I have these two questions which go as:
1) Find $x \in \{0,1...16\}$ such that $x^{13}\equiv2\pmod{17}$
2) Find $x \in \{0,1...22\}$ such that $x^{13}\equiv3\pmod{23}$
I was wondering what the best way to approach these were, I was thinking using the Euclidean Algorithm?
This is what I have so far for 1):
We know hcf$((17-1),13)=1$, so $1=13s-16t$ for some s,t. But not sure how to go further? Maybe Fermat's Little Theroem but again not sure. Thank you!
Using Fermat's little theorem the first equation is equivalent to $$x^{-3}\equiv2\mod{17}$$ $$2x^3\equiv1\mod{17}$$ $$x^3\equiv9\mod{17}$$ $$\therefore x\equiv-2\equiv15 \mod{17}$$
The second example seems like one can only solve with trial and error.
For the second one what you can do is use $$x^{13} \equiv3 \bmod 23$$ and by fermat's little theorem $$x^{22}\equiv 1 \bmod 23$$ Thus we get that $$x^4\equiv x^{22}x^4\equiv x^{26}\equiv3^2\equiv 9 \bmod23$$ cubing that we get that $$x^{12}\equiv16 \bmod23$$ and so you can see that $$x \equiv x^{13}x^{-12} \equiv 3*16^{-1} \bmod23$$ and with a little bit of trial and error you can find that $$16^{-1}\equiv13 \bmod23$$ Thus $$x\equiv 3*13\equiv 16 \bmod 23$$
Not entirely sure this isn't error free, but this general argument should lead to the right answer.
For the second example, $$ 16^{13}\equiv 3 \bmod 23. $$ Also here we can compute $16^{11}\equiv 1\bmod 23$ as descibed in the comments, so that $16^{13}\equiv 16^2\equiv 3\bmod 23$.
