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Let $m,n,k\in\mathbb{N}$ such that $n= 2k-1$ ($n$ is odd). Find $(2^m +1, 2^n -1)$, where by $(a,b)$ we mean the greatest common divisor of $a,b$.

Attempt:
Let $(2^m +1, 2^n -1)=:d$, then $$d\mid (2^m+1)\land d\mid (2^n -1)\Rightarrow d\mid (2^m + 2^n)\Leftrightarrow d\mid (2^m+2^{2k-1})$$ If $m\geq 2k-1$, that is $m=2k-1+r, r\in\mathbb{N}\cup\{0\}$, then $d\mid 2^{2k-1}(2^r+1)$ and I'm stuck. What should I do from here? How would $n$ being odd help the case?

Was told to use the euclidean algorithm here, but I don't understand what was meant by that.

References to similar problems would be appreciated.

AlvinL
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1 Answers1

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Assuming that you know the general formula: 1$$\gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1 \tag 1$$ You can apply $(1)$ for $(a,b)=(2m,n)$ and $(a,b)=(m,n)$ you will obtain:

$$\begin{align}\gcd(2^{2m}-1,2^n-1)&=2^{\gcd(2m,n)}-1=2^{\gcd(m,n)}-1=d \tag 2 \\ \gcd(2^{m}-1,2^n-1&)=2^{\gcd(m,n)}-1=d `\tag3\end{align}$$

And from here you can see if $x=2^m-1,y=2^m+1$ and $z=2^n-1$ you have: $$\gcd(xy,z)=\gcd(x,z)=d $$

You can conclude that $\gcd(y,z)=1 $

Elaqqad
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  • where did the $(a,b) = (2m,n)$ come from? – AlvinL Feb 16 '16 at 18:48
  • I mean that you can take $a=2m$ and $b=n$ in the formula $(1)$ – Elaqqad Feb 16 '16 at 18:51
  • Yes, I understand that part, but I don't understand how you came up with this substitution and furthermore $$(a,b)=(2m,n)\Rightarrow (a,b)=(m,n)$$? – AlvinL Feb 16 '16 at 18:55
  • I mean that you can take $a=2m$ and $b=n$ in the formula (1) and obtain (2). And take $a=m$ and $b=n$ in the formula (1) and obtain (3). (There is no implication between the two of them ) – Elaqqad Feb 16 '16 at 18:58
  • After fiddling around with algebra, I found answers to all of what you pointed out. Thank you for your time. – AlvinL Feb 16 '16 at 20:35