Problem - 4181
Let $p$ be an odd prime, and integer $a$ has multiplicative order of $2k$ modulo $p$, then $a^k\equiv -1\pmod{p}$.
Given $a^{2k}\equiv 1\pmod{p}$, we have $p\mid (a^k+1)(a^k-1)$. Because $p$ is an odd prime, it must hold that either $p\mid (a^k+1)$ or $p\mid (a^k -1)$ which means $a^k\equiv \pm 1\pmod{p}$. However $p^k\equiv 1\pmod{p}$ cannot hold because $2k$ is the multiplicative order. Therefore $a^k\equiv -1\pmod{p}$.