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Diophantine Equationsmedium
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The fundamental solution to x2−3y2=1x^2 - 3y^2 = 1x2−3y2=1 is (2,1)(2, 1)(2,1). Using the Brahmagupta-Fibonacci identity (x12−3y12)(x22−3y22)=(x1x2+3y1y2)2−3(x1y2+y1x2)2(x_1^2 - 3y_1^2)(x_2^2 - 3y_2^2) = (x_1x_2 + 3y_1y_2)^2 - 3(x_1y_2 + y_1x_2)^2(x12​−3y12​)(x22​−3y22​)=(x1​x2​+3y1​y2​)2−3(x1​y2​+y1​x2​)2, the next solution is obtained by composing (2,1)(2,1)(2,1) with itself. What is this next solution (x2,y2)(x_2, y_2)(x2​,y2​)?