finding a primitive root.

Question

It says for part A to Find a primitive root r of 38? Im not sure if I did it right.

I first calculated $\phi(38)=\phi(19*2)=18$. So there are 18 numbers that are relatively prime to 38. Listing them out we get 1,3,5,7,9,11,13,15,17,21,... so on. So I decided to test out 1. But $ord_1 38$ does not equal $\phi(38)$. But $ord_3 38=\phi(38)$.

So I calculated it until i found a power of 3 that was congruent to $1 \pmod {38}$ So I get $$3^1≡ 3 \pmod {38}$$ $$3^2 ≡ 9 \pmod {38}$$ $$3^3 ≡ 27 \pmod {38}$$ $$3^4 ≡ 5 \pmod {38}$$ $$ 3^5 ≡ 15 \pmod {38}$$ $$3^6 ≡ 7\pmod {38}$$ $$3^7 ≡ 21 \pmod {38}$$ $$3^8 ≡ 25 \pmod {38}$$ $$3^9 ≡ -1 \pmod {38}$$ $$3^{16} ≡ 17 \pmod {38}$$ $$3^{17} ≡ 13 \pmod {38}$$ $$3^{18} ≡ 1 \pmod {38}$$ So $ord_3 38=18=\phi(38)$.

I took the liberty of fixing the spacing in the displayed congruences. Apparently you had not learned about the \pmod macro of TeX. Take a peek at the source code now to see how it is used. — Jyrki Lahtonen, May 01 '13 at 05:21

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

1

As $\phi(19)=18$ and $ord_{19}2$ must divide $\phi(19),$ we only need to test for the powers $1,2,3,6,9,18$

$3^2=9;3^3=27\equiv-11\pmod{38};3^6\equiv(-11)^2\equiv121\equiv7\pmod{38}$

So, $3^9=3^3\cdot3^6\equiv(-11)(7)\equiv-77\equiv-1\pmod{38}$

$\implies 3 $ is a primitive root $\pmod{38}$

Alternatively, let's start with finding a primitive root of $19$

Starting with $2$, the smallest positive integer $>1$

$2^2=4,2^3=8,2^6=64\equiv7\pmod{19},$ $2^9=2^3\cdot2^6\equiv 8\cdot7\pmod{19}\equiv-1$

$\implies 2 $ is a primitive root $\pmod{19}$

Now, as $(2,38)=2>1,2$ can not be a primitive root $\pmod {38}$

But, $ord_{19}(2+19k)=ord_{19}2=18$

Now, $ord_2(2+19k)$ will be $1$ if $(2+19k)$ is odd i..e, if $k$ is odd

So, $ord_{(2\cdot19)}(2+19k)=$lcm$(1,18)=18=\phi(38)$ if $k$ is odd

Putting $k=2r+1$ where $r$ is any integer, $2+19k=2+19(2r+1)\equiv21\pmod{38}$

edited Apr 13 '17 at 12:21

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answered May 01 '13 at 05:14

lab bhattacharjee

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But $3^{18}$ doesn't work in my case? – user60887 May 01 '13 at 05:17
@user60887, "doesn't work" means? ALso, please find the edited answer – lab bhattacharjee May 01 '13 at 05:22
oh what I mean was i found $3^{18} ≡ 1 \pmod {38}$. Maybe I should ask if my calculations were correct. – user60887 May 01 '13 at 05:23
@user60887, we don't need to check that as this is proven by http://mathworld.wolfram.com/EulersTotientTheorem.html . But you missed to test for powers $6,9$ – lab bhattacharjee May 01 '13 at 05:25
yes i forgot about. So it is correct. So my r=3 then? since 3 is the primitive root. – user60887 May 01 '13 at 05:26
oops i better do that then. – user60887 May 01 '13 at 05:27
@user60887, find my 1st approach. Also, there will be $\phi(38)=18$ primitive roots $\pmod{38}$ . http://en.wikipedia.org/wiki/Primitive_root_modulo_n – lab bhattacharjee May 01 '13 at 05:30

finding a primitive root.

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