,7,11, 13, 17, 19, 3. (In each column, the roduct of the numbers with row header 1 is 1 (mod 14979)). The robability that it is rime is at least1 ( ) 1 9 =

25 58 Primality Tests and Factorization of Integers Eulerb-seudorimes Definition. An odd comosite number b is called an Euler b-seudorime if gcd(n,b) = 1 and ( ) b b (n 1)/ (mod n), n where ( b n) is the Jacobi symbol. Proosition 6.7. Let b > 1 be a fixed integer and n > 1 be an odd comosite number. If n is an Euler b-seudorime, then it is ab-seudorime. Proof. Ifb n 1 ( ) b n (mod n), then squaring, we haveb n 1 ( b n) = 1 (mod n). This shows that n is a b-seudorime. Proosition 6.8 (Solovay-Strassen rimality test). Let n > 1 be a given integer. If n asses the test ( ) b b (n 1)/ (mod n) n for b = b 1, b,...,b k, then the robability that n is a rime is at least 1 ( 1 k. ) Strongb-seudorimes Definition. Let n be an odd comosite number, and n 1 = s t where t is odd. Forb Z n,nis called a strong b-seudorime if eitherb t 1 (mod n), or there exists r,0 r < s, such that b r t 1 (mod n).

26 6.3 Probabilistic rimality tests 59 Lemma 6.9. Let b > 1 be a fixed integer and n be an odd comosite number satisfying b n 1 1 (mod n). Write n 1 = s t with t odd. Let be a rime divisor of n, and write 1 = s t with t odd. Thens s and ( ) { b 1 if s = s, = 1 if s > s Proof. Sinceb n 1 b s 1t 1 (mod n), and t,t are both odd, we have (b s 1 t ) t ( ) t ( b s 1 t b n 1 ) t ( 1) t 1 (mod n). Therefore, b s 1 t 1 (mod n). Since is a divisor ofn, we also have b s 1 t 1 (mod ). (6.1) If s < s, then s s 1 and 1 = s t is a divisor of s 1 t. By (6.1), b 1 = b s t cannot be 1 mod, contrary to Fermat s little theorem. Thus,s s. From (6.1) again, ( ) b b 1 b s 1 t (b s 1 t ) s s = ( 1) s s = { 1 if s = s 1 if s > s. Proosition Let b > 1 be a fixed integer and n > 1 be an odd comosite number. Ifnis a strong b-seudorime, then it is an Euler seudorime. Proof. We consider three cases. Case 1.b t 1 (mod n). Since t is odd, b n 1 = (b t ) s 1 1 = ( b t n ) = ( b n), and ( ) 1 = n ( ) b t = n ( ) b. n Case. b n 1 1 (mod n). Write n = as a roduct of rimes, not necessarily distinct. Let k denote the number of rimes, counting multilicity, for which s = s in Lemma??. We always have s s and ( ) ( ) b n = b = ( 1) k. On the other hand, modulo s+1, we have 1 unless is one of the k rimes for which s = s, in which case = 1 + s. Since n = 1 + s t 1 + s (mod s+1 ), we have 1+ s (1+ s ) k 1+k s (mod k+1 ). This means that k must be odd, and ( b n) = ( 1) k = 1.

27 60 Primality Tests and Factorization of Integers Case 3.b r 1t 1 (mod n) for some 0 < r < s. Since b n 1 1 (mod n), we have to show that ( b n) = 1. Again, let be a rime divisor ofn, and write 1 = s t witht odd. Analogous to Lemma??, ( ) { b 1 if s = r = 1 if s > r. Let k be the number of rimes (counting multilicity) in the roduct n = for which s = r. Then as in case, we gave ( b n) = ( 1) k. On the other hand, since n = 1+ s t 1 (mod r+1 ) andn = (1+ r ) k (mod r+1 ), it follows that it must be even, and ( b n) = 1. Remark. The converses of Proositions 6.7 and 6.10 are not true. (1) 91 is a 3-seudorime but not an Euler3-eudorime. () 561 is an Euler-seudorime but not a strong -seudorime. Proosition Letb > 1 be a fixed integer, andn 3 (mod 4). Ifnis an Euler b-seudorime, then it is a strongb-seudorime. Proof. Since n 3 (mod 4), n 1 = s t with s = 1 and t odd. Since n is an Eulerb-seudorime, b t = b n 1 ( ) b ±1 (mod n). n Now, since s = 1, this is exactly the condition for n to be a strong b-seudorime. Proosition 6.1. Ifnis an odd comosite number, then it is a strongb-seudorime for at most 5% of all 0 < b < n. Proosition 6.13 (Rabin-Miller test). Let n be an odd integer, and n 1 = s t withtodd. If for b = b 1,b,..., b k,nasses the test: either b t 1 (mod n), or there exists r, 0 r < s, such that b r t 1 (mod n), then the robability that n is rime is at least 1 ( 1 4) k. Exercise 1. Check that M 17 = andm 19 = 5487 are rimes.. Find a rime divisor of M 3 = Find a rime divisor of M 9 =

28 6.4 Carmichael numbers Consider M 47 = 47 1 = The beginning rimes of the form 94k +1 are 83, 659, 941, 119, 13, 1693, 1787, 069, 351,539,633,3761,431,4513,4889,... (a) Find two rime divisors of M 47 from this list. (b) Comletely factorize M Show that 561 is a -seudorime. 6. Show that 179 is a - and3-seudorime. 6.4 Carmichael numbers Definition. A Carmichael number is a comosite integer n for which b n 1 1 (mod n) for everyb Z n. Examle = is a Carmichael number. To rove this, we show that if gcd(b,561) = 1, b (mod 3), b (mod 11), and b (mod 17). Now, by Fermat s little theorem, b 1 (mod 3) b 560 (b ) 80 1 (mod 3), b 10 1 (mod 11) b 560 (b 10 ) 56 1 (mod 11), b 16 1 (mod 17) b 560 (b 16 ) 35 1 (mod 17). By the Chinese Remainder Theorem, b (mod 561). This shows that 561 is a Carmichael number. This examle shows that to check if n is a Carmichael number, it is enough to check if it asses the testb n 1 1 (mod n) for rime numbers b < n. Proosition If n is an odd comosite number is divisible by the square of a (rime) number, thenncannot be a Carmichael number. Proof. Suose n for a rime. Let g be a rimitive root of. Let n be the roduct of all rimes other than which divide n. By the Chinese Remainder Theorem, there is an integer b satisfying b g (mod ), b 1 (mod n ). Thengcd(b,n) = 1 andbis a rimitive root of. We claim that n is not a b-seudorime. If n is a b-seudorime, then b n 1 1 (mod n), and b n 1 1 (mod ) since n. Since order (b) = ( 1),( 1) n 1. Sincen 1 1 (mod ),n 1 is not divisible by( 1). This is a contradiction.

29 6 Primality Tests and Factorization of Integers Proosition An odd, square free, comosite number n is a Carmichael number if and only if 1 n 1 for every rime dividing n. Proof. ( ) Letnbe a Carmichael number. Suose n 1 is not divisible by 1 for some rime divisor of n. Letg be a rimitive root of. Find an integer b satisfying b g (mod ), b 1 (mod n/). Then gcd(b,n) = 1 and b n 1 g n 1 1 (mod ) since n 1 is not divisible by order (g) = 1. Hence,b n 1 1 (mod n) cannot hold. ( ) For n, writen 1 = k( 1) for some integerk. Letbbe any base with gcd(b,n) = 1. b n 1 = b k( 1) (b 1 ) k 1 (mod ). Since n is square free, it is the roduct of all rime divisors of n. If follows that b n 1 1modn, andnis a Carmichael number. Proosition A Carmichael number is the roduct of at least three distinct rimes. Proof. By roosition 6.14, a Carmichael number must be a roduct of distinct rimes. It remains to show that it cannot be a roduct of two distinct rimes. Consider n = q for distinct rimes < q. If n were a Carmichael number, then, by Proosition 6.15, modulo q 1, 1 (q 1+1) 1 n 1 0, an imossibility since 0 < 1 < q 1. Examle = is a Carmichael number, since 560 is divisible by each of,10, and16. To show that this is the smallest Carmichael number, we make use of Proosition Since the roduct of the four smallest odd rimes is > 561, it is enough to consider n = qr with rime numbers < q < r 89, and qr 561. The table below lists all such scenarios. In every row excet the bottom one,qr 1 is not divisible by one of r 1 and q 1. This shows that 561 is the smallest Carmichael number. 6.5 Quadratic sieve and factor base Fermat s factorization Given an integernwith n = m, if there is an integerx > m such thatx n = y for some y Z, y x 1, then n = x y = (x y)(x+y) is a nontrivial factorization ofn.

30 6.5 Quadratic sieve and factor base 63 Examle 6.6. (a)n = 00819; n = n = = 1681 = 41. Therefore, = = , in which both factors are rime. (b) n = ; n = = 09 = = = , in which both factors are rime. (c) n = ; n = 899. x x n (mod 10) 90 3 (mod 10) = = = , in which both factors are rime. (d) n = ; n = If x n = y for an integer y, the units digit of x must be one of 1,4, 6,9. x x n x x n x x n = = = , in which both factors are rime. Exercise (1) Factor () Factorn = 96873, given n = Factor base This is a generalization of Fermats factorization. Given an integer n with m = n, we consider integers x close to m with x n factored into small rimes. More recisely, we secify a factor base B consisting of 1 and some rimes, and restrict to those x for which the factors of x n are all in B. With enough x, we collect those for which the roduct ofx n is a square. Suose, for j = 1,,..., k, z j := x j n = y j factors into rimes in B, and that k j=1 z j = Z for an integer Z. Then, withx := k j=1 x j, X = k x j Z j=1 (mod n).

31 64 Primality Tests and Factorization of Integers If X Y (mod n), then n has a common divisor of one or both of X ±Y. This gives a nontrivial factorization ofn. Examle 6.7. n = 9509; n = 97. LetB = { 1,,5,11}. j x j z j = x j n = = = = Note thatz 1 z = ( 5 11). Therefore,( )( ) 0 (mod n); (mod n). Now,gcd(n,9435) = 37 andgcd(n,8995) = 57. This gives 9509 = 37 57, in which both factors are rime. Examle 6.8. n = 87463; n = 95. LetB = { 1,3,17}. j x j z j = x j n = = Note that x 1 x = (mod n) and = Therefore, ( )( ) 0 (mod n). This gives (mod n). Now, gcd(n,4696) = 587 and gcd(n,7450) = 149. This gives = , in which both factors are rime. Examle 6.9. Consider n = with m = n = We have Multilication gives 1964 n = (mod n), n = (mod n), n = (mod n), n = (mod n). ( ) ( ) (mod n), or (mod n). Thus, gcd(383753, ) = gcd(383753, ) = 1093 is a divisor of n. The other divisor is = , both factors are rime numbers.

32 6.5 Quadratic sieve and factor base 65 Exercise (1) FactorN = by making use of (mod N) and (mod N). () FactorN = 8833 by making use of (mod N), (mod N), (mod N).

33 66 Primality Tests and Factorization of Integers 6.6 Pollard s methods Theρ-method Let n be a given integer and f(x) Z[x]. Beginning with an integer x 0, construct a sequence of integers x 1, x, x 3,... by x k+1 = f(x k ) (mod n). If is a rime divisor of n, there is a corresond sequence z k = x k (mod ) of integers in the range [0, 1]. If n, the number of ossible values of z k is much less than the number of ossible values of x k. We exect the z k to reeat sooner. Examle Letn = 341 andf(x) = x +1 (mod n). Withx 0 = 3 we generate two sequencesx k andy k by x k+1 = f(x k ) (mod n), y k = x k (mod n). k x k y k Since341 = 11 31, we reduce x k andy k modulo 31 and obtain k x k (mod 31) y k (mod 31) Note that x 3 y 3 (mod 31). The eriod of the sequence modulo 31 is 3. We can find a factor by comuting gcd(x k y k,n). Now, gcd(x 3 y 3, 341) = 31, which is a factor of341. Let n be an integer, known to be comosite by some means. To find a divisor of n, we roceed as follows. Let f(x) be a olynomial in Z[x]; for examle, f(x) = x + 1. With an arbitrary integer x 0, form a sequence x 1, x, x 3,..., x k,... by setting x k+1 = f(x k ). Comute gcd(x k x j, n) to find a nontrivial divisor of n. In ractice, we evaluategcd(x k x t 1, n) only for t k < t+1 to see if we get a number other than1.

34 6.6 Pollard s methods The ( 1)-method Letnbe a number, known to be comosite, and with rime divisorfor which 1 has no large rime divisor. To find, we roceed as follows. (1) Choose an integerk which is a multile of all or most integers less than some bound B. For examle, k = B! or the lcm of all integers B. () Choose an integer a between and n, and comute a k mod n by successive squaring and multilication. (3) Comute d := gcd(a k 1, n). (4) Ifdis a roer divisor ofn, we are done. If not, start with a new choice ofa or k. Examle Factorization of the Fermat number F 5 = 5 +1 = k 3 k! 1 mod n gcd(3 k! 1, F 5 ) Therefore, F 5 is divisible by641. The other factor is rime. Examle 6.1. The Fermat number F 6 = 6 +1 = k 3 k! 1 mod n gcd(3 k! 1, F 6 ) Therefore, F 6 is divisible by The other factor is

35 68 Primality Tests and Factorization of Integers Examle Consider n = M 37 = 37 1 = k 3 k! 1 mod n gcd(3 k! 1, n) Examle Letn = 479. We comute k! (mod n) andgcd( k! 1 (mod n), n). k d := k! (mod n) gcd(d,n)

36 Chater 7 Pythagorean Triangles 7.1 Construction of Pythagorean triangles By a Pythagorean triangle we mean a right triangle whose side lengths are integers. Any common divisor of two of the side lengths is necessarily a divisor of the third. We shall call a Pythagorean triangle rimitive if no two of its sides have a common divisor. Let (a,b,c) be one such triangle. From the relation a +b = c, we make the following observations. 1. Exactly two of a,b,care odd, and the third is even.. In fact, the even number must be one of a and b. For if c is even, then a and b are both odd. Writing a = h+1 andb = k +1, we have c = (h+1) +(k +1) = 4(h +k +h+k)+. This is a contradiction since c must be divisible by We shall assumeaodd andbeven, and rewrite the Pythagorean relation in the form c+a c a ( ) b =. Note that the integers c+a and c a are relatively rime, for any common divisor of these two numbers would be a common divisor c and a. Consequently, each of c+a and c a is a square. 4. Writing c+a = u and c a = v, we havec = u +v anda = u v. From these,b = uv. 5. Since c andaare both odd, u andv are of different arity. We summarize this in the following theorem. Theorem 7.1. The side lengths of a rimitive Pythagorean triangle are of the form u v,uv, andu +v for relatively rime integersuandv of different arity.

37 70 Pythagorean Triangles 7. Fermat s construction of rimitive Pythagorean triangles with consecutive legs Let a, b, c be the lengths of the sides of a right triangle, c the hyotenuse. Figures (a) and (b) below, together with the Pythagorean theorem, give the following two relations (a+b c) =(c a)(c b), (7.1) (a+b+c) =(c+a)(c+b). (7.) c b b b c a c a+b c c b a a a+b c c a (a)a,b,c fromc a and c b c (b)a,b,c fromc+a andc+b Beginning with a right triangle(a,b,c), we construct a new right triangle(a,b,c ) with c a = c+b and c b = c+a. By a comarison of (8.16) and (7.), we have a +b c = a+b+c. From these, a =a+b+c, b =a+b+c, c =a+b+3c. Note that b a = b a. This construction therefore leads to an infinite sequence of integer right triangles with constant difference of legs. In articular, beginning with (3,4,5), we obtain the sequence (3,4,5), (0,1,9), (119,10,169), (696,697,985),... of Pythagorean triangles with legs differing by 1. This construction gives all such Pythagorean triangles. Note that the above construction is invertible: from a right triangle(a,b,c ) one can construct a smaller

38 7. Fermat s construction of rimitive Pythagorean triangles with consecutive legs 71 one (a,b,c) with the same difference between the legs. More recisely, a =a +b c, b =a +b c, (7.3) c = a b +3c. Since a+b+c = a +b c < a +b +c, this inverse construction does yield a smaller triangle. However, it certainly cannot lead to a strictly decreasing sequence of integer right triangles. Now,a = a +b c must be a ositive integer. Using the Pythagorean theorem, it is easy to deduce from a +b > c that4a > 3b, or a > 3(b a ). This means that from every Pythagorean triangle with legs differing by 1, there is a descent, by reeated alications of (7.3), to a minimal integer right triangle with shortest side not exceeding 3. It is clear that there is only one such triangle, namely,(3, 4, 5). This therefore shows that the above construction actually gives all Pythagorean triangles with consecutive legs.

39 7 Pythagorean Triangles Aendix: Primitive Pythagorean triles < 1000 u,v a,b,c u,v a,b,c u,v a,b,c u,v a,b,c,1 3,4,5 3, 5,1,13 4,1 15,8,17 4,3 7,4,5 5, 1,0,9 5,4 9,40,41 6,1 35,1,37 6,5 11,60,61 7, 45,8,53 7,4 33,56,65 7,6 13,84,85 8,1 63,16,65 8,3 55,48,73 8,5 39,80,89 8,7 15,11,113 9, 77,36,85 9,4 65,7,97 9,8 17,144,145 10,1 99,0,101 10,3 91,60,109 10,7 51,140,149 10,9 19,180,181 11, 117,44,15 11,4 105,88,137 11,6 85,13,157 11,8 57,176,185 11,10 1,0,1 1,1 143,4,145 1,5 119,10,169 1,7 95,168,193 1,11 3,64,65 13, 165,5,173 13,4 153,104,185 13,6 133,156,05 13,8 105,08,33 13,10 69,60,69 13,1 5,31,313 14,1 195,8,197 14,3 187,84,05 14,5 171,140,1 14,9 115,5,77 14,11 75,308,317 14,13 7,364,365 15, 1,60,9 15,4 09,10,41 15,8 161,40,89 15,14 9,40,41 16,1 55,3,57 16,3 47,96,65 16,5 31,160,81 16,7 07,4,305 16,9 175,88,337 16,11 135,35,377 16,13 87,416,45 16,15 31,480,481 17, 85,68,93 17, 4 73, 136, , 6 53, 04, 35 17, 8 5, 7, , , 340, ,1 145,408,433 17,14 93,476,485 17,16 33,544,545 18,1 33,36,35 18, 5 99, 180, , 7 75, 5, , 11 03, 396, , , 468, ,17 35,61,613 19, 357,76,365 19,4 345,15,377 19,6 35,8,397 19, 8 97, 304, 45 19, 10 61, 380, , 1 17, 456, , , 53, ,16 105,608,617 19,18 37,684,685 0,1 399,40,401 0,3 391,10,409 0, 7 351, 80, 449 0, 9 319, 360, 481 0, 11 79, 440, 51 0, 13 31, 50, 569 0,17 111,680,689 0,19 39,760,761 1, 437,84,445 1,4 45,168,457 1, 8 377, 336, 505 1, , 40, 541 1, , 67, 697 1, 0 41, 840, 841,1 483,44,485,3 475,13,493,5 459,0,509,7 435,308,533, 9 403, 396, 565, , 57, 653, 15 59, 660, 709, , 748, 773,19 13,836,845,1 43,94,95 3, 55,9,533 3,4 513,184,545 3, 6 493, 76, 565 3, 8 465, 368, 593 3, 10 49, 460, 69 3, 1 385, 55, 673 3, , 644, 75 3, 16 73, 736, 785 3, 18 05, 88, 853 3, 0 19, 90, 99 4,1 575,48,577 4,5 551,40,601 4,7 57,336,65 4,11 455,58,697 4, , 64, 745 4, 17 87, 816, 865 4, 19 15, 91, 937 5, 61, 100, 69 5, 4 609, 00, 641 5, 6 589, 300, 661 5, 8 561, 400, 689 5, 1 481, 600, 769 5, 14 49, 700, 81 5, , 800, 881 5, , 900, 949 6, 1 675, 5, 677 6,3 667,156,685 6,5 651,60,701 6,7 67,364,75 6,9 595,468,757 6, , 57, 797 6, , 780, 901 6, , 884, 965 7, 75, 108, 733 7, 4 713, 16, 745 7, 8 665, 43, 793 7, 10 69, 540, 89 7, , 756, 95 7,16 473,864,985 8,1 783,56,785 8,3 775,168,793 8,5 759,80,809 8, 9 703, 504, 865 8, , 616, 905 8, , 78, 953 9, 837, 116, 845 9, 4 85, 3, 857 9, 6 805, 348, 877 9, 8 777, 464, 905 9, , 580, 941 9,1 697,696,985 30,1 899,60,901 30,7 851,40,949 31, 957,14,965 31,4 945,48,977 31,6 95,37, Fermat Last Theorem forn = 4 Theorem 7. (Fermat). The area of a Pythagorean triangle cannot be a square. Proof. Suose to the contrary there is one such triangle, which we may assume rimitive, with side lengths (u v,uv,u + v ), u, v being relative rime of

40 7.4 Two ternary trees of rational numbers 73 different arity. The area A = uv(u v ) being a square, and no two of u, v, u v sharing common divisors, each of these numbers must be a square. We write u = a, v = b so that u v = a 4 b 4 is also a square. Since a 4 b 4 = (a b )(a +b ) and the two factors are relatively rime, we must have a b = r anda +b = s for some integersr ands. From these,a = r +s and (a) = (r +s ) = (r+s) +(r s). Thus, we have a new Pythagorean triangle(r s,r+s,a). This is a Pythagorean triangle whose area is the square of an integer: 1 (r s)(r+s) = 1 (r s ) = b. But it is a smaller triangle since b = v is a roer divisor of A = uv(u v ). By descent, beginning with one Pythagorean triangle with square area, we obtain an infinite sequence of Pythagorean triangles with decreasing areas, each of which is a square integer; a contradiction. Corollary 7.3 (Fermat Last Theorem for n = 4). The equation x 4 +y 4 = z 4 does not have solutions in nonzero integers. Proof. Suosex 4 +y 4 = z 4 for ositive integersx,y,z. The Pythagorean triangle with sides z 4 y 4,z y andz 4 +y 4 has a square area a contradiction. z y (z 4 y 4 ) = z y x 4 = (x yz), Remark. This roof actually shows that the equation x +y 4 = z 4 has no solution in nonzero integers. 7.4 Two ternary trees of rational numbers Consider the rational numbers in the oen interval (0,1). Each of these is uniquely in the form q, for relatively rime ositive integers > q. We call +q the height of the rational numbers. The rational numbers in(0,1) with odd heights can be arranged in a ternary tree with root 1, as follows. For a rational number t of odd heights, the numbers 1, t 1, and t are also in(0,1) and have odd heights. We call these the descendants +t 1+t of t and label them the left (L), middle (M), and right (R) resectively. If we write t = q, then these three descendants are, and q, and have greater q +q +q heights. Thus, the rational number 1 has left descendant, middle descendant, 5 3 and right descendant 1. 4 On the other hand, each rational number s (0,1) \ { 1, 1} with odd height 3 is the descendant of a unique rational number t, which we call its arent. In fact, s = n is m (i) the left descendant of 1 = n m if 1 < s < 1, s n