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Unformatted text preview: d ( xx ) and gcd ( a d , b d ) = 1, then b d  ( xx ) by Prop. 2.28. Therefore, xx = n b d for some n ∈ Z , so x = x + n b d . Substituting back into the equation above, we get a d n b d =b d ( yy ) or y = yn a d . Therefore, x = x + n b d and y = yn a d for some n ∈ Z , so every solution is of the required form. The last question that we have to ask is whether everything of this form is actually a solution. To determine this, we substitute. Choose any integer n . Then a ± x + n b d ² + b ³ yn a d ´ = ax + by + an b dbn a d = ax + by = d so each is actually a solution. Therefore, the complete solution is as we stated above....
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This note was uploaded on 11/25/2010 for the course MATH 135 taught by Professor Andrewchilds during the Winter '08 term at Waterloo.
 Winter '08
 ANDREWCHILDS
 Math

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