Prove that if $a = bc+1$ then $(a,b) = 1$ 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?
 A: 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, $(a,b)=1$. 
A: 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
$$
ax+by=a-bc=1
$$
A: 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.
