Question

# Let p be an odd prime. Show that the congruence $x^{4} \equiv-1(\bmod p)$ has a solution if and only if p is of the form 8k + 1.

Solution

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Let $p$ is an odd prime and we have $\phi(p) = p - 1$ is even. We will write $d = (4,p-1)$. If we look at theorem 9.17, we will know that $x^4\equiv-1$ (mod $p$) has a solution if and only if $(-1)\dfrac{\phi(p)}{d}\equiv1$ (mod $p$). Coming back from the order of $-1$ mod $p$ is 2, so we must have $2\vert\dfrac{p-1}{d}$. So, there exist $k$ such that $2k = \dfrac{p-1}{(p-1,4)}$.

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