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In previous post, I got the answer that $\gcd \left(\frac{p^2+1}{2}, \frac{p^5-1}{2} \right)=1$, where $p$ is prime number.

I am looking for more general case, that is for $p$ prime,

When is $\gcd \left(\frac{p^r+1}{2}, \frac{p^t-1}{2} \right)=1$ ?

where $t=r+s$ such that $\gcd(r,s)=1$.

I am excluding the cases $r=1=s$.

In the previous post, it was $r=2,~t=5$. So $t=5=2+3=r+s$ with $s=3$ so that $\gcd(r,s)=1$.

In this current question:

Since $\gcd(r,s)=1$, we also have $\gcd(r,t)=1$. I have the following intuition:

Case-I:

Assume $r$=even and $s$=odd so that $\gcd(r,s)=1$ as well as $\gcd(r,t)=1$. We can also assume $r$=odd and $s$=even.

I think the same strategies of previous post can be applied to show that $$\gcd \left(\frac{p^r+1}{2}, \frac{p^t-1}{2} \right)=1.$$

Case II:

The problem arises when both $r$ and $s$ are odd numbers so that $t=r+s$ is even.

If I take $r=3, ~s=5$, then $t=8$.

For prime $p=3$, $\frac{p^r+1}{2}=\frac{3^3+1}{2}=14$ and $\frac{p^t-1}{2}=\frac{3^8-1}{2}=3280$ so that the gcd is $2$ at least.

For other primes also we can find gcd is not $1$.

So I think it is possible only for Case I, where among $r$ and $s$, one is odd and another is even so that $t$ is odd.

In other word, $t$ can not be even number.

But I need to be ensured with a general method.

So the question reduces to

How to prove $\gcd \left(\frac{p^m+1}{2}, \frac{p^n-1}{2} \right)=1$ ?

provided $\gcd(m,n)=1$ and $n$ is odd number and $p$ is prime number.

Thanks

MAS
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    Just for ease of reading, it would be helpful to restate here that $p$ is prime... – abiessu Aug 06 '22 at 14:02
  • @abiessu, thanks. I have mentioned it now – MAS Aug 06 '22 at 14:04
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    $\def\A{\qquad\text{ and }\qquad}\def\m{\pmod}\def\f{\tfrac}\def\e{\equiv}$For $m$ and $n$ coprime with $n$ odd, let$$d:=\gcd(\f{p^m+1}{2},\f{p^n-1}{2}).$$For any prime $q\mid d$ the congruences$$p^m\e-1\m{q}\A p^n\e1\m{q},$$show that the order of $p$ mod $q$ divides both $2m$ and $n$, hence also $\gcd(2m,n)=1$, and so $p\equiv1\m{q}$. Then$$p^m+1\e1^m+1\e2\m{q},$$and because $q$ divides $\f{p^m+1}{2}$ this shows that $q=2$. Then if $d>1$ we must have$$p^m+1\e0\m{4}\A p^n-1\e0\m{4}.$$Because $n$ is odd the latter implies $p\e1\m{4}$. But then $p^m+1\e2\m{4}$, a contradiction. So $d=1$. – Servaes Aug 06 '22 at 16:38
  • @Thissitehasbecomeadump., thank you for the comment, though an answer as well – MAS Aug 06 '22 at 16:52
  • @Thissitehasbecomeadump. Exactly this is what I want to write but I thought m=prime I didn't read the question properly I am extremely sorry !! My mood is little heavy today so sorry agin for being rude. –  Aug 06 '22 at 17:18
  • @Thissitehasbecomeadump., just one to make sure. Since we have no restriction on $m$, so it can be odd or even both. In other word, the gcd $d$ is independent of $m$, provided $(m,n)=1$. Am I right ? – MAS Aug 07 '22 at 13:28
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    @MAS That is correct. Note that the argument also doesn't require $p$ to be prime; it works for any odd integer. – Servaes Aug 08 '22 at 02:01
  • @Thissitehasbecomeadump., thanks a lot – MAS Aug 08 '22 at 02:13

1 Answers1

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Suppose that $$\gcd\left(\frac{p^m + 1}{2}, \frac{p^n - 1}{2}\right) \neq 1.$$ There are two cases:

  1. Both $p^m + 1$ and $p^n - 1$ are multiples of $4$.
  2. Both $p^m + 1$ and $p^n - 1$ are multiples of the same odd prime $q$.

In case 1, $p^m + 1 \equiv 0 \pmod 4$ implies that $p^m \equiv 3 \pmod 4$, so that $p \equiv 3 \pmod 4$ and $m$ is odd. But then $p^n \equiv 3 \pmod 4$ as well, since $n$ is odd. This contradicts the fact that $p^n - 1$ is a multiple of $4$.

In case 2, let $r$ be the order of $p$ modulo $q$. Since $p^m \equiv -1 \pmod q$, $r$ must be even (as $r \mid 2m$, but $r \not \mid m$). But this makes it impossible for $p^n \equiv 1 \pmod q$, since $n$ is odd and thus cannot be a multiple of $r$.

Since neither case is possible, it must be that $\gcd\left(\frac{p^m + 1}{2}, \frac{p^n - 1}{2}\right) = 1$.

Joseph Camacho
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