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I know that if $a$ and $b$ have a highest common factor $h$ then you can write $h=ax+by$ for some $x,y \in \mathbb{Z}$ but how about if you can write $h=ax+by$ for some $x,y \in \mathbb{Z}$ then can you write $h=\gcd(a,b)$?

I.e. Is $h=\gcd(a,b) \iff h=ax+by,~ x,y \in \mathbb{Z}$

Bart Michels
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Jared
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4 Answers4

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What you ask is not quite true, but something related is true.

Let $a,b$ be integers with highest common factor (more commonly called greatest common divisor) $d$ then the set $\{x a + y b\colon x, y \in \mathbb{Z} \}$ is exactly the same as $\{zd \colon z \in \mathbb{Z}\}$ so the set of multiples of $d$.

So, if you can write something as $ax + by$ you at least know it is divisible by the highest common factor, or put the other way round the highest common factor must divide that number. Especially, when you can write $1$ in this way you know the highest common factor is $1$ (as $1$ has no positive divisor other than $1$).

quid
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  • so does the proof provided by ahmed here:http://math.stackexchange.com/questions/1125271/proof-with-3-coprime-numbers/1125283?noredirect=1#comment2296111_1125283 work??? – Jared Jan 29 '15 at 18:38
  • Yes it works. (Incidentally, the upvote on that answer is mine.) Note that I specifically address this at the end of my answer when I say "when you can write $1$ in this way you know the highest common divisor is $1$." It is an "if and only if" in a certain way, but not in the one you gave. What is true si two numbers $a,b$ are coprime if and only if you can write $1= ax + by$. The main reason is that a divisor common to both $a,b$ will divide any combination $ax+by$. Indeed highest common divisors are the only divisors that can be expressed in this form. – quid Jan 29 '15 at 19:00
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General Bezout Identity $\,\ h = ax\! +\! by\,$ for some $\,x,y\in\Bbb Z \iff d\!:=\!(a,b)\mid h$

Proof $\ \ (\Rightarrow)\ \ $ If $\ h = \color{#c00}ax\!+\!\color{#c00}by\ $ then $\,d\mid \color{#c00}{a,b}\,\Rightarrow\, d\mid h$

$(\Leftarrow)\ \ $ If $\,d\mid h \,$ then $\,h = kd\ $ for $\,k\in\Bbb Z.\,$ By Bezout there are $\,m,n\in\Bbb Z\,$ such that

$$\ d\!=\!(a,b) = a m\! +\! b n \ \overset{\times\ k}\Rightarrow\, h \,=\, a (km) + b(kn)\quad{\bf QED} $$

Remark $ $ Said more succinctly in ideal or group language it is simply

$$ \bbox[8px,border:1px solid #c00]{a\,\Bbb Z + b\, \Bbb Z\, =\, (a,b)\,\Bbb Z}\qquad\qquad\qquad$$

Bill Dubuque
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  • so does the proof provided by ahmed here:http://math.stackexchange.com/questions/1125271/proof-with-3-coprime-numbers/1125283?noredirect=1#comment2296111_1125283 work? – Jared Jan 29 '15 at 18:37
  • @Jared It works for that question, not this one. – Bill Dubuque Jan 29 '15 at 18:42
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What you can say is that if you can write $h=ax+by$ and if $h$ is a (positive) common factor of $a$ and $b$, then $h=\gcd(a,b)$.

paw88789
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My first answer handled the case of coprime integers (I misread the question).

To answer your question, the other way of course doesn't work, there are a lot of counter examples you can find easily:

3 - 1 = 2; 5 + 3 = 8; 10 - 7 = 3; etc.