prove that a congruence is solvable

**comssa** · November 14th 2009, 11:08 AM

For all primes p, show that $x^8 \equiv 16 \mbox{ (mod p)}$ is solvable.

From a theorem, the question is equivalent to showing that
$\displaystyle{16^{\frac{p-1}{(8, p-1)}} \equiv 1 \mbox{ (mod p)}}$

And then I am stuck.

**chiph588@** · November 14th 2009, 02:45 PM

There are four options for $(8,p-1)$ since $\{1,2,4,8\}$ is the exhaustive list of factors of $8$ .

If $(8,p-1) = 1$ , it's pretty easy to see $16^{p-1} \equiv 1 \mod{p}$ .

If $(8,p-1) = 2, \; 16^{\frac{p-1}{2}} = 4^{p-1} \equiv 1 \mod{p}$ .

If $(8,p-1) = 4, \; 16^{\frac{p-1}{4}} = 2^{p-1} \equiv 1 \mod{p}$ .

If $(8,p-1) = 8, \; 16^{\frac{p-1}{8}} = 2^{\frac{p-1}{2}}$ .
In this case, $p \equiv 1 \mod{8}$ and hence $2$ is a square modulo $p$ , i.e. $\left(\frac{2}{p}\right) = 1$ . Therefore we can choose $x$ such that $x^2 \equiv 2 \mod{p}$ . So $2^{\frac{p-1}{2}} \equiv (x^2)^{\frac{p-1}{2}} \equiv x^{p-1} \equiv 1 \mod{p}$ .

Therefore $x^8 \equiv 16 \mod{p}$ is always solvable.

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