Bézout's identity and Diophantine Equation

haki

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?

17x+6y +3z =73

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

fresh_42

Mentor
2018 Award
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

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

"Bézout's identity and Diophantine Equation"

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