Is "a and b being relatively prime" the only situation where na+mb=1 for some n,m?

Asked Oct 19 '16 at 01:53

Active Oct 19 '16 at 02:27

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Suppose a and b are two integers such that gcd(a, b)=1.

First, is there any simple argument showing why there must be some integers $n$, $m$ such that $na+mb=1$?

Second, is this the sufficient condition for us to say there exist $m$ and $n$ such that $na+mb=1$?

Thank you!

asked Oct 19 '16 at 01:53

Norman

3

For both questions, yes. See Bézout's identity. – Cave Johnson Oct 19 '16 at 01:55
See here for a conceptual proof of Bezout's identity for the gcd. It is sufficient since if $,na+mb = 1,$ then $,d\mid a,b,\Rightarrow, d\mid na+mb = 1,,$ so $,\gcd(a,b) = 1.,$ – Bill Dubuque Oct 21 '16 at 15:06

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