LDETheorem - d x-x and gcd a d b d = 1 then b d | x-x by...

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MATH 135 Winter 2009 Linear Diophantine Equation Theorem 2.31 (i) The LDE ax + by = c has a solution if and only if gcd( a, b ) | c . (ii) If gcd( a, b ) = d = 0, and x = x 0 , y = y 0 is a particular solution, then the complete solution is x = x 0 + n b d y = y 0 - n a d for all n Z Proof: (i) “= Suppose that there are integers x , y with ax + by = c . Now gcd( a, b ) | a and gcd( a, b ) | b so, by Prop. 2.11.ii, gcd( a, b ) | ax + by = c . =” Suppose d = gcd( a, b ) | c , say c = qd for some q Z . By the EEA, we can find a solution ( x 0 , y 0 ) to the equation ax + by = d , ie. ax 0 + by 0 = d . Then aqx 0 + bqy 0 = qd or a ( qx 0 ) + b ( qy 0 ) = c , so x = qx 0 , y = qy 0 is a solution to ax + by = c , ie. ax + by = c has a solution. (ii) Let ( x 0 , y 0 ) be the particular solution and ( x, y ) an arbitrary solution. We show that ( x, y ) is of the correct form. Then ax 0 + by 0 = d and ax + by = d . Subtracting these equations, we get a ( x - x 0 ) + b ( y - y 0 ) = 0 a ( x - x 0 ) = - b ( y - y 0 ) a d ( x - x 0 ) = - b d ( y - y 0 ) We recall, though, that a d , b d Z and gcd ( a d , b d ) = 1 by Prop. 2.27. Also, since b d divides into the right side, then b d must divide into the left side.
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Unformatted text preview: d ( x-x ) and gcd ( a d , b d ) = 1, then b d | ( x-x ) by Prop. 2.28. Therefore, x-x = 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 ( y-y ) or y = y-n a d . Therefore, x = x + n b d and y = y-n 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 so-lution. To determine this, we substitute. Choose any integer n . Then a ± x + n b d ² + b ³ y-n a d ´ = ax + by + an b d-bn a d = ax + by = d so each is actually a solution. Therefore, the complete solution is as we stated above....
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