Lemma (x, y, z) is a Pythagorean triple iff (y, x, z) is a Pythagorean triple.

Chapter Pythagorean Triples.1 Introduction. The Pythagorean triples have been known since the time of Euclid and can be found in the third century work Arithmetica by Diophantus [9]. An ancient Babylonian table (1600 B.C) contains a list of Pythagorean triples including (4961, 6480, 8161) (about 1000 years before Pythagoras). Definition.1.1. A triple (x, y, z) of positive integers satisfying x +y = z is called a Pythagorean triple. A Pythagorean triple (x, y, z) is primitive if gcd(x, y, z) = 1. Lemma.1.1. (x, y, z) is a Pythagorean triple iff (y, x, z) is a Pythagorean triple. Remark.1.1. If (x, y, z) is a Pythagorean triple, then (nx, ny, nz) is a Pythagorean triple for all n N and conversely. Thus if we know a Pythagorean triple, then we can have infinitely many Pythagorean triples. Proposition.1.1. Pythagorean triples which are in A.P. are 3n, 4n, 5n where n N. Proof. Let x n, x, x + n be a Pythagorean triple. (x n) + x = (x + n) i.e. x = 4xn and so x = 4n. Hence the triple is 4n n, 4n, 4n + n i.e. 3n, 4n, 5n. The contents in this chapter are published in SRTMUs Research Journal of Science, Vol.,3, No., 014, ISSN: 77-8594 (Print),pp 1-7. 14

CHAPTER. PYTHAGOREAN TRIPLES 15 For any Pythagorean triple (x, y, z) how many entries of (x, y, z) are even? Can z (hypotenuse of the Pythagorean triple (x, y, z)) be the only even entry? For solving such questions, we need to observe that square of an even number is even and square of an odd number is odd. Hence we can not have all x, y, z be odd, since the sum or difference of any two odd number is even. In Pythagorean triple (6, 8, 10) all the entries are even. Also in (3, 4, 5) single entry is even. But exactly two entries from the triple (x, y, z) can not be even since the sum or difference of two even numbers is even. Thus one may observe that there can be Pythagorean triple with (i) one even number or (ii) three even numbers. In case (i), the hypotenuse z can not be even. If in Pythagorean triple (x, y, z), x = m + 1, y = n + 1 are odd and z = l is even, then we have, x + y = z (m + 1) + (n + 1) = (l) 4m + 4m + 1 + 4n + 4n + 1 = 4l 4(m + m + n + n) + = 4l which is not possible. Hence if in the Pythagorean triple (x, y, z), only one number is even then hypotenuse z is not even. Theorem.1.1. The infinite sequence of all Pythagorean triples (x, y, z) where y, z are consecutive is of the form, ( (x, y, z) = n, n 1, n + 1 ) (.1.1) where integer n > 1 is odd.

CHAPTER. PYTHAGOREAN TRIPLES 16 Proof. Consider n = m + 1 an odd integer, then equation (.1.1) becomes, ( (x, y, z) = m + 1, (m + 1) 1, (m + 1) + 1 ) = (m + 1, m + m, m + m + 1) we check this triple is a Pythagorean triple. Therefore consider x + y = (m + 1) + (m + m) = 4m + 4m + 1 + 4m 4 + 8m 3 + 4m = 4m 4 + 8m 3 + 8m + 4m + 1 = (m + m + 1) = z Thus the following table gives the Pythagorean triples (x, y, z) with y + 1 = z Table.1: (x, y, z) is a Pythagorean triple with y + 1 = z. x 3 5 7 9 11 13 15 17 19 1 3 5 7 9 31 y 4 1 4 40 60 84 11 144 180 0 60 31 364 40 460 z 5 13 5 41 61 85 113 145 184 1 61 313 365 41 461 x 33 35 37 39 41 43 45 47 49 51 53 y 54 59 664 740 80 904 99 1084 1180 180 1384 z 55 593 665 741 81 905 993 1085 1181 181 1385 At this stage one may ask the natural question as Are these the only Pythagorean triples? Remark.1.. Plato (430-349 BC) found Pythagorean triples (x, y, z) with y+ = z namely (4n, 4n 1, 4n + 1), n N.

CHAPTER. PYTHAGOREAN TRIPLES 17 To check the triples (x, y, z) = (4n, 4n 1, 4n + 1) are Pythagorean, consider x + y = (4n) + (4n 1) = 16n + 16n 4 8n + 1 = 16n 4 + 8n + 1 = (4n + 1) = z Table.: Pythagorean triples ( Plato). 4n 4 8 1 16 0 4 8 3 36 40 44 48 5 56 4n 1 3 15 35 63 99 143 195 57 35 401 485 575 675 843 4n + 1 5 17 37 65 101 145 197 59 37 403 487 577 677 845 4n 60 64 68 7 76 80 84 88 9 96 100 104 4n 1 899 963 1155 195 1443 1599 1763 1935 115 303 499 703 4n + 1 901 965 1157 197 1445 1601 1765 1937 117 305 501 705 From table (.), one may observed that in the list of Pythagorean triples, x is a multiple of 4 and y + = z. Also observe that, y is one less than perfect square and z is one more than perfect square [17], [6]. Lemma.1.. Let (x, y, z) is a Pythagorean triple. Then the following statements are equivalent. (a) d is a common divisor of x, y and z. (b) d is a common divisor of x and y. (c) d is a common divisor of x and z. (d) d is a common divisor of y and z. Proof. Clearly (a) implies (b), (c), (d). Now consider (b), i.e. d x and d y

CHAPTER. PYTHAGOREAN TRIPLES 18 d x, d y and so d x + y, i.e. d z and hence d z. Thus (b) implies (a), (c), (d) and so on. This proves the lemma (.1.) and from it we conclude that: If (x, y, z) is a Pythagorean triple, then gcd(x, y, z) = d iff gcd(x, y) = gcd(x, z) = gcd(y, z) = d. Thus if one of gcd(x, y), gcd(x, z), gcd(y, z) is one, then Pythagorean triple is primitive. In table (.1), if d = gcd(y, z) = gcd(y, y + 1), then d y, d y + 1 d y + 1 y and hence 1 = d =gcd(y, z). Hence Pythagorean triples in table (.1) are primitive. In table (.), if d =gcd(y, z) = gcd(y, y + ), then d and as both y, z are odd, we have only d = 1. Hence Pythagorean triples in table (.) are primitive. Lemma.1.3. If (x, y, z) is a primitive, then x and y have different parity, that is, one of them is even and other is odd. Proof. If x, y are both even, then (x, y, z) is not primitive and if x, y are both odd, then x = m 1, y = n 1 and z = x + y = [(m + n m n) + 1], which is impossible as the square of a number cannot be double of an odd number. Note: 1) For x and y with different parity, we have z as an odd number. ) In a Pythagorean triple (x, y, z) with gcd(x, y, z) = 1 and x is even, then 4 x, 1 xy, 60 xyz, 3 z. Since here y = a 1, z = b 1 for some a, b N and so x = z y = 4(b + b a a), b + b a a is even i.e. 8 x and so 4 x..1.1 Primitive Pythagorean Triples. Theorem.1. (Euclid). [37], [39] All the solutions of the Pythagorean equation x + y = z satisfying the condition x, y, z N, x is even, gcd(x, y, z) = 1 are given by x = st, y = s t, z = s + t (.1.)

CHAPTER. PYTHAGOREAN TRIPLES 19 for integers s and t such that s > t > 0, gcd(s, t) = 1 and s t(mod ). Theorem (.1.) also gives all primitive Pythagorean triples. So all Pythagorean triples can be given by formulas x = nst, y = n(s t ), z = n(s + t ), where n, s and t are arbitrary positive integers such that s > t, s and t have no prime factores in common and one of s or t is odd and the other is even. Table.3: Primitive Pythagorean triples (x, y, z). s 1 1 3 4 1 5 4 1 3 5 7 4 8 t 3 4 4 5 5 6 6 7 7 8 8 8 8 9 9 9 x = st 4 1 8 4 0 40 1 60 8 56 16 48 80 11 36 7 144 y = s t 3 5 15 7 1 9 35 11 45 33 63 55 39 15 77 65 17 z = s + t 5 13 17 5 9 41 37 61 53 65 65 73 89 113 85 96 145 Theorem.1.3. In any primitive Pythagorean triple (x, y, z), the numbers 3, 4 and 5 divides x, y, z either separately or jointly in different ways. Proof. From theorem (.1.), we have x = st, y = s t, z = s + t with s t(mod ) i.e. exactly one from s, t is odd. If 3 s or 3 t, then 3 x. Consider 3 s and 3 t, then s ±1(mod 3) and t ±1(mod 3) s t 0(mod 3) 3 s t i.e. 3 y. Now s t(mod ), so exactly one of s, t is even and hence 4 st i.e. 4 x. From above 1 xy. If 5 s or 5 t, then 5 x. When neither s nor t is divisible by 5, they can only be ±1 or ±(mod 5) either s t 0(mod 5) or s + t 0(mod 5) that is either 5 y or 5 z. From above 60 xyz. Considering all cases for z, it is clear that it is never divisible by 3.

CHAPTER. PYTHAGOREAN TRIPLES 0 Note: Based on the divisibility of x, y, z by 3 and 5, Primitive Pythagorean triples divided in the following six classes [38]. Class A: 3 x and 5 z Divisible by x y z 3 5 Example: (3, 4, 5), (4, 7, 5) Class B: 5 x and 3 y Divisible by x y z 3 5 Example: (5, 1, 13), (35, 1, 37) Class C: 3 x and 5 x Divisible by x y z 3 5 Example: (15, 8, 17), (45, 58, 53) Class D: 3 y and 5 z Divisible by x y z 3 5 Example: (7, 4, 5), (13, 84, 85)

CHAPTER. PYTHAGOREAN TRIPLES 1 Class E: 3 x and 5 y Divisible by x y z 3 5 Example: (1, 0, 9), (9, 40, 41) Class F: 3 y and 5 y Divisible by x y z 3 5 Example: (11, 60, 61), (91, 60, 109) From theorem (.1.), we know that for generating the primitive Pythagorean triples one needs a pair of relatively prime positive integers with different parity. We can do this by taking an odd positive integer and write it as a pair of relatively prime numebrs with different parity which generate a primitive Pythagorean triples. In the following table, we have for a given odd number which is a sum of pair of relatively prime numbers with different parity generates a primitive Pythagorean triples and one can find the class of its.

CHAPTER. PYTHAGOREAN TRIPLES Table.4: For given an odd number obtained from coprime pair, one can generate Pythagorean triples and find its calss. Odd number Co-prime pair Primitive Pythagorean Triple Class 3,1 3,4,5 A 5 3, 5,1,13 B 4,1 15,8,17 C 7 4,3 7,4,5 D 5, 1,0,9 E 6,1 35,1,37 B 9 5,4 9,40,41 E 7, 45,8,53 C 8,1 63, 16, 65 A 11 6,5 11,60,61 F 7,4 33,56,65 A 8,3 55,48,73 B 9, 77,36,85 D 10,1 99,0,101 F Now consider the mapping w from primitive Pythagorean triples (x, y, z) into the corresponding above classes as w((x, y, z)). This map w plays a very important role in Cryptographic applications [45]. The function w((x, y, z)) for some of the primitive Pythagorean triples is shown in the following table.

CHAPTER. PYTHAGOREAN TRIPLES 3 Table.5: The values of the map w((x, y, z)) Primitive Pythagorean Triples (x, y, z) w((x, y, z)) 3 4 5 A 5 1 13 B 7 4 5 D 9 40 41 E 11 60 61 F 13 84 85 D 15 8 17 C 17 144 145 D 19 180 181 F 1 0 9 E.1. Rational Points on the Unit Circle and Pythagorean Triples. Consider the unit circle a + b = 1. A point (x, y) on the unit circle is a rational point if x, y Q and x + y = 1. Similarly, (x, y) is an integer point on the unit circle if x and y are integers with x + y = 1. One may observed that, the points ( 1, 0), (0, 1), (1, 0) and (0, 1) are the only integer points on the unit circle. So for a given point (x, y) can one check it to be rational or not on the unit circle? That is for a given x = 1 ( 1 ) 3 1, we have on unit circle y = ± = ± ( 1 3 ) So, ± is not a rational point on the unit circle a + b = 1. ( 3 One may observe that the points 5, 4 ( 5, 5) 13, 1 ) ( 8 and 13 17, 15 ) are the rational 17 points on the unit circle. Consider the line y = m(x + 1) with slope m and passes through ( 1, 0) and (x, y) a point on the unit circle. If (x, y) is a rational point on the unit circle, then the slope m = y 0 x ( 1) = y x + 1.

CHAPTER. PYTHAGOREAN TRIPLES 4 For x 1, m is a quotient of two rational numbers y and x + 1 and hence m is rational. ( 3 For m = 0.5, we have (x, y) = 5 5), 4 and ( 3 ) ( 4 ) + = 1 3 + 4 = 5 which is a Pythagorean triple. 5 5 ( x Therefore if (x, y, z) is a Pythagorean triple, then the point z, y ) is a rational point z on the unit circle a + b = 1, since x + y = z x z + y z = 1 and as x, y, z are integers, x z, y both are rational numbers. z Now if (x, y) is the point of intersection of the line y = m(x + 1) and the unit circle a + b = 1, in the first quadrant and slope m is a rational number, then one can show that x, y are also rationals. That is Consider x + y = 1 x + (m(x + 1)) = 1 x + m x + m x + m 1 = 0 x = m ± 4m 3 4(m 4 1) (1 + m ) = m ± 4 (1 + m ) x = m 1 1 + m, 1 m 1 + m x = 1, 1 m 1 + m = m ± 1 1 + m Therefore in the first quadrant x = 1 m m and y = m(x + 1) = 1 + m 1 + m So x, y are the rational numbers. Since m Q, so m = p q, q 0 ( ) x = 1 p 1 m 1 + m = q ( p 1 + q ( ) y = m 1 + m = p q ( 1 + p q ) = q p p + q and ) = pq p + q So x + y = 1 ( q p ) ( pq ) + = 1 p + q p + q

CHAPTER. PYTHAGOREAN TRIPLES 5 (q p ) + (pq) = (p + q ) (pq, q p, p + q ) are Pythagorean triples. This gives us a one-to-one correspondence between the rational points on the unit circle in the first quadrant and the Pythagorean triples [54]..1.3 Representing Pythagorean triples as Gopal-Hemachandra Numbers. For any pair g, e, consider the sequence g, e, g + e, g + e, g + 3e, 3g + 5e,. Such sequence is considered as a GH sequence. With the help of GH sequence consider two numbers f = g + e, and h = e + f = g + e, then the members of the four tuple (g, e, f, h) with these conditions are called GH numbers [38]. When g = 1 = e one obtain the Fibonacci sequence. (z x)f Theorem.1.4. In the GH four tuple (g, e, f, h) if x = gh, y = and z = e eh + fg, then (x, y, z) is a Pythagorean triple. If the four tuple (g, e, f, h) has no common factors and g is odd, then (x, y, z) is a primitive Pythagorean triple. The values of y and z may also be written as y = ef, z = e + f. Proof. By definition of GH numbers, we have f = g + e, and h = e + f = g + e Therefore the values of x, y, z interms of g, e, f, h becomes z = eh + fg = e(g + e) + (g + e)g = eg + e + g + ge = e + ge + g y = (z a)f e x = gh = g(g + e) = g + ge = (e + ge + g g ge)(g + e) e = e g + e 3 e = eg + e

CHAPTER. PYTHAGOREAN TRIPLES 6 x + y = (g + ge) + (eg + e ) = g 4 + 4g 3 e + 4g e + 4e g + 8e 3 g + 4e 4 = (e + ge + g ) = z (x, y, z) is a Pythagorean triple. If g is even, so h = g + e is also even which gives z to be even. But by definition, y is also even. Therefore from lemma.1., (x, y, z) is not a primitive. So that if g is odd, then Pythagorean triple (x, y, z) is primitive. Now y = (z x)f e = eg + e = e(g + e) = ef and z = e + ge + g = e + e + ge + g = e + (e + g) = e + f ( f = g + e) Note: 1) One may observed that the Pythagorean triple (x, y, z) is obtained from GH four tuple (g, e, f, h) in such a way that, x is a product of outer two numbers, y is double the product of two inner numbers and z is the sum of squares of the two inner numbers. ) The GH four tuple can be written in the form distinct odd integers with s > t. Example.1.1. For GH four tuple (1, 1,, 3), we have ( t, s t, s + t, s ) where s, t are

CHAPTER. PYTHAGOREAN TRIPLES 7 z = 3 + = 1 + = 5, x = 3 1 = 3 and y = = 4. That is (3, 4, 5) is a Pythagorean triple corresponding to the GH four tuple (1, 1,, 3)..1.4 New Pythagorean Triple Algorithm. From theorem (.1.), one can observed that for any pair of relatively prime positive integer s, t with different parity have unique fundamental solution (x, y, z). But New Pythagorean Triple Algorithm [46] redefined this concept that: for any pair of relatively prime positive integer s, t with different parity have atleast two fundamental solutions namely (x 1, y 1, z 1 ) and (x, y, z ). That is x 1 = p + pq, y 1 = q + pq, z 1 = p + q + pq and x 1 = p pq, y 1 = q pq, z 1 = p + q pq. Also theorem.1. provides the third solutions as x 3 = pq, y 3 = p q, z 3 = p + q. This New Pythagorean Triple Algorithm is used for producing the key for data encryption and decryption. Remark.1.3. The (x, y, z) primitive Pythagorean triples where 60 x are given by (60n, 900n 1, 900n + 1), such triples are infinitely many. In this list of primitive Pythagorean triples following primitive Pythagorean triples are not included where 60 x : (60, 11, 61), (180, 19, 181), (40, 19, 41), Form table 1, and 3, we obtain (infinitely) many (x, y, z) primitive Pythagorean triples where 4 x, 3 y, and 5 z as:(4, 3, 5), (56, 33, 65), (364, 7, 365), (54, 33, 55), (16, 63, 65), (64, 963, 965), (76, 1443, 1445), (104, 703, 705), Also there are infinitely many primitive Pythagorean triples (x, y, z) where 4 x and y or z is a prime number as: (4, 3, 5), (40, 9, 41), (84, 13, 85), (8, 15, 17), (1, 35, 37), (0, 99, 85), (4, 7, 5)

CHAPTER. PYTHAGOREAN TRIPLES 8. Some Properties of Pythagorean Triples [6]. Lemma..1. Neither 1 nor can appear in any Pythagorean triple, but that every integer k 3 can appear. Proof. 1 and are not sums of two positive integers squares so for any Pythagorean triple (x, y, z) we must have z 3. Now x < x + 1 < (x + 1) for any x N, so x + 1 is not a perfect square which gives 1 / {x, y, z}. i.e. x 1, y 1, and so x, y. If y =, then x < x + y = x + 4 < (x + 1), since x, and thus x + y is not a perfect square, so y. Similarly x. Thus {x, y, z} {3, 4, 5, 6, }. Let k N, k 3 any. If k is odd, then k = l + 1 with l 4 and l + k = (l + 1) ; if k is even, then k = 4l with l 4 and (l 1) + k = (l + 1). Thus for any integer k 3, we have a Pythagorean triple (x, y, z) such that k {x, y, z}. Result..1. There is no Pythagorean triple where numbers are in G. P. Proof. If (a, ar, ar ) is a Pythagorean triple, then so is (1, r, r ), a contradiction, since 1 is not a number in any Pythagorean triple. Result... For any integer n > 1, n 1 and n + 1 are not perfect squares. (otherwise 1 is a member in a Pythagorean triple, a contradiction). Lemma... For any k N, k 3, there are only finitely many Pythagorean triples containing k. Proof. Let (x, y, z) be a Pythagorean triple. If z = k(k N, k 3), then x +y = k gives 3 x, y k 1 and x y, giving only finitely many possibilities. If x = k, then k = z y z (z 1) = z 1, so 3 y < z < k + 1 only finitely many possibilities for y, z. Lemma..3. There is no Pythagorean triple of type (x, nx, z)[(x, y, ny)]. giving Otherwise showing n + 1 [or n 1] is a perfect square or a Pythagorean triple containing 1, a contradiction. In particular if (x, y, z) is a Pythagorean triple, then x y z, x z.

CHAPTER. PYTHAGOREAN TRIPLES 9 Result..3. There is no Pythagorean triple with numbers in H. P. Proof. Suppose there is a Pythagorean triple with numbers in H. P. ab Then there is a primitive Pythagorean triple a, a + b, b. a + 4a b (a + b) = b, i.e. a (a + b) + (ab) = b (a + b) and gcd(a, b) = 1. (a(a + b), ab, b(a + b)) is a Pythagorean triple. Now a is a common divisor of a(a + b) and ab, (a + b) is a common divisor of a(a + b) and b(a + b) a(a + b) divides b(a + b), i.e. a b, i.e. gcd(a, b) = a 3, a contradiction to gcd(a, b) = 1. So the supposition there exists a Pythagorean triple with numbers in H. P. is wrong. Example..1. From lemma (..3) we can show that, 5 etc. are irrational numbers. If is rational, then there exist p, q N with p q =, i.e. p = q = q + q, showing (q, q, p) is a Pythagorean triple, a contradiction. Hence is an irrational number. If 5 is rational, then there exist a, b N with 5 = a b, i.e. b = 5a = (a) + a, showing (a, a, b) is a Pythagorean triple, a contradiction. Using above technique we get following as irrational numbers: 10, 17, 6, 37, 50, 65, 8, 101, Example... 3 is an irrational number. If 3 is rational number, then there exist m, n N with 3 = m n, i.e. m = 3n and so m + n = (n), showing (m, n, n) is a Pythagorean triple, a contradiction. Hence 3 is an irrational number. Using the above technique we get following as irrational numbers: 8, 15, 4, 35, 48, 63, 80, 99,. Remark..1. Let (x, y, z) be a primitive Pythagorean triple. Then 4 divides x or y, 1 xy, 60 xyz, 4 z. Hence if (x, y, z) is a Pythagorean triple and d z where ( x d {, 3, 4, 6, 1}, then d, y d d), z is alos a Pythagorean triple and for any d = gcd (x, y), ( x d, y d, z d) is a primitive Pythagorean triple.

CHAPTER. PYTHAGOREAN TRIPLES 30 Example..3. 7 is an irrational number. Suppose 7 is a rational number. Then there exist m, n N with gcd(m, n) = 1 and 7 = m m = 7n = (4n) (3n), i.e. (3n) + m = (4n) n ( 3n (3n, m, 4n) is a Pythagorean triple, i.e. 4, m ) 4, n is a Pythagorean triple. 4 3n, 4 m i.e. 4 n, 4 m, a contradiction to gcd(m, n) = 1. So the supposition 7 is a rational number is wrong. Using the above technique we get following as irrational numbers: 11, 1, 0, 3, 44, Example..4. 13 is an irrational number. Suppose 13 is a rational number. Then there exist m, n N with gcd(m, n) = 1 and 13 = m n m = 13n = (3n) + (n), i.e. (n, 3n, m) is a Pythagorean triple, and gcd(n, 3n) = n (, 3, m ) is a Pythagorean triple, a contradiction, since is not a member of any n Pythagorean triple. So the supposition 13 is a rational number is wrong. Using the above technique we get following as irrational numbers: 0, 9, 40, 53, 68, 85, Note: There is only one Pythagorean triple that involves consecutive numbers is (3, 4, 5). For any integer n 4, ( n(n 3) ) + (3n 1) = (n + 1) 3. So x + y = z 3 has infinitely many solutions in the set of positive integers. Theorem..1. Let x 1 = 3, y 1 = 4, z 1 = 5 and let x n, y n, z n for n =, 3, 4, be defined recursively by x n+1 = 3x n + z n + 1 y n+1 = 3x n + z n + z n+1 = 4x n + 3z n + Then (x n, y n, z n ) is a Pythagorean triple. Moreover if (x, y, z) is a Pythagorean triple

CHAPTER. PYTHAGOREAN TRIPLES 31 with y = x + 1, then (x, y, z) = (x n, y n, z n ) for some n N. Proof. Result is true for n = 1. Assume it is true for n, i.e. x n + y n = z n and x n + 1 = y n Now x n+1 + yn+1 = (3x n + z n + 1) + (3x n + z n + ) = (3x n + z n ) + 5 + 6(3x n + z n ) = 18x n + 8(x n + yn) + 4x n z n + 5 + 18x n + 1z n = 6x n + 8yn + 4x n z n + 5 + 18x n + 1z n zn+1 = (4x n + 3z n + ) = (4x n + 3z n ) + 4(4x n + 3z n ) + 4 = 16x n + 9(x n + yn) + 16x n + 1z n + 4x n z n + 4 = 5x n + 9yn + 16x n + 1z n + 4x n z n + 4 x n+1 + yn+1 zn+1 = x n yn + x n + 1 = (x n + 1) yn = 0 x n+1 + yn+1 = zn+1, showing that the result is true for n + 1. Hence by the principle of mathematical induction, result is true for all n N. Table.6: Primitive Pythagorean triples (x n, y n, z n ) with y n = x n + 1 n 1 3 4 5 6 7 8 9 10 x n 3 0 119 696 4059 3660 137903 803760 4684659 7304196 y n 4 1 10 697 4060 3661 137904 803761 4684660 7304197 z n 5 9 169 985 5741 33461 19505 1136689 665109 38613965

CHAPTER. PYTHAGOREAN TRIPLES 3 Example..5. To find all Pythagorean triples such that one member is 17. Let (x, y, z) be a Pythagorean triple and 17 {x, y, z}. If z = 17, then 17 = 4 + 1 so s = 4, t = 1 and x = st = 8, y = s t = 15. If y = 17, then 17 = 9 8 and so s = 9, t = 8 and x = st = 144, z = s + t = 145. Thus there are only two Pythagorean triples containing 17, which are (8, 15, 17) and (17, 144, 145). Example..6. The equation 3 x + 4 y = 5 z has the unique solution x = y = z = in the set of positive integers..3 Congruent Numbers. An integer that is the area of a right angled triangle with rational sides is called a congruent number.such a triangle was known to Arab Mathematicians of the twelfth century and rediscovered by Euler in the Eighteenth century [9]. Theorem (.1.) gives us as many congruent numbers as we want. i.e. for s, t N, with (s, t) = 1 of different parity, we have n = 1 (st)(s t ) So n is a congruent number. The following table gives us the different congruent numbers which arises from Pythagorean triples (x, y, z). Table.7: Congruent numbers arises from Pythagorean triples. s t x y z n 1 4 3 5 6 3 1 5 13 30 1 4 8 15 17 60 7 8 45 53 630 8 9 144 17 145 14

CHAPTER. PYTHAGOREAN TRIPLES 33 If n N is a congruent number, which is area of a right angled triangle with sides of length as rational numbers a, b, c with a + b = c, n = ab and if k N, k > 1 such that k n n, then is also a congruent number since it is area of right angled k triangle with sides of length a k, b k, c k. Thus if n = k m N, where m is a square free, then n is a congruent number iff m is a congruent number. Therefore, it is sufficient to consider n N as square free about discussion whether it is a congruent number or not. If (a, b, c) is a Pythagorean triple with a + b + c = ab, then (a 4)(b 4) = 8. For millennia it has remain unsolved problem to find an algorithm for checking whether a given integer is congruent number or not. shown that such an algorithm can be devised. In recent times, Tunnel has Theorem.3.1 (Tunnell s Theorem [64]). Suppose n is a positive integer and (x, y) denotes a rational point on the elliptic curve y = x 3 n x with x equal to the square of a rational with an even denominator. Then n is a congruent number. Example.3.1. For a right angled triangle with sides x = 3, y = 4 and z = 5, its area is xy = 6, so 6 is a congruent number. Also the right angled triangle with sides 10 7, 7 10, 101 has area 6. 70 Consider the triangle with sides 3, 0 3, 41. This is a right angled triangle and 6 its area = 1 3 0 = 5, so 5 is a congruent number. 3 Consider the triangle with sides 35 1, 4 5, 3337. This is a right angled triangle with 60 area 7, so 7 is a congruent number. Theorem.3.. [15]. 1 is not a congruent number. This theorem was open problem, stated first by Leonardo (15) and proved first by Fermat after four centuries. 1 is not a congruent number is equivalent to no congruent number be a square. Note that and 3 are not congruent numbers. Hence following numbers are not congruent numbers: 1,, 3, 4, 8, 9, 1, 16, 18, 5, 3, 36, 49, 50,. Following numbers are congruent numbers:

CHAPTER. PYTHAGOREAN TRIPLES 34 5, 6, 7, 0, 4, 8, 30, 41, 45, 54, 63, 65,. Note: As we have infinitely many Pythagorean triples so does infinitely many congruent numbers. It is not difficult to produce lots of congruent numbers. But the Table.8: Congruent numbers with side lengts of right angled trianlges Congruent Number n Sides of the Triangle 3 5, 0 3, 41 6 4 14 9, 7 4, 65 36 15 4, 15, 17 7 5 1, 1, 30 5, 1, 13 34 4, 17 36, 145 36 40 41 3, 13 0, 881 8 65 97 65, 1, 6 6 difficulty is to determine a given number n to be a congruent number or not. For the above method of determining a congruent number, if we cannot find a triangle with area n, which does not imply that n is not a congruent number. Example.3.. [1] Bastien in 1914 found right angled triangle with sides x = 71104064578955010000 11817143185779451900, y = 396778060334950039 11817143185779451900 and hypotenuse z = 015446949760001961 with area 101. This is known as the smallest solution in terms of sizes of numerator and denominator corresponding to the 11817143185779451900 congruent number. Example.3.3. [1] n = 157 is a congruent number. The simplest right angled triangle with area 157 has rational sides as

CHAPTER. PYTHAGOREAN TRIPLES 35 x = 6803984878643505117540 411340519771614938303, y = 411340519771614938303 1666555693714761309610.3.1 Elliptic Curves from Congruent Numbers. Elliptic curves are the curves given by a cubic polynomial in two variables, together with the point at infinity. These curves have a property of group law and are related with both the congruent number problem and to some questions in algebraic geometry, Cryptography [4]. In this thesis we have discussed the formulation of elliptic curve with the help of congruent number problem [8], [0], [43]. By the definition of congruent number, we have n is a congruent number is equivalent to the existence of rational numbers x, y, z so that z = x + y (.3.1) and n = 1 xy (.3.) Multiplying equation (.3.1) by 4 and addiing and substracting it from eqaution (.3.), we get z + 4n = x + y + xy = (x + y) and z 4n = x + y xy = (x y) ( z ) ( x + y ) + n = (.3.3) ( z ) ( x + y ) and + n = (.3.4) ( x y ) ( z ) 4 (4.3.3)and(.3.4) = n 4 v = u 4 n (.3.5)

CHAPTER. PYTHAGOREAN TRIPLES 36 where v = x y, u = z 4 Therefore multiplying (.3.5) by u, we get (uv) = u 6 n u ( z ) Let x = u z(x y ) = and y = uv = which gives us 8 y = x 3 n x (.3.6) Thus equation (.3.6) tells us that, the right angled triangle with rational sides and area equals n produces a rational solution to the equation (.3.6). Equation (.3.6) is an example of a curve known as elliptic curve.