# prove that a congruence is solvable

• Nov 14th 2009, 11:08 AM
comssa
prove that a congruence is solvable
For all primes p, show that $\displaystyle x^8 \equiv 16 \mbox{ (mod p)}$ is solvable.

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

And then I am stuck.(Worried)
• Nov 14th 2009, 02:45 PM
chiph588@
There are four options for $\displaystyle (8,p-1)$ since $\displaystyle \{1,2,4,8\}$ is the exhaustive list of factors of $\displaystyle 8$.

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

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

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

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

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