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Given any two values of $u$ and $v$, there exists values for $a$ and $b$ that would result in the greatest common factor between $u$ and $v$.

$$\forall u,v \in \mathbb Z,\exists a,b \in \mathbb Z, au + vb = GCF(u, v)$$

This also effectively proves that any values of $u,v,a,b$ can be used, and will still be a multiple of the GCF, but I don't know how to prove the statement.

F.Carette
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Celentia
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  • Hi, and welcome to MSE! Please try to use Mathjax to format your questions/answers. I submited an edit on your post to include it. Please also note that you will get more answers if you include what you tried to do to solve your question. – F.Carette Oct 11 '18 at 09:35

1 Answers1

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Consider the set $S = \{au + bv: a,b \in \mathbb Z \}$. $S$ is a set of integers so it has a smallest positive member - call this $k$. So

$\nexists \space n \in S, 0 < n < k$

We know that $GCF(u,v)|u$ and $GCF(u,v)|v$, so $GCF(u,v)$ is a factor of all members of S. In particular $GCF(u,v)|k$.

If we can also show thet $k|GCF(u,v)$ then we must have $k=GCF(u,v)$. so $GCF(u,v) \in S$.

Note that if $m$ and $n$ are in $S$ then any linear combination of $m$ and $n$ i.e. $am + bn; a,b \in \mathbb Z$ is also in $S$.

Suppose $\exists n \in S$ with $n$ not a multiple of $k$. Then we can find $p$ and $r$ such that $n=pk+r$ with $0\lt r \lt k$. But then $r=n-pk$ is a linear combination of $n$ and $k$ and so $r$ is a member of $S$. But this contradicts the definition of $k$ as the smallest positive member of $S$. Therefore all members of $S$ are multiples of $k$.

How does this help you to show that $k|GCF(u,v)$ ?

gandalf61
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  • Simpler: we can use subtractive descent: $,n>k>0\in S,\Rightarrow, n-k\in S\ $ (vs. $,n\bmod k,$ above), which amounts to using the extended Euclidean algortihm for the GCD in subtractive vs. mod(remainder) form. I explain this more structurally here. This is a special case of using descent by the division algorithm in order to prove that Euclidean domains are PIDs. – Bill Dubuque Oct 11 '18 at 16:42