Pythagorean Triples Over Gaussian Integers

Transcription

1 International Journal of Algebra, Vol. 6, 01, no., Pythagorean Triples Over Gaussian Integers Cheranoot Somboonkulavui 1 Department of Mathematics, Faculty of Science Chulalongkorn University Bangkok 10330, Thailan rung [email protected] Ajchara Harnchoowong Department of Mathematics, Faculty of Science Chulalongkorn University Bangkok 10330, Thailan [email protected] Abstract This paper investigates the unique factorization of primitive Pythagorean triples over the Gaussian integers. Moreover, we show the isomorphisms between the groups of Pythagorean triples with ifferent operations an the multiplicative group of the quotient fiel of Gaussian integers. Mathematics Subject Classification: 11R04, 11R7 Keywors: free abelian group, gaussian integer, pythagorean triple, unique factorization 1 Introuction Let K be a number fiel with ring of integers R. A triple (a, b, c) of elements of R is sai to be a Pythagorean triple if a + b = c.forr = Z, E. Eckert [3] efine an operation, aition, by (a 1,b 1,c 1 )+(a,b,c )=(a 1 a b 1 b,a 1 b + b 1 a,c 1 c ) so that the set of Pythagorean triples of natural numbers an (1, 0, 1) with + is a free abelian group. P. Zanaro an U. Zannier [6] generalize the omain from Z to any ring of integers R such that i/ R. R. Beauregar an E. Suryanarayan [1] consiere the set of Pythagorean triples over Z an efine *by(a 1,b 1,c 1 ) (a,b,c )=(a 1 a,b 1 c + b c 1,b 1 b + c 1 c ). The well-known 1 Supporte by Chulalongkorn University Grauate Scholarship to Commemorate the 7 n Anniversary of His Majesty King Bhumibol Aulyaej

2 56 C. Somboonkulavui an A. Harnchoowong representation of Pythagorean triples in Number Theory resulte in properties an a unique factorization theorem of primitive Pythagorean triples. The set of equivalence classes of Pythagorean triples is a free abelian group which is isomorphic to the multiplicative group of positive rationals. N. Sexauer [5] investigate solutions of the equation x +y = z on unique factorization omains satisfying some hypotheses. Later, K. Kubota [4] characterize Pythagorean triples in an arbitrary unique factorization omain. Where R is the Gaussian integers, James T. Cross [] isplaye a metho for generating all Pythagorean triples. Each equivalence class of primitive Pythagorean triples is mappe from a certain pair of Gaussian integers. Inspire by R. Beauregar an E. Suryanarayan s work, this paper escribe the unique factorization of primitive Pythagorean triples when R is the Gaussian integers. The group of equivalence classes of Pythagorean triples over Z[i] is isomorphic to Q[i] an its quotient group is free abelian. In Gaussian integers, i = 1 comes in hany when we show the relation between two operations. The last theorem escribes that the group of Pythagorean triples whose first components are non-zero with operation * is isomorphic to the group of Pythagorean triples whose thir components are non-zero with the operation + efine above. The Semigroup of Pythagorean Triples Over Gaussian Integers Let PT be the set of all Pythagorean triples in the ring of Gaussian integers where their first components are non-zero; i.e., PT = {(a, b, c) a, b, c Z[i] with a 0;a + b = c }. Define the operation * on PT by (a 1,b 1,c 1 ) (a,b,c )=(a 1 a,b 1 c + b c 1,b 1 b + c 1 c ). (1) Proposition.1. The set PT uner the operation * is a commutative monoi with the ientity element (1, 0, 1). Proof. The proof is straightforwar an left to the reaer. K. Kubota etermine the representation of Pythagorean triples in a unique factorization omain [4]. We applie the theorem to the ring of the Gaussian integers. Proposition.. If (a, b, c) PT, then there are f, u, v, Z[i] where is a factor of relatively prime to f an u ± v such that a = fuv, b = f(u v ), an c = f(u + v ). ()

3 Pythagorean triples over Gaussian integers 57 Definition.3. A Pythagorean triple (a, b, c) is sai to be primitive if the components a, b, c have no common ivisor. Corollary.4. If (a, b, c) PT is primitive, then there exist u, v, Z[i] where is a factor of an u ± v such that a = uv, b = u v, an c = u + v. (3) Proof. From Proposition., if f is not a unit, then (a, b, c) is not primitive. Parity makes things much easier in Z. James T. Cross use δ := 1 + i to efine even an o Gaussian integers an gave a proof of the following lemma []. Lemma.5. Z[i]/ <δ>={[0], [1]}. Hence [0] an [1] are the resiue classes of 0 an 1 in Z[i]/ <δ>, respectively. Definition.6. Let a be a Gaussian integer. We say that a is even or o accoring as a is in the resiue class etermine by 0 or 1. It follows that all elementary properties of even an oness hol. example, the sum of an even Gaussian integer an an o one is o. For Lemma.7. If (a, b, c) PT is primitive, then only one of a, b, c is even an the others are o. Proof. Suppose that two of a, b, c are even. Since a + b = c, all a, b, c are even. This contraicts the fact that (a, b, c) is primitive. A significant ifference between the set of integers an Gaussian integers is i. This number is the key to the next lemma which plays important role in several following theorems. The proof is straightforwar. Lemma.8. (a, b, c) PT if an only if (c, bi, a) PT. The notation a b will be use when a an b are associates. Note that if, then 1, δ or δ. Proposition.9. For each primitive triple (a, b, c) in PT, either a, b or c is a multiple of δ 3.

4 58 C. Somboonkulavui an A. Harnchoowong Proof. By Lemma.7, only one of a, b, c is even an the others are o. If a is even, by Corollary.4, there exist Gaussian integers u, v, where an u ± v such that a = uv, b = u v, an c = u + v. Case1 : u is even an v is o. Then u v is o. Since b is o, we have 1. Hence a uv an thus a is ivisible by δ 3. Case : u is o an v is even. This is similar to the above case. Case3 : u an v are o. Both u v an u + v are ivisible by δ. Therefore, u v. Since b is o, it follows that an a uv. Hence a is o, a contraiction. Case4 : u an v are even. If δ u or δ v, then a is ivisible by δ 3 an we are one. Suppose that δ u an δ v. Thus u = δu 1 an v = δv 1 where u 1,v 1 are o Gaussian integers. Since b =(u v )/ = δ (u 1 v 1 )/ an u 1 v1 is even, b is even. This is a contraiction. For the case that b is even, we can prove in a similar way. When c is even by Lemma.8, (c, bi, a) PT an the above proof shows that c is ivisible by δ 3. From Proposition.9 an Lemma.7, there are no Gaussian integers b 1,b,c 1,c such that (δ, b 1,c 1 ) an (,b,c ) are primitive. However, every o prime appears in specific forms of primitive Pythagorean triples. Proposition.10. Let p be an o prime in the Gaussian integers (i.e., p δ). If p occurs as a component of a primitive Pythagorean triple in PT, then it must be one of the following forms: (i) (p, ± p 1 (ii) (± p 1 (iii) (± p +1, ± p +1,p,± p +1, ± p 1 ) an (p, ± p +1 ) an (± p +1 i, p) an (± p 1 i, ± p 1 i) i, p, ± p 1 i) i, ± p +1,p). Proof. It is easy to verify that each triple liste is an element in PT. First we will show that case (i) is the only way in which p can occur as the first component of a primitive Pythagorean triple. Let (p, b, c) PT. By Corollary.4, there exist Gaussian integers u, v, where an u ± v such that p = uv, b = u v, an c = u + v. Since p is o, p /. Therefore, p u or p v. Ifp u, then v 1 an follows from p =uv/. It can be seen that there exist exactly 16 combinations

5 Pythagorean triples over Gaussian integers 59 that satisfy the conitions that p u an v 1. Upon substituting each of these combinations into the formulas for b an c, we obtain 4 possible forms as follows: (p, (p 1)/, (p +1)/), (p, (p 1)/, (p +1)/), (p, (p +1)i/, (p 1)i/) an (p, (p +1)i/, (p 1)i/). If p v, then u 1. Substituting the 16 combinations that satisfy the conition into the formulas given in Corollary.4, we obtain 4 formulas where each of the mile components has the ifferent sign from the 4 previous formulas. Case (ii) can be prove similarly an case (iii) follows from Lemma.8. Since each Gaussian integer has the unique factorization up to units, this fact effects the unique factorization of each Pythagorean triple. We then introuce units an irreucible elements in PT. Definition.11. (a, b, c) PT is calle a unit if there exists (, e, f) PT such that (a, b, c) (, e, f) =(1, 0, 1). Lemma.1. All units in PT are (±1, 0, ±1), (±1, ±i, 0), (±i, 0, ±i) an (±i, ±1, 0). Proof. If (1,b,c) PT, then there exist u, v, Z[i] where an u ±v such that 1= uv, b = u v, an c = u + v by Corollary.4. This implies that, u 1, v 1 an all triples satisfying these conitions are (1, 0, ±1) an (1, ±i, 0). Since the first component of a unit in PT must associate 1, we are one. Definition.13. Let (a, b, c), (, e, f) PT. If there exists a unit (x, y, z) PT such that (a, b, c) = (, e, f) (x, y, z), we say that (a, b, c) associates (, e, f) enote by (a, b, c) (, e, f). For example, (3, 4, 5) (3, 5i, 4i) since (3, 5i, 4i) =(3, 4, 5) (1,i,0). Definition.14. A non-unit (a, b, c) PT is sai to be irreucible provie that: whenever (a, b, c) =(u, v, w) (x, y, z) we will have (u, v, w) or (x, y, z) is a unit. For example, (1 + i, +i, 1+i) is irreucible but (1, 5, 13) = (3, 4, 5) (4, 3, 5) is not. Furthermore, every triple in case (i) of Proposition.10 is irreucible because the prime p cannot be factore. Proposition.15. For each positive integer k 3, δ k occurs as the first component of a primitive Pythagorean triple in PT as follows an in no other way: (δ k, ±(δ k 4 +1), ±(δ k 4 1)) an (δ k, ±(δ k 4 1)i, ±(δ k 4 +1)i). Moreover, these triples are irreucible. Proof. Let (δ k,b,c) PT be primitive. By Lemma.7, b an c must be o. Then δ k = uv, b = u v, an c = u + v

6 60 C. Somboonkulavui an A. Harnchoowong for some Gaussian integers u, v, where an u ± v by Corollary.4. Case1 : u is even an v is o. Then u v is o. Since b is o, 1 an δ k uv. Hence v 1 an u δ k. These conitions give rise to 4 possible forms as follows: (δ k,δ k 4 +1, δ k 4 1), (δ k, (δ k 4 + 1), (δ k 4 1)), (δ k, (δ k 4 1)i, (δ k 4 +1)i) an (δ k, (δ k 4 1)i, (δ k 4 +1)i). Case : u is o an v is even. Similarly, 1, u 1 an v δ k. We obtain another 4 possible forms where the mile components have ifferent signs from the previous case: (δ k, (δ k 4 +1),δ k 4 1), (δ k,δ k 4 +1, (δ k 4 1)), (δ k, (δ k 4 1)i, (δ k 4 +1)i) an (δ k, (δ k 4 1)i, (δ k 4 +1)i). Case3 : u an v are o. Since δ k =uv/ an k 3, this is a contraiction. Case4 : u an v are even. Since b =(u v )/ is o, an u, v cannot be both ivisible by δ.ifu δ, then v δ k 1 an the result is the same as in case. For v δ, we have u δ k 1 an the result is the same as in case1. Now suppose that (δ k,b,c)=(δ i,b 1,c 1 ) (δ j,b,c ) where b 1,b,c 1,c Z[i] an i, j N. Since (δ k,b,c) is primitive, (δ i,b 1,c 1 ) an (δ j,b,c ) are primitive. By Lemma.7, b 1,b,c 1,c are o. Then c = b 1 b + c 1 c is even, a contraiction. Hence (δ k,b,c) is irreucible. The unique factorization of any primitive Pythagorean triple (a, b, c) is reflecte by the unique factorization of a, its first component. The following theorem shows how a primitive Pythagorean triple can be factore into a prouct of irreucible triples. We use the usual integer-exponent notation. For example, A = A A for A PT. Proposition.16. (Unique factorization theorem) Let A =(a, b, c) PT be primitive an a = δ s 0 p s ps k k, where p i are istinct o primes, s i are non-negative integers an s 0 1,. Then A has the unique (up to orer of factors an the multiplication of factors by units) factorization where P 0 A = P 0 P s P s k k { (1, 0, 1) if a is o, (δ s 0, ±(δ s0 4 +1),δ s0 4 1) if a is even an P i (p i, ± p i 1, p i +1 ) for i 1. The choice of ± epening on (a, b, c). Proof. There exist Gaussian integers u, v, where an u ± v such that a = uv, b = u v, an c = u + v.

7 Pythagorean triples over Gaussian integers 61 If a is o, then. We obtain (a, b, c) (uv, u v, u + v )=(u, u 1, u +1 ) (v, 1 v, 1+v ) where the two triples on the right-han sie are elements in PT. Mathematical inuction implies the factorization in this case. In case that a is even, b an c are o by Lemma.7. The parity of u an v can be ivie into 4 cases as follows: Case1 : u is even an v is o. Since b =(u v )/ is o, 1. Let u = δ k n where n is an o Gaussian integer an k N. Then (a, b, c) (δ k, (δ k 4 +1), (δ k 4 1))*(nv, (n v )/, (n + v )/). Since n an v are o, (nv, (n v )/, (n + v )/) PT is primitive. We then factor (nv, (n v )/, (n + v )/) as in the o case. Case : u is o an v is even. This is similar to the above case. Case3 : u an v are o. Then δ 3 oes not ivie a. This is a contraiction. Case4 : u an v are even. Thus u = δm an v = δn for some m, n Z[i]. Since b =(u v )/ =(δ m δ n )/ is o, we have an b m n. This means that m an n must have the ifferent parity. Then (a, b, c) (mn, m n,m + n ) which can be factore as in case1 or case. From the property that (x, y, z) (x, y, z) =(x, 0,x ) for all (x, y, z) PT, the choice ± of the term P s i i of A cannot vary, otherwise A woul not be primitive. Since a etermines the first components of all factors of A, we assume that A = P 0 P s P s k k = Q 0 Q s Q s k k where for each i, P i an Q i are irreucible triples with ientical first components. Now if P =(x, y, z), we efine P =(x, y, z). If P j Q j for some j, it can be cancelle by multiplying P j on both sies of the equation. Repeating this process until we have P x0 P x1... P xm = Q x0 Q x1... Q xm which is a factor of A an P xi oes not associate Q xi. Propositions.10 an.15 an Lemma.1 show that P xi Q x i. Then, by multiplying the above equation by each of the Q x i, we have (P x0 P x1... P xm ) =(r, 0,r ) where the Gaussian integer r is the prouct of the first components of P xi.it follows that P x0 P x1... P xm (r, 0,r) which contraicts primitivity. This completes the proof. Observe that (l, 0,l) (a, b, c) =(la, lb, lc). We will use the notation l(a, b, c) for (l, 0,l) (a, b, c) in the next proposition which inicates how (δ k, ±(δ k 4 + 1),δ k 4 1) can be generate from (δ 3, ±(δ +1),δ 1).

8 6 C. Somboonkulavui an A. Harnchoowong Proposition.17. If k 3 is an integer, then δ k 6 (δ k,δ k 4 +1,δ k 4 1) (δ 3,δ +1,δ 1) k an δ k 6 (δ k, (δ k 4 +1),δ k 4 1) (δ 3, (δ + 1),δ 1) k. Proof. It is trivial when k = 3. For k > 3, δ (δ k,δ k 4 +1,δ k 4 1) (δ 3,δ +1,δ 1) (δ k 1,δ k 6 +1,δ k 6 1). Mathematical inuction gives the esire result. When mile components have ifferent signs the proof is similar. Example.18. For the primitive triple (96 + 7i, i, i) an i = δ 6 3 (1 + i), Proposition.16 provies (96 + 7i, i, i) = (δ 6,δ 8 +1,δ 8 1) (3, 5i, 4i) (1 + i, i, 1 +i). By Proposition.17, δ 6 (96 + 7i, i, i) can be written as (1, i, 0) (δ 3,δ +1,δ 1) 4 (3, 5i, 4i) (1 + i, i, 1+i). 3 The Group of Primitive Pythagorean Triples Over Gaussian Integers Definition 3.1. Let (a, b, c), (, e, f) be Pythagorean triples in Z[i]. We say that (a, b, c) is equivalent to (, e, f) if there exists a nonzero element k Q[i] such that (a, b, c) =(k, ke, kf). Denote the equivalence class of (a, b, c) by [a, b, c]. Since Z[i] is a UFD, the set of all equivalence classes of Pythagorean triples may be consiere as the set of all primitive Pythagorean triples. For this reason, let PPT = {[a, b, c] a, b, c Z[i] with a 0;a + b = c } be the set of all equivalence classes of Pythagorean triples in Z[i] where first components are not zero. Define operation * as in (1). Note that the set of all Pythagorean triples, PT, with the operation * is just a commutative monoi whereas the set of all equivalence classes, PPT, uner * is an abelian group. Proposition 3.. (PPT,*) is an abelian group. The ientity element in PPT is [1, 0, 1], an the inverse of [a, b, c] is [a, b, c]. Next we investigate a free abelian group, making use of the subgroup H := {[1, 0, 1], [1, 0, 1], [1,i,0], [1, i, 0]} of PPT. Proposition.16,.17 an 3. give the following corollary. Corollary 3.3. (PPT/H,*) is a free abelian group which is generate by the set of [a, b, c]h with a = δ 3 or a is an o prime.

9 Pythagorean triples over Gaussian integers 63 We establishs an isomorphism between PPT an the multiplicative group of the quotient fiel Q[i] of Z[i]. Proposition 3.4. (PPT,*) is isomorphic to (Q[i], ). Proof. Define ϕ : (PPT,*) (Q[i], ) byϕ([a, b, c]) = (b + c)/a. It is clear that ϕ is well-efine. Let [a 1,b 1,c 1 ], [a,b,c ] PPT. Then ϕ([a 1,b 1,c 1 ] [a,b,c ]) = ϕ([a 1 a,b 1 c + b c 1,b 1 b +c 1 c ]) =(b 1 b +c 1 c +b 1 c +b c 1 )/a 1 a =((b 1 +c 1 )/a 1 ) ((b +c )/a ) =ϕ([a 1,b 1,c 1 ]) ϕ([a,b,c ]). To show that ϕ is injective, let [a, b, c] PPT be such that ϕ([a, b, c]) = 1. Hence (b + c)/a = 1, i.e., a = b + c. Since a + b = c, we obtain b(b + c) =0. Then b =0orb + c = 0. But b + c = a which is not 0, then b = 0 an a = c. Therefore, [a, b, c] =[a, 0,a]=[1, 0, 1] as esire. Now let u/v Q[i] where u, v Z[i] \{0}. Choose a =uv, b = u v,c= u + v Z[i]. Then ϕ([a, b, c]) = (b + c)/a =(u + v + u v )/uv = u/v. This implies that ϕ is an isomorphism. In orer to make PPT a fiel, we a [0, 1, 1] into PPT an efine operation aition by using the isomorphism ϕ between (PPT,*) an (Q[i], ). The mapping φ : PPT {[0, 1, 1]} Q[i] given by { ϕ([a, b, c]) if [a, b, c] PPT, φ([a, b, c]) = 0 if [a, b, c] =[0, 1, 1] is both injective an surjective. Define operation on PPT {[0, 1, 1]} by [a 1,b 1,c 1 ] [a,b,c ]=φ 1 (φ([a 1,b 1,c 1 ]) + φ([a,b,c ])). It is not ifficult to show the following proposition. Proposition 3.5. (PPT {[0, 1, 1]},, *) is a fiel. The set of Pythagorean triples with operation + efine by (a 1,b 1,c 1 )+(a,b,c )=(a 1 a b 1 b,a 1 b + b 1 a,c 1 c ) (4) was also stuie in term of its structure. In [6], P. Zanaro an U. Zannier escribe on a ring of integers R such that i/ R. In our case where i Z[i], let PPT = {[a, b, c] a, b, c Z[i] with c 0;a + b = c }. With operation + in (4), PPT can be mae into a group. The next proposition will show that (PPT,*) an (PPT, +) are isomorphic. Hence (PPT, +) is isomorphic to (Q[i], ) as well. Proposition 3.6. (PPT,*) is isomorphic to (PPT, +).

10 64 C. Somboonkulavui an A. Harnchoowong Proof. We nee the fact that [a, b, c] PPT if an only if [c, bi, a] PPT. Define λ : (PPT,*) (PPT, +) by λ([a, b, c]) = [c, bi, a]. Let [a 1,b 1,c 1 ], [a,b,c ] PPT. Then λ([a 1,b 1,c 1 ] [a,b,c ]) = λ([a 1 a,b 1 c + b c 1,b 1 b +c 1 c ]) =[b 1 b +c 1 c, (b 1 c +b c 1 )i, a 1 a ]=[c 1 c b 1 ib i, c 1 b i+c b 1 i, a 1 a ] =[c 1,b 1 i, a 1 ]+[c,b i, a ]=λ([a 1,b 1,c 1 ]) + λ([a,b,c ]). The rest of the proof comes from the above fact. Let us remark that Definition 3.1, Proposition 3., 3.4 an 3.5 can be generalize to any ring of integers through similar proofs. Proposition 3.6 is also true for the ring of integers R in case that i R. References [1] Raymon A. Beauregar an E. R. Suryanarayan, Pythagorean Triples: The Hyperbolic View, The College Mathematics Journal, 7 (May, 1996), pp [] James T. Cross, Primitive Pythagorean Triples of Gaussian Integers, Mathematics Magazine, 59 (Apr., 1986), pp [3] Ernest J. Eckert, The Group of Primitive Pythagorean Triangles, Mathematics Magazine, 57 (Jan., 1984), pp. -7. [4] K. K. Kubota, Pythagorean Triples in Unique Factorization Domains, The American Mathematical Monthly, 79 (May, 197), pp [5] N. E. Sexauer, Pythagorean triples over gaussian omains, The American Mathematical Monthly, 73 (Oct., 1966), pp [6] P. Zanaro an U. Zannier, The group of pythagorean triples in number fiels, Annali i Matematica pura e applicata (IV), CLIX (1991), pp Receive: August, 011