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Brahmagupta 3 All letters will denote integers A point in the plane (m, n) is a lattice point if m, n ∈ Let a ∧ b = 0 are 1 . Then the lattice solutions to ax + by = all of the form (kb, −ka ) where k ∈ If ax + by = c has a solution, then a ∧ b is a divisor of c If ax + by = c has a solution (m, n) , then infinitely many as given by (m, n) + k ( a−∧bb , a∧a b ) (2,3),(3,0),(−1,1) are, ( 12 ,0) is not 4x + 7 y = 0 has solutions (4, −7),(8, −14),(−4,7),(−8,14),... Thus 4 x + 6 y = 7 could not have a solution 4x + 6 y = 8 has (2,0) + k (−3, 2) for solutions Actually, it is algorithmic: 1 has a solution a∧b = 1 if and only if ax + by = Kuttaka Stage Rs Qs 0 2155 1 138 2 m am + bn n 1 0 2155 15 0 1 138 85 1 1 -15 85 3 53 1 -1 16 53 4 32 1 2 -31 32 5 21 1 -3 47 21 6 11 1 5 -78 11 7 10 1 -8 125 10 8 1 10 13 -203 1 ax + by = c has a solution when and only when a ∧ b is a a b c x+ y= divisor of c . Equivalent to . a ∧b a ∧b a ∧b 12508 x − 58823 y = 177 is solvable since 12508 ∧ 58823 = 59 and 177= 3 × 59 There were 63 equal piles of plantain fruit put together and 7 single fruits. They were divided evenly among 23 travelers. How many plantains did each traveler receive? Start with equation 23 = x 63 y + 1 . By Kuttaka this leads to y = 4 x = 11 . Thus solution to 23x = 63 y + 7 is given x = 77 and y = 28 . Another solution is x = 14 y = 5 Let d be a common divisor of a and b . Then d= a ∧ b if and only if d can be written as a linear combination of them, namely, if and only if ax + by = d has a solution. Sylvester’s Theorem. Let a and b be positive integers with a ∧ b = 1 . Then c = ab − a − b is the largest value of c for which the equation ax + by = c does not have a nonnegative integer solution. Moreover, half of values below c have unique solutions, and the sum of all the c ’s for which there is no solution is: Consider a = 30 and b = 24 . Of their common divisors, 1, 2, 3 and 6, 30 x + 24 y = 6 is the only one to have a solution. ( a − 1)(b − 1)(2ab − a − b − 1) 12 Consider a = 5 and b = 11 . So the last c for which there is no solution to 5 x + 11b = c is c = 39 . Indeed we have no solutions for 1, 2, 3, 4, 6, 7, 8, 9, 12, 13, 14, 17, 18, 19, 23, 24, 28, 29, 34,and 39. These 20 numbers add up to 310.