3
$\begingroup$

I'm interested in knowing if $2^x \equiv -1$ mod $24n-1$ has a solution. I'm also interested in knowing in general how to find out which elements of the group $(\textbf{Z}/(24n-1)\textbf{Z})^{\times}$ generate $-1$.

One thing that I considered was using Jacobi symbols. $\bigg(\dfrac{2}{24n-1}\bigg) = 1$, but this doesn't help me if $24n - 1 = pq$, $p$ and $q$ prime, and $\bigg(\dfrac{2}{p}\bigg)$ = $\bigg(\dfrac{2}{q}\bigg) = -1$, as is the case in mod 95.

Another thing that I considered was using the Chinese Remainder Theorem to construct an isomorphism between $(\textbf{Z}/(24n-1)\textbf{Z})^{\times}$ and $(\textbf{Z}/p_1^{k_1})^{\times} \times (\textbf{Z}/p_2^{k_2})^{\times} \times \dots (\textbf{Z}/p_r^{k_r})^{\times}$ where $24n-1 = p_1^{k_1}p_2^{k_2}\dots p_r^{k_r}$. I've figured out that by doing this, I can show that $(\textbf{Z}/(24n-1)\textbf{Z})^{\times}$ is isomorphic under some bijection $\phi$ to $\{1,g\} \times A$, where $A$ is an abelian group, and $\{1,g\}$ is $C_2$. I can construct $\phi$ such that $\phi(-1) = (g,e_A)$, but I don't see a way of showing whether or not $(g,e_A) \in <\phi(2)>$.

I would be able to move forward if I could somehow analyze which subgroups of $(\textbf{Z}/(24\textbf{Z}-1))^{\times}$ contain $-1$, but I don't see how to do this.

Any help would be appreciated. Thank you.

1 Answers 1

1

Write $m=24n-1=p_1^{a_1}\cdots p_k^{a_k}$.

We know that $x$ must be odd, since if $2^x+1$ is divisible by $m$ with $x$ even, then $m\not\equiv 3\pmod{4}$.

Since $x$ is odd, we know that for each $p_i$ that $2$ is a square modulo $p_i$ iff $-1$ is a square modulo $p$. This means that either $p_i\equiv 3\pmod{8}$ or $p_i\equiv 1\pmod{8}$.

So you can't even get $m=8n-1$.