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Is this true, that for all integer $n>0$, but $n=1$ and $n=6$, there exists a prime $p$ such that $n=ord_p(2)$, where $ord_p(2)$ is the multiplicative order of $2$ modulo $p$?

If not, what is the next $n$, larger than $6$, for which there is no prime $p$ such that $n=ord_p(2)$ ?

If yes, can it be proven in an elementary way? and what makes $6$ special?

$p$ must be a divisor of $2^{ord_p(2)}-1$. The prime divisors of $2^6-1$ are $3$ and $7$, but neither $ord_3(2)(=2)$ nor $ord_7(2)(=3)$ is equal to $6$. For all other $n$ that I have tested, so far, there is a prime divisor $p$ of $2^n-1$ so that $n=ord_p(2)$.

René Gy
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1 Answers1

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What you are looking for is known as a special case of Zsigmondy's theorem. The theorem says that for $a,b>0$ and $n\geq2$, $a^n-b^n$ has at least one prime divisor that does not divide $a^k-b^k$ for all $k<n$, except in the cases $2^6-1^6$ and the case where $n=2$ and $a+b$ is a power of $2$.

The case $a=2$ and $b=1$ is also known as Bang's theorem.

Here is an Elementary proof of Zsigmondy's theorem. Understanding the proof requires some knowledge of cyclotomic polynomials, a good introduction to them is available here. I don't think the special case you're asking for has a shorter proof, but I could be wrong.

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Bart Michels
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