Given $b\mid (ac-1)$ show that $(a,b) = 1$

Question

How would one show that if $b|(ac-1)$, then $(a,b) = 1$

Answer 1

You have $(a,b)=1$ if and only if there are integers $m$ and $n$ such that $am+bn=1$. But you know that there is an integer $k$ such that $bk=ac-1$ and$$bk=ac-1\iff ac-bk=1.$$

Answer 2

Suppose $(a,b)=d$. Then $d\mid b$ and $b\mid ac-1$ implies $d\mid ac-1$. But $d\mid a$, so $d\mid 1$, hence $d=1$.

Answer 3

It should be a basic fact at your fingertips that $\gcd(n, n-1) = 1$ for all $n$. (This is because if $d|n$ and $d|n-1$ then $d|n-(n-1) =1$ so $d|1$.)

So $\gcd(ac -1,ac) = 1$ and $ac-1$ and $ac$ have no factors (other than $1$) in common.

But $b$ (and all its factors) is a factor of $ac-1$ so $b$ has no factors (other than $1$) in common with $ac$ or with any of its factors including $a$.

....

Put another way. Let $\gcd(a,b) =d$. Then $d|b$ and $b|ac-1$ so $d|ac-1$. But $d|a$ and $a|ac$ so $d|ac$. So $d$ is a common divisor of $ac-1$ and $ac$. So $d$ is a divisor of $ac -(ac -1) = 1$. So $d = 1$.

Answer 4

$b|ac-1$ implies that there exists a $k \in \mathbb{N}$ such that $bk=ac-1$ which implies

$1=ac-bk$.

Therefore, there exist two integers $c$ and $-k$ which when multiplied by $a$ and $b$ respectively, the sum gives $1$.

We know that there exist two integers $x,y$ for all integers $a,b$ such that $ax+by$ has the least positive value equal to $\gcd(a,b)$.

Since a linear combination of $a$ and $b$ can be equal to $1$, this implies $\gcd(a,b) =1$.