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How to calculate the great common factor between $2^{2n+1}-1$ and $2^{4m+2}+1$, where $n$ and $m$ are positive numbers.

We guess that: the great common factor is $1$.

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    $\begingroup$ This question is from studying the order of the finite non-abelian simple group $^2B_2(q)$ $\endgroup$
    – C. Simon
    Commented May 16, 2023 at 15:13
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    $\begingroup$ If that is the context, maybe you should say something about it in your question. Questions of the form "Prove that ...." just sound like homework questions. $\endgroup$ Commented May 16, 2023 at 15:14

2 Answers 2

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Suppose that $p$ is a prime number that divides $2^{2n+1}-1$. This means that $2^{2n+1}\equiv 1\pmod{p}$. Consequently, the order of $2$ modulo $p$ must be an odd number dividing $2n+1$.

On the other hand, if $q$ is a prime number that divides $2^{4m+2}+1$, then this means $2^{4m+2}\equiv -1\pmod{q}$. Hence $2^{8m+4}\equiv 1\pmod{q}$. Thus, the order of $2$ modulo $q$ must be of the form $4k$ for some $k\mid (2m+1)$.

Thus, your two numbers cannot share any prime divisors.

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In general we have $$\gcd(2^a + 1, 2^b - 1) = \frac{2^{\gcd(2a,b)}-1}{2^{\gcd(a,b)}-1},$$ which is evaluates to $1$ for $a=4m+2$ and $b=2n+1$.

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