How you get from $\gcd(a,b)= 1 \Longleftrightarrow ax+by=1 \Longrightarrow a \nmid b$?
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HINT $\ $ If $\rm\:a\:|\:b\:$ then $\rm\:a\:$ is a common divisor of both $\rm\:a,\:b\:,\:$ hence $\rm\:a\ |\ a\:x+b\:y\: =\: gcd(a,b) = 1\:,\:$ i.e. all common divisors must divide the gcd, i.e. the gcd is the divisibly-greatest common divisor, so $\rm\ c\:|\:a,b\iff c\:|\:gcd(a,b)\:.\:$ Adjoining the (necessary) hypothesis $\rm\ a\nmid 1\ $ concludes the proof.