-4 是 mod p 二次剩余, 则 -4 是 mod p 四次剩余

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素数 $p,$ 若 $-4$ 是$\bmod p$ 二次剩余, 则 $-4$ 是$\bmod p$ 四次剩余
证明:
$p=2$显然不满足条件. 设$p>2,$
设$-4≡x^2\pmod p,$
则$(1+x / 2)^2=1+x+x^2 / 4≡x\pmod p,$
所以$-4≡(1+x / 2)^4\pmod p.$

source

A Problem Seminar, D. J. Newman
38. Show that the number 16 is a perfect 8th power $\operatorname{mod}p$ for any prime $p$.
If 2 is a perfect square (quadratic residue) mod $p$, then we are done, and similarly if $-2$ is a perfect square. The point is that if $a$ and $b$ are both not perfect squares, then $a \cdot b$ is. Hence we may assume that $-2 \cdot 2=-4$ is a perfect square, say $x^2\pmod p$, and we may assume $p$ is odd. But then $(1+x / 2)^2=1+x+x^2 / 4≡x\pmod p$ so that $-4$ is a perfect 4 th power for $1+x / 2$), and again we are done. (A curious fact emerges here, that if $-4$ is a perfect square, then it is automatically a perfect fourth power.)

问题:
除了$-4$还有其它的数满足这个性质吗