# Showing $x^8\equiv 16 \pmod{p}$ is understandable for all tops $p$

I'm still making my means along in Niven is Intro to Number Theory, and also the title trouble is offering me a little problem near completion, and also I was wishing a person can aid get me via it.

Currently $x^8\equiv 16\pmod{2}$ is understandable with $x\equiv 0\pmod{2}$, so I think $p$ is a weird prime. From a theory previously in the message,

If $p$ is a prime and also $(a,p)=1$, after that the harmony $x^n\equiv a\pmod{p}$ has $(n,p-1)$ remedies or no remedy according as $a^{(p-1)/(n,p-1)}\equiv 1\pmod{p}$ or otherwise.

So given that $(16,p)=1$, the trouble lowers to revealing that $16^{(p-1)/(8,p-1)}\equiv 1\pmod{p}$ holds for all $p$. I keep in mind that $(8,p-1)$ can just take values $2,4,8$. For $2$, the above equivalence is after that $4^{p-1}\equiv 1\pmod{p}$, which holds true by Fermat is little Theorem. For $4$, it is after that $2^{p-1}\equiv 1\pmod{p}$, which once more holds by FlT. Nonetheless, the instance where $(8,p-1)=8$ is tossing me off. At ideal I see that $16^{(p-1)/8}\equiv 2^{(p-1)/2}\pmod{p}$, yet I'm not exactly sure just how to show this is conforming to $1$ modulo $p$. Possibly there is an extra classy means to do it without considering instances. Many thanks for any kind of understanding.

I generally set this as a workout when educating Number Theory. My tip is to ask the pupils: what are the remedies of $z^8=16$ in the intricate numbers?

**HINT ** $\rm\ \ \ x^8 - 16\ =\ (x^2 - 2)\: (x^2 + 2)\: (x^4 + 4)\:.\ \:$ If the first 2 variables have no origins in $\rm\ \mathbb Z/p\ $ after that $\:2, -2\:$ are nonsquares so their item $-4\: $ is a square, so $\rm\: i = \sqrt{-1} \in \mathbb Z/p\:$. Hence the 3rd variable has an origin given that $\rm\ x^4 + 4\ $ has origins $\rm\: \pm 1\pm i\:$.

One means is to make use of the Legendre symbol identification $2^{(p-1)/2} \equiv (\frac{2}{p}) \equiv (-1)^{(p^2-1)/8} \pmod p$ (for weird tops p), remembering that if $(8,p-1)=8$ after that $p \equiv 1 \pmod 8$ .

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