Malaysian Journal of Mathematical Sciences 10(S February: 15-35 (016 Secial Issue: The 3 rd International Conference on Mathematical Alications in Engineering 014 (ICMAE 14 MALAYSIAN JOURNAL OF MATHEMATICAL SCIENCES Journal homeage: htt://einsem.um.edu.my/journal Solvability and Number of Roots of Bi-Quadratic Equations over adic Fields Saburov, M. and Ahmad, M. A. K. Deartment of Comutational & Theoretical Sciences Kulliyyah of Science, International Islamic University Malaysia E-mail: msaburov@gmail.com Corresonding author ABSTRACT Unlike the real number field R, a bi-quadratic equation x 4 + 1 = 0 is solvable over some adic number fields Q, say = 17, 41,. Therefore, it is of indeendent interest to rovide a solvability criterion for the bi-quadratic equation over adic number fields Q. In this aer, we rovide solvability criteria for the bi-quadratic equation x 4 + ax = b over domains Z, Z \ Z, Q \ Z, Q, where >. Moreover, we also rovide the number of roots of the bi-quadratic equation over the mentioned domains. Keywords: Bi-quadratic equation, adic number, solvability criterion.
Saburov, M. and Ahmad, M. A. K. 1. Introduction The field Q of adic numbers which was introduced by German mathematician K. Hensel was motivated rimarily by an attemt to bring the ideas and techniques of the ower series into number theory. Their canonical reresentation is analogous to the exansion of analytic functions into ower series. This is one of the manifestations of the analogy between algebraic numbers and algebraic functions. For a fixed rime, it is denoted the field of adic numbers by Q which is a comletion of the rational numbers Q with resect to the non-archimedean norm : Q R given by x = { k x 0, 0, x = 0, (1 here, x = k m n with r, m Z, n N, (m, = (n, = 1. A number k is called a order of x and it is denoted by ord (x = k. Any adic number x Q can be uniquely reresented in the following canonical form (Borevich and Shafarevich, 1986 x = ord(x ( x 0 + x 1 + x + where x 0 {1,, 1} and x i {0, 1,, 1}, i 1, We resectively denote the set of all adic integers and units of Q by Z = {x Q : x 1}, Z = {x Q : x = 1}. Any adic unit x Z has the following unique canonical form x = x 0 + x 1 + x + where x 0 {1,, 1} and x i {0, 1,, 1}, i N. Any nonzero x Q has a unique reresentation x = x x, where x Z. A number a 0 Z is called an r th ower residue modulo if the following congruent equation x r a 0 (mod ( is solvable in Z. 16 Malaysian Journal of Mathematical Sciences
Solvability and Number of Roots of Bi-Quadratic Equations Proosition 1.1 (Rosen (011. Let be an odd rime, a 0 Z, with (a 0, = 1 and d = (r, 1. The following statements hold true: 1. a 0 is the r th ower residue modulo iff a 1 d 1 (mod.. If a 1 d 0 1 (mod then the congruent equation ( has d number of distinct (non-congruent solutions in Z. Throughout this aer, we always assume that > is an odd rime unless otherwise mentioned. Let us consider the following quadratic equation x a 0 (mod (3 Due to Proosition 1.1, the congruent equation (3 is solvable iff a 1 0 1 (mod and it has two distinct non-congruent solutions in the set {1,,, 1}. It is clear that one solution of the congruent equation (3 is less than and another solution is greater than. Definition 1.1. We denote by a 0 (res. a 0 the solution of quadratic congruent equation (3 which is less than (res. greater than. Remark 1.1. Due to the definition, a 0 exists if and only if a 1 0 1 (mod. Moreover, a 0 and a 0 {1,,, 1}. Let us now consider the following quadratic equation over Q x = a (4 where a Q is a nonzero adic number. Let a = a a with a = a 0 + a 1 + a + such that a 0 {1,,, 1}, a i {0, 1,,, 1}, i N. We know that the quadratic equation (4 is solvable in Q iff a 1 0 1 (mod and log a is even. Moreover, it has two distinct solutions x + and x such that x + a 0 (mod and x a 0 (mod. Malaysian Journal of Mathematical Sciences 17
Saburov, M. and Ahmad, M. A. K. Definition 1.. We denote the solution x + (res. x of the quadratic equation equation (4 by a (res. a. Remark 1.. By the definition, for the given nonzero a Q, a exists if and only if a 1 0 1 (mod and log a is even. Moreover, a is a solution of the quadratic equation (4 such that ( a a 0 (mod and a is a solution of the quadratic equation (4 such that ( a a 0 (mod. Unlike the real number field R, the bi-quadratic equation x 4 + 1 = 0 is solvable over some adic number fields Q such as = 17, 41,. Therefore, it is of indeendent interest to rovide a solvability criterion of a bi-quadratic equation x 4 + ax = b over adic number fields Q. In this aer, we rovide solvability criteria and the number of roots of the bi-quadratic equations x 4 + ax = b over the domains Z, Z \ Z, Q \ Z, Q. It is worth of mentioning that the similar roblems for cubic equations were studied in refs. (see Mukhamedov and Saburov (013, Mukhamedov et al. (013, Mukhamedov et al. (014, Saburov and Ahmad (014, and Saburov and Ahmad (015. Alications of those results were demonstrated in refs. (see Mukhamedov and Akin (013a and Mukhamedov and Akin (013b.. The Main Strategies Obviously, we can aly the substitution method for the bi-quadratic equation. However, this method is not efficient. Let us exlain it in a detail. We consider the bi-quadratic equation where a, b Q. x 4 + ax = b. (5 If we let y = x then we may have the following quadratic equation y +ay = b. In its own turns, in order to get a solvability criterion for the bi-quadratic equation, we have to solve the following quadratic equations x = a± a +4b. To check whether one of the last two quadratic equations has a root or not, we have to verify that log a± a +4b is even number and the first adic ( a± digit of the adic unit a +4b is a quadratic residue. This is a tedious work. However, our aim is to rovide a solvability criterion for the bi-quadratic equation in terms of a, b. Therefore, we suggest another aroach to fulfill our aim. 18 Malaysian Journal of Mathematical Sciences
Solvability and Number of Roots of Bi-Quadratic Equations Remark.1. It is worth of mentioning that the solvability of a bi-quadratic equation is comletely different from the solvability of a quadratic equation obtained by the substitution method. For examle, this quadratic equation y y + = 0 is solvable, i.e., y = is a unique solution. However, this bi-quadratic equation x 4 x + = 0 is not solvable because of the fact that does not exist. We know that, by definition, two adic numbers are close when their difference is divisible by a high ower of. This roerty enables adic numbers to encode congruence information in a way that turns out to be owerful tools in the theory of olynomial equation. In fact, Hensel s lifting lemma allows us to lift a simle solution of a olynomial equation over the finite field F u to the unique solution of the same olynomial equation over the ring Z of adic integer numbers. However, that solution cannot be any more lifted u to the field Q of adic numbers. At this oint, we are aiming to study the relation between solutions of the olynomial equations over Q and Z. We shall show that, indeed, any solution of any bi-quadratic equation over Q (or some secial sets can be uniquely determined by a solution of another bi-quadratic equation over Z. Consequently, in some sense, it is enough to study bi-quadratic equations over Z. Whenever a, b 0, let a = a a and b = b b where a Z, b Z with a = a 0 + a 1 + a + a 3 3 +, b = b 0 + b 1 + b + b 3 3 + Remark.. We use the notation r a whenever the monomial equation x r = a is solvable in Q. The solvability criterion for the last monomial equation was given in ref. Mukhamedov and Saburov (013. Namely, for >, there exists 4 a if and only if log a is divisible by 4 and a 1 d 0 1 (mod, where d = ( 1, 4, a = a 0 + a 1 +, and a = a a. If ab = 0 then the bi-quadratic equation (5 can be easily studied. Proosition.1. Let ab = 0. The following statements hold true: (i Let a = 0 = b. Then (5 has a solution x 0 = 0 of multilicity-4. (ii Let a 0 = b. If a then (5 has solutions x ± = ± a and x 0 = 0 of multilicity-. Otherwise, (5 has only solution x 0 = 0 of multilicity-. Malaysian Journal of Mathematical Sciences 19
Saburov, M. and Ahmad, M. A. K. (iii Let a = 0 b. The bi-quadratic equation (5 is solvable over Q if and only if 4 b. In this case, if 1 (mod 4 then (5 has 4 distinct solutions and if 3 (mod 4 then (5 has distinct solutions. The roof is straightforward. What it follows, we always assume that ab 0. Let A Z be any subset. We introduce the following set Z A := { x Q : log x A }. It is easy to check that Z A = i A S i(0, where S i(0 = {x Q : x = i } is the shere with the radius i. Proosition.. Let be any rime, a, b Q with ab 0, and A Z be any subset. The bi-quadratic equation (5 is solvable in the set Z A iff there exists a air (y, k Z A such that y is a solution of the following bi-quadratic equation y 4 + A k y = B k (6 where A k = a k and B k = b 4k. Moreover, in this case, a solution of the bi-quadratic equation (5 has the form x = y k. Proof. Let x Q and x = k. Then x Z is a solution of the bi-quadratic A equation (5 if and only if y = x x Z is a solution of the bi-quadratic equation (6. This comletes the roof. Here, we list frequently used domains in this aer. 1. If A 1 = {0} then Z A1 = Z ; 0 Malaysian Journal of Mathematical Sciences
Solvability and Number of Roots of Bi-Quadratic Equations. If A = N (all negative natural numbers then Z A = Z \ Z ; 3. If A 3 = N then Z A3 = Q \ Z ; 4. If A 4 = Z then Z A4 = Q. Corollary.1. Let be any rime, a, b Q with ab 0, and A i Z be a subset given as above, i = 1, 4. The bi-quadratic equation (5 is solvable in the set Z Ai iff there exists (y, k Z A i such that y is a solution of the following bi-quadratic equation where A k = a k and B k = b 4k. y 4 + A k y = B k Consequently, it is enough to study solvability of the bi-quadratic equation (5 over Z, where a, b Q with ab 0. Proosition.3. Let be any rime and a, b Q with ab 0. If the bi-quadratic equation (5 is solvable in Z then either one of the following conditions holds true: (i a = b 1; (ii b < a = 1; (iii a < b = 1. Proof. Let x Z be a solution of (5. Since ab 0, one can get that b = x 4 + ax max{1, a }, a = ax = b x 4 max{1, b }, 1 = x 4 = b ax max{ a, b }. Thus, if a b then max{ a, b } = 1 and if a = b then a = b 1. This yields the claim. Malaysian Journal of Mathematical Sciences 1
Saburov, M. and Ahmad, M. A. K. This roosition gives necessary conditions for solvability of the bi-quadratic equation over Z. To get the solvability criteria, we need Hensel s lifting lemma. Lemma.1 (Hensel s Lemma. Let f(x be olynomial whose coefficients are adic integers. Let θ be a adic integer such that for some i 0 we have f(θ 0 (mod i+1, f (θ 0 (mod i, f (θ 0 (mod i+1. Then f(x has a unique adic integer root x 0 which satisfies x 0 θ (mod i+1. 3. The Solvability Criteria Throughout this section we always assume that >. We resent a solvability criterion of the bi-quadratic equation over A x 4 + ax = b (7 where A { Z, Z \ Z, Q \ Z, Q }. Let a, b Q with ab 0 and D = a + 4b. We then have that a = a a, b = b b and D = D D whenever D 0, where a = a 0 + a 1 + a + b = b 0 + b 1 + b + D = d 0 + d 1 + d + where a 0, b 0, d 0 {1,, 1} and a i, b i, d i {0, 1,, 1} for any i N. Throughout this aer, the notation (a b means that there exists either a or b, the notation (a b means that there exists only a or b, and the notation (a b means that there exists both a and b. In this case, it is clear that {(a b } = {(a b } {(a b }. 3.1 The Solvability Criterion over Z Theorem 3.1. The bi-quadratic equation (7 is solvable in Z if and only if either one of the following conditions holds true: Malaysian Journal of Mathematical Sciences
Solvability and Number of Roots of Bi-Quadratic Equations 1. a < b = 1,. b < a = 1, 4 b ; a ; 3. a = b > 1, ab ; ( D 4. a = b = 1 D, a ; 5. a = b = 1 = D, D, ( a+ D a D Proof. Due to Proosition.3, if the bi-quadratic equation (7 is solvable in Z then either one of the following conditions must be satisfied: a = b 1 or b < a = 1 or a < b = 1. We shall study case by case. 1. Let a < b = 1. In this case, we want to show that the bi-quadratic equation (7 is solvable in Z if and only if there exists b or equivalently 1 (4, 1 b0 1 (mod. Only if art. Let x Z be a solution of the bi-quadratic equation (7. Since a < 1, it yields that x 4 b (mod or x 4 0 b 0 (mod. This means (4, 1 that b 0 is the forth ower residue modulo, i.e., b0 1 (mod. 1 (4, 1 If art. Let b0 1 (mod. Then there exists x {1,, 1} such that x 4 b 0 (mod. Let us consider the following bi-quadratic function f a,b (x = x 4 + ax b. It is clear that f a,b ( x = x 4 + a x b x 4 b x 4 b 0 0 (mod, f a,b( x = 4 x 3 + a x 4 x 3 0 (mod. Then due to Hensel s Lemma, there exists x Z such that f a,b (x = 0 and x x (mod. Since x 0 (mod, we have that x Z.. Let b < a = 1. We show that the bi-quadratic equation (7 is solvable in Z if and only if there exists a or equivalently ( a 0 1 1 (mod. Only if art. Let x Z be a solution of the bi-quadratic equation (7. Since b < 1, it yields x 0(x 0 + a 0 0 (mod. We have that x 0 a 0 (mod. So, a 0 is the quadratic residue modulo, i.e, ( a 0 1 1 (mod. 1 Malaysian Journal of Mathematical Sciences 3
Saburov, M. and Ahmad, M. A. K. If art. Let ( a 0 1 1 (mod, so there exist x a 0 (mod where ( x, = 1. We consider the following bi-quadratic function f a,b (x = x 4 + ax b. It is clear that f a,b ( x = x 4 + a x b x ( x + a a 0 a 0 0 (mod, f a,b( x = 4 x 3 + a x x 3 + x( x + a x 3 0 (mod. Then due to Hensel s Lemma, there exists x Z such that f a,b (x = 0 and x x (mod. Since x 0 (mod, we have that x Z. 3. Let a = b > 1. We show that the bi-quadratic equation (7 is solvable in Z if and only if there exists ab or equivalently (a 0 b 0 1 1 (mod (because of ab = a. Since a = b = k for some k N, the solvability of the following two bi-quadratic equations are equivalent x 4 + ax = b, k x 4 + a x = b. (8 Moreover, any solution of the first bi-quadratic equation is a solution of the second one and vise versa. On the other hand, the second bi-quadratic equation is suitable to aly Hensel s lemma. Therefore, we study the second bi-quadratic equation instead of the first one. Only if art. Let x Z be a solution of the bi-quadratic equation k x 4 + a x = b. We then have that k x 4 + a x b (mod a 0 x 0 b 0 (mod (a 0 x 0 a 0 b 0 (mod. This means that a 0 b 0 is a quadratic residue modulo or (a 0 b 0 1 1 (mod. If art. Let (a 0 b 0 1 1 (mod. Then there exists (a 0 x a 0 b 0 (mod where ( x, = 1. We consider the following bi-quadratic function g a,b (x = k x 4 + a x b. We get that g a,b ( x = k x 4 + a x b a 0 x b 0 0 (mod, g a,b( x = 4 k x 3 + a x a 0 x 0 (mod. Then due to Hensel s Lemma, there exists x Z such that g a,b (x = 0 and x x (mod. Since x 0 (mod, we have that x Z. This shows that the bi-quadratic equation (7 is also solvable in Z. Now let us consider the bi-quadratic equation (7 in the form of (x + a = D. (9 4 Malaysian Journal of Mathematical Sciences
Solvability and Number of Roots of Bi-Quadratic Equations It is clear that if a = b = 1 then D = a + 4b 1. 4. We consider the case a = b = 1 D. Let D = 0. We have that x + a = 0. The last equation is solvable over Q if and only if there exists a. Since a = 1, we have that log a = 0 is even. In this case, it can be easily checked that the solution x of the biquadratic equation (7 is in Z. Therefore, if a = b = 1 and D = 0 then the bi-quadratic equation (7 is solvable in Z if and only if ( a 0 1 1 (mod. Let 0 < D < 1. Let y = x + a. The bi-quadratic equation (9 can be reduced to the quadratic equation y = D. The last quadratic equation is solvable in Q iff log D is even and d 1 0 1 (mod or equivalently there exists D. If we let log D = k and d 1 0 1 (mod where k > 0 then we may have the following quadratic equation x = a ± D = a ± k D. (10 The last quadratic equation (10 is solvable over Q iff log a± D and [ ( a 0 ± k d 0 ] 1 1 (mod. Let us find the value of a± D where D = a + 4b. We know that D a = D a D + a = 4b = 1 is even Since D ± a 1, due to the revious equality, we have that D ± a = 1. a± Therefore, log D = 0 is always even. Since k > 0, we have that [ ( a 0 ± k d 0 ] 1 ( a 0 1 1 (mod. This is equivalent to say that there exists a. In this case, we can easily checked that the solution x of the bi-quadratic equation (7 is in Z. Therefore, if a = b = 1 D and there exists D then the bi-quadratic equation (7 is solvable in Z if and only if there exists a. 5. We consider the case a = b = 1 = D. Malaysian Journal of Mathematical Sciences 5
Saburov, M. and Ahmad, M. A. K. Then we have that [ ( d 0 a 0 ] 1 1 (mod or [ ( d 0 a 0 ] 1 1 (mod. We can easily checked that the solution x of the bi-quadratic equation (7 is in Z. Therefore, if a = b = 1 = D and there exists D then the bi-quadratic equation (7 is solvable in Z if and only if [ ( d0 a 0 ] 1 1 (mod or [ ( d 0 a 0 ] 1 1 (mod (this is a+ equivalently to say that there exists D a or D. This comletes the roof. 3. The Solvability Criterion over Z \ Z Theorem 3.. The bi-quadratic equation (7 is solvable in Z \ Z if and only if either one of the following conditions holds true: 1. a < b < 1,. b < a < 1, 4 b ; a ; 3. a > b, a > b, 4. D < a = b < 1, 5. D = a = b < 1, ab ; ( D a ; D, ( a+ D a D Proof. Let x Q be a nonzero adic number and x = k where k Z. Due to Corollary.1, x is a solution of the bi-quadratic equation (7 in Z \ Z if and only if y = k x is a solution of the following bi-quadratic equation in Z for some k N where A k = a k, B k = b 4k. y 4 + A k y = B k (11 It is clear that A k = a, B k = b and A k = k a, B k = 4k b. Let D = a + 4b and D k = A k + 4B k. Then D k = 4k D and D k = D, D k = 4k D whenever D k 0 (or equivalently D 0. We know that, due to Theorem 3.1, the bi-quadratic equation (11 is solvable in Z if and only if either one of the following conditions holds true: I. A k < B k = 1, 4 B k ; 6 Malaysian Journal of Mathematical Sciences
Solvability and Number of Roots of Bi-Quadratic Equations II. B k < A k = 1, III. A k = B k > 1, Ak ; Ak B k ; IV. A k = B k = 1 D k, Dk, Ak ; V. A k = B k = 1 = D k, ( A D k, k + D k A k D k We want to describe all adic numbers a, b Q for which at least one of the conditions I V should be satisfied for some k N. 1 1. Let us consider the condition I: A k < B k = 1 and 4 B k (or equiva- (4, 1 lently b0 1 (mod. We have that log 4k b = 0. Hence, we obtain that 4k = log b. It is clear that k N if and only if log b is divisible by 4 and b < 1. Moreover, one has that A k = k a < 1, where k = 1 4 log b, if and only if a < b. Therefore, if a (4, 1 < b < 1, 4 log b and b0 1 (mod (or equivalently a < b < 1 and 4 b then the condition I is satisfied with k = 1 4 log b N.. Let us consider the condition II: B k < A k = 1 and A k (or equivalently ( a 0 1 1 (mod. We have that A k = 1 if and only if k = log a. It is clear that k N if and only if log a is even and a < 1. Besides that, we have that B k < 1, where k = 1 log a, if only if b < a. Hence, if b < a < 1, log a and ( a 0 1 1 (mod (or equivalently b < a < 1 and a then the condition II is satisfied with k = 1 log a N. 3. Let us consider the condition III: A k = B k > 1 and A k B k ; (or equivalently (a 0 b 0 1 1(mod. We have that log ( 4k b = log ( k a > 0. Hence, we obtain that b k = log b log a. It is clear that k N if and only if log a (equivalently log ab is even and b < a. Moreover, one has that log ( k a > 0, where k = 1 log b a, if and only if a > b. Therefore, if a > b, a > b, and ab then the condition III is satisfied with k = 1 log b a N. 1 Malaysian Journal of Mathematical Sciences 7
Saburov, M. and Ahmad, M. A. K. 4. Let us consider the condition IV: A k = B k = 1 D k, D k, and Ak. We have that A k = 1 and B k = 1, simultaneously, if and only if k = log a and 4k = log b. This means that a = b and log a is even. It is clear that k N if and only if a < 1, b < 1, log a is even and a = b. From here, if log a is even then log b is also divisible by 4. We know that log D k = 4k + log D. So, log D k is even if and only if log D is even. Besides that, we have that D k < 1 if and only if D < a = b, where k = 1 log a = 1 4 log b. Therefore, if D < a = b < 1, D, and a then the condition V is satisfied with k = 1 log a = 1 4 log b N. 5. Let us consider the condition V: A k = B k = 1 = D k, D k, ( A k + D k A k D k. We have that A k = 1 and B k = 1, simultaneously, if and only if k = log a and 4k = log b. This means that a = b and log a is even. It is clear that k N if and only if a < 1, b < 1, log a is even, and a = b. Besides that, we have that D k = 1 if and only if D = a = b, where k = 1 log a = 1 4 log b. From here, if log a is even then log b and ( A log D are also divisible by 4. In this case, k + D k A k D k ( a+ is equivalent to D a D. Therefore, if D = a = b < 1, ( a+ D, D a D then the condition V is satisfied with k = 1 log a = 1 4 log b N. This comletes the roof. 3.3 The Solvability Criteria over Q \ Z and Q The roof of the following theorem is similar to the roof of Theorem 3.. Theorem 3.3. The bi-quadratic equation (7 is (A Solvable in Q \ Z iff either one of the following conditions holds true: 1. a < b, b > 1,. a > b, a > 1, 4 b ; a ; 8 Malaysian Journal of Mathematical Sciences
Solvability and Number of Roots of Bi-Quadratic Equations 3. a > b, a < b, ab ; ( D 4. D < a = b, b > 1, a ; ( 5. D = a = b > 1, D, a+ D a D. (B Solvable in Q iff either one of the following conditions holds true: 1. a < b, 4 b ; ( a. a > b, ab ; 3. a = b > D, ( D a ; 4. a = b = D, ( D, a+ D a D. Remark 3.1. In the adic analysis, the field Q should be treated in a comletely different way from the field Q for >. In a forthcoming aer, we are aiming to study the bi-quadratic equation in Q. 4. The Number of Roots In this section, we resent the number N A (x 4 + ax b of roots (including multilicity of bi-quadratic equation where x 4 + ax = b (1 A { Z, Z \ Z, Q \ Z, Q }. Theorem 4.1. Let the bi-quadratic equation (1 be solvable in A where A { Z, Z \ Z, Q \ Z, Q }. Then the following statements hold true: = N Z (x 4 + ax b = 4, a < b = 1, 4 b, ( 1 (mod 4 D 4, a = b = 1 > D, a ( a+ D 4, a = b = 1 = D, D,, a < b = 1, 4 b, 3 (mod 4, b < a = 1, a, a = b > 1, ab, a = b = 1 = D, D, ( a+ D a D a D Malaysian Journal of Mathematical Sciences 9
Saburov, M. and Ahmad, M. A. K. = N Z\Z (x4 + ax b = 4, a < b < 1, 4 ( b, 1 (mod 4 a 4, b < a < 1, ab ( D 4, D < a = b < 1, a ( a+ D 4, D = a = b < 1, D,, a < b < 1, 4 ( b, 3 (mod 4 a, b < a < 1, ab, a > b, a 1, ab, D = a = b < 1, D, a D ( a+ D a D = N Q\Z (x 4 + ax b = 4, a < b, b > 1, 4 b (, 1 (mod 4 a 4, a > b, a < b, ab ( D 4, D < a = b, b > 1, a 4, D = a = b > 1, ( a+ D, D a D, a < b, b > 1, 4 b (, 3 (mod 4 a, a > b, a < b, ab, a b, a > 1, a, D = a = b > 1, D, ( a+ D a D = N Q (x 4 + ax b = 4, a < b, 4 ( b, 1 (mod 4 a 4, a > b, ab ( D 4, a = b > D, a 4, a = b = D, ( a+ D, D, a < b, 4 ( b, 3 (mod 4 a, a > b, ab, a = b = D, ( a+ D, D a D a D 30 Malaysian Journal of Mathematical Sciences
Solvability and Number of Roots of Bi-Quadratic Equations Proof. Let us discuss the case Z. As we refer to Theorm 3.1, the bi-quadratic equation (1 is solvable in Z if and only if one of the conditions 1 5 should be satisfied. We want to verify the number of solutions for every case. 1 Condition 1 : a < b = 1, 4 b. This means that b 4, 1 0 1 (mod. In this case, the number of solutions of bi-quadratic equation (1 is the same as the number of solutions of the equation x 4 0 b 0 (mod. If 1 (mod 4 then the last congruent equation has 4 solutions and if 3 (mod 4 then the last congruent equation has solutions. Therefore, if a < b = 1, there exists 4 b and 1 (mod 4 then the bi-quadratic equation (1 has 4 solutions in Z and if a < b = 1, there exists 4 b and 3 (mod 4 then the bi-quadratic equation (1 has solutions in Z. Condition : b < a = 1, a. In this case, the number of solutions of the bi-quadratic equation (1 is the same as the number of solutions of the congruent equation x 0 + a 0 0 (mod. The last equation has two solutions because of ( a 0 1 1 (mod. Therefore, if b < a = 1 and there exists a then the bi-quadratic equation (1 has two solutions in Z. Condition 3 : a = b > 1, ab. In this case, the number of solutions of the bi-quadratic equation (1 is the same as the number of solution of the equation a 0 x 0 b 0 (mod. The last congruent equation has two solutions because of (a 0 b 0 1 1 (mod. Therefore, if a = b > 1 and there exists ab then the bi-quadratic equation (1 has two solutions in Z. Now, let us consider the bi-quadratic (1 in the form of (x + a = D. (13 Condition 4 : a = b = 1 > D, D, a. Let D = 0. We have that (x + a = 0. It is clear that x + a = 0 and it has two solutions because of ( a 0 1 1 (mod. We can easily verify that these solutions are in Z. Therefore, if a = b = 1, D = 0 and there exists a then the bi-quadratic equation (1 has four solutions (two solutions of multilicity- in Z. Let 0 < D < 1. The bi-quadratic equation (13 has the same number of solutions as the total number of solutions of equations x + a = D and x + a = D. Each of the equations have two solutions because of ( a ± 1 D ( a 0 1 1 (mod. It is easily can be checked that these four Malaysian Journal of Mathematical Sciences 31
Saburov, M. and Ahmad, M. A. K. solutions are in Z. Hence, if a = b = 1 > D 0, there exist D and a then the bi-quadratic equation (1 has four solutions in Z. Condition 5 : a = b = 1 = D, a+ D a or D. ( a+ D D a D, i.e., there exist both and there exists either a+ D and Let a D. The bi-quadratic equation (13 has the same number of solutions as the total number of solutions of equations x +a = D and x +a = D. The total number of solutions of the last two equations is four because each of them has two solutions. We can easily verify that these solutions are in Z. Therefore, if a = b = 1 = D, there exist D and bi-quadratic equation (1 has for solutions in Z. Let ( a+ D a± D then the a D a+, i.e., there exists only one of D a and D. The bi-quadratic equation (13 has the same number of solutions as the total number of solutions of equations x + a = D and x + a = D. The total number of solutions of the last two equations is two because only one of them is solvable (each equation has two solutions if it is solvable. It is also can be checked that these solutions are in Z. Therefore, if a = b = 1 = D, there exists D and there exists only one of a+ D a and D then the bi-quadratic equation (1 has two solutions in Z. Let us turn to the case Z \ Z. Due to the Theorem 3., the bi-quadratic equation (1 is solvable in Z \ Z if and only if one of conditions 1 5 holds true. We want to find the number of solutions in every case. Condition 1 : a < b < 1 and 4 b. The number of solutions of the bi-quadratic equation (1 in Z \ Z is the same as the number of solutions of the following bi-quadratic equation in Z, y 4 + a b y = b. It is clear that a b < b = 1 and there exists 4 b or equivalently 1 (4, 1 b0 1 (mod. In this case, we have two distinct solutions in Z if 3 Malaysian Journal of Mathematical Sciences
Solvability and Number of Roots of Bi-Quadratic Equations 3 (mod 4 and four distinct solutions in Z if 1 (mod 4. Thus, if a < b < 1, there exists 4 b and 3 (mod 4 then the bi-quadratic equation (1 has two distinct solutions in Z \ Z and if a < b < 1, there exists 4 b and 1 (mod 4 then the bi-quadratic equation (1 has four distinct solutions in Z \ Z. Conditions 3 : b < a < 1, a and a > b, a > b, ab. Now, let us define the following sets A = { (a, b Q : b < a < 1, a }, { B = (a, b Q : a > b, a > b, } ab, ( a S = {(a, b Q : b < a < 1, ab }, { B 1 = (a, b Q : b < a < 1, } ab, B = {(a, b Q : a > b, a 1, ab }, ( a S 1 = {(a, b Q : b < a < 1, ab }, ( a S = {(a, b Q : b < a < 1, ab }. One can easily checked that B = B 1 B, S = S 1 S = A B 1 and A B = S B Firstly, let us consider the set S 1. In this case, the number of solutions of the bi-quadratic equation (1 in Z \ Z is the same as the total number of solutions of the following bi-quadratic equations in Z, y 4 + a y = a b (14 z 4 + a b a z = b b a (15 It is clear that a b < a = 1, a b = b ( b a > 1, ab. a a We have already discussed that each bi-quadratic ( equation given above has two a solution in Z. Thus, if b < a < 1 and ab then the biquadratic equation (1 has four distinct solutions in Z \ Z. Next, let us consider the set S. In this case, the number of solutions of the bi-quadratic equation (1 in Z \ Z is the same as the total number of solutions of the bi-quadratic equation (14 and (15 in Z. It is clear that Malaysian Journal of Mathematical Sciences 33
Saburov, M. and Ahmad, M. A. K. a b < a = 1, a b = b b > 1 and a a ( a ab. In this case, as we have already discussed, that only one ( of the equations (14 and a (15 is solvable. Therefore, if b < a < 1 and ab then the bi-quadratic equation (1 has two distinct solutions in Z \ Z. Lastly, let us consider the set B. In this case, the number of solutions of the bi-quadratic equation (1 in Z \Z is the same as the number of solutions of the bi-quadratic equation (15 Z. It is clear that a b = b b > 1 and ab. We have already discussed that bi-quadratic (15 equation has two solution in Z. Therefore, if a > b, a 1 and ab then the bi-quadratic equation (1 has two distinct solutions in Z \ Z. Condition 4 : D < a = b < 1, D and a. In this case, the number of solutions of the bi-quadratic equation (1 in Z \Z is the same as the number of solutions of the following bi-quadratic equation in Z, y 4 + a y = b. We can see that D < a = b = 1, D and a. Then, the last equation has four solutions in Z.Therefore, if D < a = b < 1, D and a then the bi-quadratic equation (1 has four solutions in Z \Z. It is worth of mentioning that if D = 0 then bi-quadratic equations has two solutions of multilicity- in Z \ Z. Condition 5 : D = a = b < 1, D, a a ( a+ D a D. In this case, the number of solutions of the bi-quadratic equation (1 in Z \Z is the same as the number of solutions of the following bi-quadratic equation in Z, y + a y = b. We can see clearly that D = a = b = 1 and D. Let ( a+ D solutions in Z. Let a D ( a+ D equation has two solutions in Z.. In this case, the last equation has four. In this case, the last a D Similarly, one can rove in the cases Q \ Z and Q. This comletes the roof. 34 Malaysian Journal of Mathematical Sciences
Solvability and Number of Roots of Bi-Quadratic Equations Acknowledgments This work has been suorted by ERGS13 05 0058 and IIUM EDW B 13 09 0914. The first Author (M.S. is also grateful to the Junior Associate scheme of the Abdus Salam International Centre for Theoretical Physics, Trieste, Italy. References Borevich, Z. I. and Shafarevich, I. R. (1986. Number Theory. Academic Press, London, th edition. Mukhamedov, F. and Akin, H. (013a. On -adic otts model on the cayley tree of order three. Theor. Math. Phys., 176(3:167 179. Mukhamedov, F. and Akin, H. (013b. Phase transitions for -adic otts model on the cayley tree of order three. J. Stat. Mech., (07. Mukhamedov, F., Omirov, B., and Saburov, M. (014. On cubic equations over adic fields. International Journal of Number Theory, 10(5:1171 1190. Mukhamedov, F., Omirov, B., Saburov, M., and Masutova, K. (013. Solvability of cubic equations in adic integers, > 3. Siberian Mathematical Journal, 54(3:501 516. Mukhamedov, F. and Saburov, M. (013. On equation x q = a over Q. Journal of Number Theory, 133(1:55 58. Rosen, K. H. (011. Elementary Number Theory and Its Alications. Pearson, Boston, 6th edition. Saburov, M. and Ahmad, M. A. K. (014. Solvability criteria for cubic equations over Z. AIP Conference Proceedings, 160:79 797. Saburov, M. and Ahmad, M. A. K. (015. Solvability of cubic equations over Q 3. Sains Malaysiana, 44(4:635 641. Malaysian Journal of Mathematical Sciences 35