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Show that $16$ is a $8th$ power $\mod{}$ $p$ for any prime number $p$.

I have no idea how to approach this.

I tried,

$$a^8\equiv16\pmod{p}$$

$$(a^4+4)(a^4-4)\equiv 0 \pmod{p}$$

$$a^4 \equiv \pm4\pmod{p}$$

$$a^4 \equiv 2,p-2\pmod{p}$$

But this doesn't seem to lead anywhere.

Guy
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