Help showing that $2^{p}f(x) - 1$ and $1 + 3f(x)$ are coprime Coprimality of numbers is a new concept for me. I've been reading up on it and I don't quite grasp it yet. Is there a way to show that given $2^{p}f(x) - 1$ and $1 + 3f(x)$ are coprime for all values of x where $p \geq 1$ and is an integer and $f(x) \geq 1$ and is an integer for all values of $x$?
I can see that for any value of $f(x)$, one side will be even and the other odd. But I don't know if that's enough on it's own.
Edited for clarity. The question I have is that I’m working with an equation of the form
$$2^{q}(2^{p}f(x) - 1) = 3(1 + 3f(x))$$
This has the form $p_{1}^{a_{1}} \times g(x) = p_{2}^{a_{2}} \times h(x)$, where $p_{1}$ and $p_{2}$ are primes and $a_{1}$ and $a_2$ are positive integers. I’m trying to show in this case that
$$2^{p}f(x) - 1 = 3$$
$$2^{q} = 1 + 3f(x)$$
But this is only true if $2^{p}f(x) - 1$ and
$1 + 3f(x)$ are coprime. Is there a way to prove that they are coprime?
 A: If $\ f(x)\ $ is an integer and $\ f(x)\ge1\ $, then the only non-negative integers $\ p, q\ $ for which the equation
$$
2^q\big(2^pf(x)−1\big)=3(1+3f(x))
$$
can be satisfied are $\ p=q=2\ $, in which case $\ f(x)=1\ $.  So $\ 2^pf(x)-1=3 $ and $\ 1+3f(x)=2^q\ $ are indeed relatively prime, although this is just an incidental consequence of the fact that the only integer solutions of the equation
$$
2^q\big(2^pr−1\big)=3(1+3r)
$$
with $\ p\ge0$, $\ q\ge0\ $, and $\ r\ge1\ $ are $\ p=q=2\ $ and $\ r=1\ $.
Rewriting the above equation as
$$
r=\frac{2^q+3}{2^{p+q}-9}\ ,
$$
the condition that $\ r\ge1\ $ tells us that $\ p+q\ge4\ $, and
$$
12\ge2^q(2^p-1)\ .
$$
If $\ p\ne0\ $, the only other non-negative values of $\ p\ $ and $\ q\ $ which can satisfy these two inequalites are $\ p=1, q=3\ $ or $\ p=q=2\ $. The first of these gives
$$
r=\frac{11}{7}\ ,
$$
which isn't an integer.
If $\ p=0\ $ then
$$
r=\frac{2^q+3}{2^{q}-9}=1+\frac{12}{2^q-9}\ ,
$$
with $\ q\ge4\ $. But $\ q=4\ $ gives $\ r=\frac{19}{7}\ $, and $\ q>4\ $ gives $\ 1<r<2\ $ and it's therefore impossible for $\ r\ $ to be an integer in this case.
Thus, under the conditions $\ p\ge0, q\ge0\ $ and $\ r\ge1\ $, this Diophantine equation has the unique solution $\ r=1, p=q=2\ $.
