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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.

Martin Sleziak
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Abhishek Kumar
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    See http://math.stackexchange.com/questions/1465046/to-show-congruence – lab bhattacharjee Oct 05 '15 at 06:09
  • @labbhttacharjee how does it relate to this question ? – Abhishek Kumar Oct 05 '15 at 06:13
  • we calculated $3^4,3^8\pmod{17}$ there, right? – lab bhattacharjee Oct 05 '15 at 06:14
  • The $n$-th power of $a$ is $a^n$ as usual. – Empiricist Oct 05 '15 at 06:16
  • got it....., thanks @lab bhattatacharjee – Abhishek Kumar Oct 05 '15 at 06:18
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    The long way: Calculate the powers of $3$ modulo $17$, one after the other. We get $3$, $9$, $27$ which mod $17$ simplifies to $10$. Now multiply by $3$, reduce mod $17$. We get $30$, which is $13$, Multiply by $3$, reduce modulo $17$. We get $5$. Multiply by $3$, reduce (nothing happens). We get $15$. Continue. But there is a shortcut. If we find that $3^8$ is not congruent to $1$, we know all numbers from $1$ to $16$ will occur as residues of powers of $3$. And $3^8$ can be found in three squarings modulo $17$. – André Nicolas Oct 05 '15 at 06:23

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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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Martin Sleziak
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