<s>[INST] if $4^{\alpha} \equiv k+1 \pmod{2k+1}$ prove there is no $\beta$ where $4^{\beta} \equiv k\pmod{2k+1}$. Suppose that $3 \nmid 2k+1$ and there is $\alpha$ with $4^{\alpha} \equiv k+1 \pmod{2k+1}$ where $0 \leq \alpha \leq k$. I want to prove that there is no $\beta$, $0\leq \beta \leq k$ such that $4^{\beta} \equiv k \pmod{2k+1}$. I have used the contradiction method but there was no result, it will be nice if you help me about it with your ideas, thanks.
 [/INST] Let $m$ be the order of $2$ modulo $2k+1$, $m = \min \bigl\{ n\in \mathbb{N}\setminus \{0\} : 2^n \equiv 1 \pmod{2k+1}\bigr\}$.
If $2^{2\alpha} = 4^\alpha \equiv k+1 \pmod{2k+1}$, then $2^{2\alpha+1} \equiv 1 \pmod{2k+1}$, so $m\mid (2\alpha + 1)$, hence $m$ is odd.
If on the other hand $2^{2\beta} = 4^\beta \equiv k \pmod{2k+1}$, then $2^{2\beta + 1} \equiv -1 \pmod{2k+1}$ and $2^{2(2\beta+1)} \equiv 1 \pmod{2k+1}$. So then $m \mid 2(2\beta+1)$ but $m\nmid (2\beta+1)$, and that implies $2\mid m$, i.e. $m$ is even.
Thus at most one of the congruences $4^\alpha \equiv k+1 \pmod{2k+1}$ and $4^\beta \equiv k \pmod{2k+1}$ can be solvable for any $k$.
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