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Given the Diophantic Equation $$1188x +63y =26$$ Task: Find integer solution(s)

I found that $$1188x +63y =26 \Longleftrightarrow 132x+7y = \frac{26}{9}$$ One can easily see that LHS $\in\mathbb Z$ but RHS $\notin \mathbb Z$ for all choices of $x,y\in\mathbb Z$, therefor an integer-solution cannot exist.

Is this a valid proof?

More general: If I have a Diophantic Equation like $$ax+by=c$$ And there is an $d$ with $d\,\vert\, a$ and $d\,\vert\, b$, but $d\,\not\vert\, c$, is it true that an integer solution cannot be found in this case, or did I miss something?

user127.0.0.1
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1 Answers1

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Yes. Simpler, it is clear by casting nines, that $\,9\,$ divides the LHS but not the RHS, contradiction.

Generally by Bezout's Identity, $\,ax+by=c\,$ is solvable $\iff \gcd(a,b)\mid c.\,$ Indeed, the direction $\,(\Rightarrow)\,$ is clear, and $\,(\Leftarrow)\,$ follows by scaling the Bezout identity for $\,\gcd(a,b)\,$ by $\,c/\gcd(a,b).$

Said structurally $\ c\in a\,\Bbb Z + b\,\Bbb Z\, =\, \gcd(a,b)\,\Bbb Z\iff \gcd(a,b)\mid c.\,$ The underlying innate structure becomes clearer after one learns about ideals.

Bill Dubuque
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