<s>[INST] Find powers of 3 mod 17 Show that powers of $3\ (\text{mod}\ 17)$ are $3,9,10,13,15,11,16,14,8,7,4,12,2,6,1$.
What is meaning of power of something, how do I proceed this question? Just provide me hint to get start in this problem.
 [/INST] Let us try a very naive approach. I should add that this is precisely the method suggested in  André Nicolas's comment.
You start by $3^0=1$ and multiply both sides by $3$ repeatedly
$$3^0 \equiv 1 \pmod{17}\\
3^1 \equiv 3 \pmod{17}\\
3^2 \equiv 9 \pmod{17}$$
The next result would be $3^3 \equiv 27 \pmod{17}$. But $27\equiv10\pmod{17}$. 
$$
3^3 \equiv 10 \pmod{17}\\
3^4 \equiv 13 \pmod{17}\\
$$
We can also use $3^4 \equiv -4 \pmod{17}$ if it makes computation easier. In this case we either multiply $-4$ or $13$ by $3$. But in both case we get $-12 \equiv 5 \pmod{17}$ and $39 \equiv 5 \pmod{17}$.
$$
3^5 \equiv 5 \pmod{17}\\
3^6 \equiv 15 \pmod{17}\\
3^7 \equiv 11 \pmod{17}\\
3^8 \equiv 16 \pmod{17}\\$$
We continue this way until we get $1$ and the sequence starts to repeat. Or we could notice that $3^8\equiv-1\pmod{17}$, so from this point on we can repeat the same numbers we already have but with the minus sign.
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