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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?

Bill Dubuque
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Tiny Tim
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    Sorry, what's the question? It's obviously not true that $2^pn-1$ and $1+3n$ are relatively prime for all $n$. Is that what you were asking? – lulu Aug 27 '22 at 14:33
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    Take $f(x)=2$ and $p=2$ – InanimateBeing Aug 27 '22 at 14:33
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    Voting to close the question as it is not clear what you are asking. If you can, please edit your post for clarity. – lulu Aug 27 '22 at 14:40
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    I'm guessing this question derives from a misinterpretation of one originally about polynomials. If $\ f(x)\in\mathbb{R}[x]\ $, then $\ 2^pf(x)-1\ $ and $\ 1+3f(x)\ $ will be relatively prime in that ring. – lonza leggiera Aug 27 '22 at 15:21
  • @lulu I’ve edited the question for clarity – Tiny Tim Aug 27 '22 at 17:39
  • @lonzaleggiera I've edited the question to provide a little more information and hopefully be a little more clear. How do I prove that the two functions will be relatively prime? Is there a formula or method I can use to prove this statement? – Tiny Tim Aug 27 '22 at 21:19
  • @AndrewChin Yes, I think it does. I didn't know that things could be done this way. That's incredibly helpful! – Tiny Tim Aug 28 '22 at 02:30

1 Answers1

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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\ $.

lonza leggiera
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