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Given integers $x$, $y$ and $c$. Can I always find integers a and b such that $ax + by = c$?

This is my exam question. My answers goes like this:

$ax + by = c$

Lets take $x = 2, y = 0$ and $ c = 3$.

so $2a = 3 => a = 3/2 = 1.5$

Therefore $a ∉ ℤ $

Similarly for b. Hence statment is false. Is my counter example correct? Please help. Thanks.

2012ssohn
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mathguy
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  • Your example is indeed a counter-example. – user259242 Nov 05 '16 at 04:22
  • You can skip the "similarly for $b$" part (which doesn't really work, anyway). The question asks whether there is always a solution in which both $a$ and $b$ are integers. If $a$ is not an integer then the statement that both $a$ and $b$ are integers is false regardless of what $b$ is. – David K Nov 05 '16 at 04:29

3 Answers3

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The statement is false, for what refers to the word "always". In fact, the statement is true only if the GCD(x,y) is c. This is the Bézout (1730-1783) identity, and you can find extensive treatments regarding it in the Web. It can be demonstrated, for example, using the Euclidean Algorithm to find the GCD of x and y.

Regarding your counterexample, it eludes the main point, because the GCD of (2,0) is (in my opinion) 2, not c =3, and therefore the two integers a and b, thanks to Bézout, cannot be found. Indeed you can see that if c were = 2, you would have your solution: a =1 and b = any integer.

user464227
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Yes, your counter-example is valid and thus goes part of the way toward showing that the statement in your question is false.

user259242
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For the more general case, let $x,y$ be arbitrary integers, and let $z = \gcd(x,y)$. Then, if $z$ does not evenly divide $c$, there is no pair of integers $(a,b)$ that satisfy $ax+by=c$. This is because in $\bmod z$, the left side of the equation is congruent to $0$ while the right side is not.

Your example is - at least, technically - not wrong, though one might argue that you have trivialized the problem, since setting $y = 0$ removes the $by$ term, and of course $\gcd(x,0)$ is undefined since there is no such thing as a GCD of $x$ and $0$ (see this post). Another counterexample would be something like $(x,y,z) = (21,14,5)$. Here, $\gcd(21,14) = 7$, and $7$ does not divide $5$, and therefore there is no pair of integers $(a,b)$ that satisfy the equation.

Community
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2012ssohn
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  • You say 'trivializing' as if its a bad thing. In mathematics, the simpler we can make a proof, the better. – user259242 Nov 05 '16 at 04:27
  • I thought it would be more useful to give a general instruction in how to approach questions of this type - yes, technically setting $(x,y) = (0,0)$ and $c \ne 0$ will always work, but this answer must surely be more useful when the OP advances in mathematics and comes across Diophantine equations and such. I wanted to teach him to fish, not give him one. – 2012ssohn Nov 05 '16 at 04:35
  • I think the strategy of trying always to simplify is best in any level of mathematics. Diophantine equations are notoriously difficult. It would be quite foolish to approach them by attempting to make them even harder! – user259242 Nov 05 '16 at 04:37