If $x^5 ≡ 2 $(mod $17$) then $x ≡ 15 $(mod $17$)

Question

How am I suppose to reduce the $x^5$ Should I write the entire table ?

Answer 1

FLT $x^{16}\equiv 1 \mod 17$

So $(x^5)^3x \equiv 8x \equiv 1 \mod 17$

So $16x \equiv -x \equiv 2 \mod 17$

So $x \equiv -2 \equiv 15 \mod 17$

Answer 2

You probably know that $x^{p-1}\equiv 1 \pmod p$ for non-zero $x$. So in our case the exponent we have, $5$, is relatively prime to $p-1=17-1=16$. So there exist integers $a,b$ such that $5a+16b=1$. (E.g. $a=-3,b=1$) does So $x\equiv x^{5a+16b}\equiv x^{5a}(x^{16})^b\equiv x^{5a}\equiv (x^5)^a\equiv 2^{-3}\equiv 2^{16-3}\equiv 2^{13} \pmod {17}$.

Now $2^{13}$ reduces to $15$ modulo 17.