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Unformatted text preview: Math 465 Solution Set 8 Ae Ja Yee 1. Find all integer solutions to the Diophantine equation 40 x + 25 y = 600 . Solution. The equation is equivalent to 40 x ≡ 600 (mod 25) , 8 x ≡ 120 (mod 5) , 3 x ≡ (mod 5) . Thus x ≡ (mod 5) , or equivalently x = 5 k, y = 24 8 k 2 2. Find all integer solutions of the Diophantine equation 8 x 1 4 x 2 + 6 x 3 = 6 Solution. Let y = 2 x 2 + 3 x 3 . Then 8 x 1 + 2 y = 6 , which is equivalent to 8 x 1 ≡ 6 (mod 2) , 4 x 1 ≡ 3 (mod 1) , · x 1 ≡ (mod 1) Thus x 1 = k Then, 4 x 2 + 6 x 3 = 6 8 k 4 x 2 ≡ 6 8 k (mod 6) , 2 x 2 ≡ 3 4 k (mod 3) , x 2 ≡  k (mod 3) Thus, x 2 = k + 3 l, x 3 = 1 2 k + 2 l 2 1 3. Let a 1 x 1 + a 2 x 2 + ··· + a n x n = b be a linear Diophantine equation in the n ( ≥ 2) variables x 1 , x 2 , . . . , x n , and let d be the gcd of a 1 , a 2 , . . . , a n . (a) If d b , prove that the equation have no solutions....
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This note was uploaded on 07/23/2008 for the course MATH 465 taught by Professor Yee during the Spring '08 term at Penn State.
 Spring '08
 YEE
 Math, Number Theory

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