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Is Bezout's lemma enough to confirm the GCD of a number?

So suppose we have $$ax+by=z$$ does this mean $$\gcd(a,b)=z$$

Bill Dubuque
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Juan Kerr
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  • No, of course not. Say if hcf of a and b is 1, then there are integers $(x,y)$ such that $ax+by=1$, but then there are also integers $(2x,2y)$ such that $a\cdot (2x)+b\cdot (2y)=2$. This doesn't mean the hcf is 2. However, if z is the smallest positive integer that is of the form $ax+by$, then it is the hcf. – Sawarnik May 20 '15 at 18:54
  • The HCF of $a$ and $b$ is unique, but there are any number of linear combinations of $a$ and $b$. – Jonathan Hebert May 20 '15 at 19:04
  • Note: I changed HCF to GCD in the question. – Bill Dubuque Aug 11 '19 at 13:19

3 Answers3

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I note that $\mathrm{hcf}$ is usually denoted as $\gcd$ (greatest common divisor). Now from $$ax+by = z$$ we can only see that $\gcd(a,b) \mid z$ (i.e. $z$ is a multiple of $\gcd(a,b)$) and by Bezout's lemma there exist some $(x,y)$ such that $z=\gcd(a,b)$.
You can't infer anything else about $\gcd(a,b)$ from this, not even if you force $\gcd(x,y)=1$. For example $x=y=1$ gives $z=a+b$. There is not much you could say about the $\gcd$ of some numbers only knowing their sum.

AlexR
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Consider the following: $$ 1\cdot 2 + 1\cdot 3 = 5,$$ and so $$\gcd (1,1) = 5.$$ You might try almost any combination of numbers and check for yourself the validity of your statement.

davidlowryduda
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A common divisor $\,c\mid a,b\,$ is greatest if it can be written in linear form $\,c = j a + k b,\,$ since then $\,d\mid a,b\,\Rightarrow\, d\mid ja+kb = c,\,$ so $\, d\le c.\,$ Conversely, by Bezout, we know every greatest common divisor can be expressed as such a linear combination.

But only the least positive such linear combination of $\,a,b\,$ equals the gcd. Indeed, such linear combinations are preserved by scaling by arbitrary integers, so any multiple of the gcd is also such a linear combination. Further, one can show that the integers of the form $\ ja + kb \ $ are precisely all integer multiples of the gcd, i.e. $\ a \Bbb Z + b\Bbb Z\, =\, \gcd(a,b)\Bbb Z.\,$ Said equivalently, there exist integers $\,x,y\,$ such that $\ a x + by = c \iff\gcd(a,b)\mid c,\,$ see here.

Bill Dubuque
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