hw09-solutions - MATH4430 Introduction to Number Theory...

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MATH4430: Introduction to Number Theory Homework 9 Due: Friday, April 15 D. Geraghty 1. Let a and b be positive integers with gcd( a, b ) = 1. Let x 0 and y 0 be integers with ax 0 + by 0 = 1 . (We can find such integers by applying Euclid’s algorithm.) Show that the general solution to the equation: ax + by = 1 is given by: x = x 0 + bt y = y 0 - at for t an arbitrary integer. Solution: Let x, y be another solution. Then after subtracting the two equations from each other we get: a ( x - x 0 ) + b ( y - y 0 ) = 0 and hence a ( x - x 0 ) = - b ( y - y 0 ) . We deduce that b divides the left hand side. But gcd( a, b ) = 1, so this means: b | x - x 0 . Hence x - x 0 = bt for some integer t . (Hence x = x 0 + bt .) Now plugging in, we get: abt = - b ( y - y 0 ) and canceling b gives: y = y 0 - at . So all solutions are of the required form. And conversely, any x , y given by the formula of the question are a solution to the equation. 1
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2. Let a and b be positive integers with gcd( a, b ) = 1. Show that there exist positive numbers x, y, z, w with ax - by = 1 and bz - aw = 1 . Solution: Let x 0 and y 0 be integers with ax 0 + by 0 = 1 (we can get these from Euclid). Since a and b are positive, one of x 0 and y 0 is positive and the other negative. Let’s suppose x 0 > 0 and y 0 < 0. We can then take x = x 0 and y = - y 0 . Now we need to find a z and w . Well, use thing the previous question: we know that a ( x 0 + bt ) + b ( y 0 - at ) = 1 for all t . All we need to do is take t to be a negative number with | t | large enough that x 0 + bt < 0 and y 0 - at > 0 . Then take z = y 0 - at and w = - ( x 0 + bt ). 3. As in the previous questions, let a and b be positive integers with gcd( a, b ) = 1. Let a b = CF( q 0 , . . . , q n ) be its continued fraction expansion. Let x = B n - 1 and y = A n - 1 . Show using the results we proved in class that: ax - by = ( - 1) n - 1 .
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