Bézout's identity and Diophantine Equation

  • Thread starter haki
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I am having problems with one exam question.

Does this diophantine equation have a solution(s)

12a+21b+33c=6

as far as I know this is not a linear equation, and what I read online says that Bezout identity only applies for linear diophantine equations.

The solution says gcd(12,21,33) = 3, 6|3 the above equation has infinitely many solutions. Is that correct? Am I correct to assume then that

12a+21b+33c+24d = 6 again has infinite solutions aswell?

How about this equation

17x+6y +3z =73

since gcd(17,6,3) = 1 and 73 | 1 this has also infinitely many solutions?
 

fresh_42

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Bézout's Lemma says, that the greatest common divisor ##d## of numbers ##a_1,\ldots,a_n## can always be written as ##d= s_1a_1+\ldots +s_na_n##. This immediately applies to all of your examples.

An example where it does not work is ##2x+4y+6z = 7##.
 

WWGD

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You can always reduce the case ax+by+cz=d to ax+by=d by letting z=0.
 

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