Bézout's identity and Diophantine Equation

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SUMMARY

The discussion centers on the application of Bézout's identity to Diophantine equations, specifically the equation 12a + 21b + 33c = 6. It is established that since the greatest common divisor (gcd) of the coefficients (12, 21, 33) is 3, and 6 is divisible by 3, this equation has infinitely many solutions. The participant also correctly infers that the modified equation 12a + 21b + 33c + 24d = 6 retains infinite solutions. In contrast, the equation 2x + 4y + 6z = 7 does not have solutions as 7 is not divisible by the gcd of the coefficients, which is 1.

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

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