<s>[INST] The equation $x^k = -1$ over finite fields Let $k$ be a positive integer and let $F_q$ be a finite field with $q$ elements, and consider the equation $X^k + 1 = 0$. When does this equation has a solution in $F_q$?  For example if $k =2$ and $q$ is a prime congruent to $3 \pmod{4}$ the equation has no solution in $F_q$. Does the same hold for other values of $k$ too?
 [/INST] If $q$ is even then $x = 1$ is a solution, so assume $q$ is odd. Write $k = 2^a b$ where $b$ is odd. We can solve $x^k=-1$ iff we can solve $y^{2^a} = -1$: if $x^k = -1$ then $(x^b)^{2^a} = -1$, and conversely if $y^{2^a} = -1$ then $y^k = (-1)^b = -1$. Now $y^{2^a} = -1$ if and only if $y$ has order $2^{a+1}$, and because $F_q^\times$ is cyclic of order $q-1$ such an element exists if and only if $2^{a+1} \mid q-1$.
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