Prove that if $a = bc+1$ then $(a,b) = 1$

Question

I started by saying that there exists an $x,y$ such that $1 = ax +by$ but I really don't know where else to go with this. Any hints?

Answer 1

If $d$ is any common divisor of $a$ and $b$ then it also divides their difference $a-b$. Then also the difference of $a-b$ and $b$ thus $a-2b$. Continuing like this, we get that $d$ is a common divisor of all integers $a-\kappa b$, for any integer $\kappa$. Thus, $d$ will be a divisor of $a-bc=1$. Consequently, $d=1$. Thus, $(a,b)=1$.

Answer 2

More generally, if $a_1a_2...a_n =b_1b_2...b_m\pm 1 $, then $\gcd(a_i, b_j) = 1 $ for every choice of $i$ and $j$.

The proof is identical to the others here: If $k$ divides both $a_i$ and $b_j$, then $k$ divides both $a_1a_2...a_n$ and $b_1b_2...b_m$ so that it divides their difference which is $\pm 1$.

Even more generally, if $a_1a_2...a_n =b_1b_2...b_m\pm c $, then $\gcd(a_i, b_j) $ divides $c$ for every choice of $i$ and $j$.

The proof is identical.

Answer 3

Whenever you find two integers $x$ and $y$ such that $1=ax+by$, you can conclude that $\gcd(a,b)=1$. Indeed, if some integer $d>0$ divides both $a$ and $b$, we have $$ a=du,\quad b=dv $$ so $$ 1=ax+by=d(ux+vy) $$ and therefore $d\mid 1$, hence $d=1$.

If you choose $x=1$ and $y=-c$, you have $$ ax+by=a-bc=1 $$