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Construction of Bhaskara Pairs CONSTRUCTION OF BHASKARA PAIRS RICHARD J. MATHAR Abstract. We construct integer solutions {a, b} to the coupled system of diophantine quadratic-cubic equations a2 + b2 = x3 and a3 + b3 = y2 for fixed ratios a/b. 1. Pair of Coupled Nonlinear Diophantine Equations 1.1. Scope. Following a nomenclature of Gupta we define [4, 4.4]: § Definition 1. (Bhaskara pair) A Bhaskara pair is a pair a,b of integers that solve the system of two nonlinear Diophantine equations of F{ermat} type: (1) a2 + b2 = x3 a3 + b3 = y2 ∧ for some pair x, y . { } Remark 1. Lists of a and b are gathered in the Online Encyclopedia of Integer Sequences [14, A106319,A106320]. The symmetry swapping a and b in the equations indicates that without loss of information we can assume 0 a b, denoting the larger member of the pair by b. We will not look into solutions≤ ≤ where a or b are rational integers (fractional Bhaskara pairs). The two equations can be solved individually [1, 5, 2]. Algorithm 1. Given any solution a,b , further solutions as6,bs6 are derived by multiplying both a and b by a sixth{ power} of a common integer{ s, multiplying} at the same time on the right hand sides x by s4 and y by s9. Definition 2. (Fundamental Bhaskara Pair) A fundamental Bhaskara pair is a arXiv:1703.01677v1 [math.NT] 5 Mar 2017 Bhaskara pair a,b where a and b have no common divisor which is 6-full— meaning there is{ no} prime p such that p6 a and p6 b. | | Although fundamental solutions are pairs that do not have a common divisor that is a non-trivial sixth power, individually a or b of a fundamental pair may contain sixth or higher (prime) powers. Example 1. The following is a fundamental Bhaskara pair with 26 a, 26 ∤ b: a = 26 54 313 613, b = 54 313 613 83, x = 53 13 312 | 612, and y =3 ×56 ×7 31×5 615. × × × × × × × × × × Date: March 7, 2017. 2010 Mathematics Subject Classification. Primary 11D25; Secondary 11D72. Key words and phrases. Diophantine Equations, Modular Analysis. 1 2 RICHARD J. MATHAR 2. Trivial Solutions 2.1. Primitive Solutions. A first family of solutions is found by setting a = 0. This reduces the equations to (2) b2 = x3 b3 = y2. ∧ x3 must be a perfect cube, so in the canonical prime power factorization of x3 all exponents of the primes must be multiples of three. Also in the canonical prime power factorization of b2 all exponents must be even. So the first equation demands that the exponents on both sides must be multiples of [2, 3] = 6. Definition 3. Square brackets [., .] denote the least common multiple. Parenthesis (., .) denote the greatest common divisor. In consequence all b must be perfect cubes. Likewise the second equation de- mands that the exponents of b3 and of y2 are multiples of 6. In consequence all b must be perfect squares. Uniting both requirements, all b must be perfect sixth powers. And this requirement is obviously also sufficient: perfect sixth powers [14, A001014] generate Bhaskara pairs: 6 Theorem 1. All integer pairs 0,n , n Z0, are Bhaskara pairs. The associated right hand sides are x = n4, y ={ n9. } ∈ 2.2. Bhaskara Twins. Definition 4. (Bhaskara Twins) Bhaskara twins are a Bhaskara pair where a = b. According to Definition 1 the Bhaskara twins [14, A106318] solve (3) 2a2 = x3 2a3 = y2. ∧ Working modulo 2 in the two equations requires that x3 and y2 are even, so x and y must be even, say x =2α, y =2β. So (4) a2 =4α3 a3 =2β2. ∧ The first equation requires by the right hand side that in the canonical prime power factorization of both sides the exponents of the odd primes are multiples of 3 and that the exponent of the prime 2 is 2 (mod 3). By the left hand side of the first equation it requires that all exponents≡ are even. So the exponents of the odd primes are multiples of 6, and the exponent of 2 is 2 (mod 6). So from the first equation a =21+3×33×53× , which means a is twice≡ a third power. ··· Definition 5. The notation 3 in the exponents means “any multiple of 3.” × The second equation in (4) demands by the right hand side that the exponents of the odd primes are even and that the exponent of 2 is 1 (mod 2). Furthermore by the left hand side all exponents are multiples of 3.≡ This means all exponents of the odd primes are multiples of 6, and the exponent of the prime 2 is 3 (mod 6) So from the second equation a = 21+2×32×52× , which means a must≡ be twice a perfect square. Uniting both requirements, ···a must be twice a sixth power. Obviously that requirement is also sufficient to generate solutions: 6 6 Theorem 2. The Bhaskara Twins are the integer pairs 2n , 2n , n Z0. The associated free variables are x =2n4, y =4n6. { } ∈ CONSTRUCTING BHASKARA PAIRS 3 k 1+ k2 1+ k3 k 1 2 2 1 2 5 32 2 3 2 5 22 7 3 4 ×17 5 ×13 22 5 2 13 2 32× 7 5 6 × 37 × 7 ×31 2 3 × × Table 1. Prime factorizations of 1 + k2,1+ k3 and k 3. Rational Ratios of the two Members 3.1. Prime Factorization. The general solution to (1) is characterized by some ratio a/b = u/k 1 with some coprime pair of integers (k,u) = 1. Cases where u and k are not≤ coprime are not dealt with because they do not generate new solutions. If k were not a divisor of b, a = ub/k would require that k is a divisor of u to let a be integer, contradicting the requirement that u and k are coprime. Algorithm 2. We only admit the denominators k b. | Theorem 1 and 2 cover the solutions of the special cases u = 0 or u = 1. Introducing the notation into (1) yields (5) (1 + u2/k2)b2 = x3 (1 + u3/k3)b3 = y2; ∧ (6) (u2 + k2)b2 = k2x3 (u3 + k3)b3 = k3y2. ∧ Define prime power exponents ci, di, bi, xi and yi as follows by prime power factorizations, where pi is the i-th prime: 2 2 ci (7) u + k = pi , Yi 3 3 di (8) u + k = pi , Yi bi (9) b = pi , Yi ki (10) k = pi , Yi xi (11) x = pi , Yi yi (12) y = pi . Yi In (7), u2 + k2 is the sum of two squares [14, A000404]. Because u and k are coprime, these u2 +k2 are 2, 5, 10, 13, 17, 25, 26, 29, 34, 37, 41,..., numbers whose prime divisors are all p 1 (mod 4) with the exception of a single factor of 2 [14, A008784][12, Thm. 2.5][≡9, Thm. 3]: Lemma 1. (13) c 0, 1 . 1 ∈{ } 4 RICHARD J. MATHAR (14) p 1 (mod 4), if c > 0 p 3. i ≡ i ∧ i ≥ 6 2 2 3 Example 2. If u = 2 , k = 83 as in Example 1, u + k = 5 13 , so c3 = 1, c =3, and u3 + k3 =32 72 31 61, so d =2, d =2, d =1× , d =1. 6 × × × 2 4 11 18 The uniqueness of the prime power representations in (6) requires for all i 1 ≥ (15a) ci +2bi = 2ki +3xi, (15b) di +3bi = 3ki +2yi, for unknown sets of bi, xi,yi and known ci, di, ki (if u/k is fixed and known). For some i—including all i larger than the index of the largest prime factor of [u2 + 2 3 3 k ,u + k , k] once u/k is fixed—we have ci = di = ki = 0. For these (16a) 2bi = 3xi (16b) 3bi = 2yi The first equation requires 2 x and 3 b . The second equation requires 3 y | i | i | i and 2 bi. The combination requires 6 bi. The absence of the i-th prime allows to multiply| b by a sixth (or 12th or 18th...)| power of the i-th prime. These factors are of no interest to the construction of fundamental Bhaskara pairs. In practice we use the Chinese Remainder Theorem (CRT) for all i, whether the ci or di are zero or not [13, 7]. Multiply (15a) by 3 and (15b) by 2, (17) 3c +6b =6k +9x 2d +6b =6k +4y i i i i ∧ i i i i such that the two factors in front of the bi are the same, and work modulo 9 in the first equation and modulo 4 in the second: (18a) 6b 6k 3c (mod 9); i ≡ i − i (18b) 6b 6k 2d (mod 4). i ≡ i − i Because 9 and 4 are relatively prime, the CRT guarantees that an integer 6ai exists. Furthermore the result will always be a multiple of 6 (hence ai an integer), because from (18a) the equations read modulo 3 we deduce that 6ai is a multiple of 3, and from (18b) read modulo 2 that 6ai is a multiple of 2: Algorithm 3. For each ratio a/b = u/k, the prime power decompositions of u2 +k2 3 3 bi and u + k generate a unique exponent bi of the prime power pi of a conjectured solution b. We compute 6bi (mod 9 4) by any algorithm [11], so bi is determined (mod 6). The values of b k that× result from the CRT for the three relevant values of c i − i i and the two relevant di establish Table 2.
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