Number theory - Primitive root of $338$

Question

Im having problem $338$ root. I know it has a root because $13^2\times2=338$ but what is the correct way to find it??

Answer 1

We know from here and here, if $ord_pa=d, ord_{(p^2)}a= d$ or $pd$

So, if $a$ is a primitive root of $13^2\iff ord_{13^2}a=\phi(13^2)=13(13-1)$

$ord_{13}a$ must be $13-1=12\implies a$ must be a primitive root of $13$

Starting with $2,$ the minimum natural number that is co-prime with $13$

$2^2=4,2^3=8,2^4=16\equiv3\pmod{13},2^6=64\equiv-1\pmod{13}\implies ord_{13}=12$

$\implies 2$ a primitive root of $13$

Now, $2^6=13\cdot5-1$ $\implies 2^{12}=(2^6)^2=(13\cdot5-1)^2=13^2\cdot5^2+1-2\cdot13\cdot5\not\equiv1\pmod{13^2}$

So, $ord_{13^2}2\ne 12\implies ord_{13^2}2=12\cdot13=\phi(13^2)$

$\implies 2$ a primitive root of $13^2$

Now, $\phi(2\cdot13^2)=\phi(2)\phi(13^2)=13(13-1)$

But $2$ can not be a primitive root of $2\cdot13^2=338$ as $(2,338)=2>1$

Now, we know $ord_nb=ord_n(b+k\cdot n )\implies ord_{13^2}(2+k\cdot13^2)=13\cdot12$ where $k$ is any integer.

Again $ord_2(2+k\cdot13^2)=1$ if $(2+k\cdot13^2)$ is odd

Again if $ord_{m_1}a=d_1, ord_{m_2}a=d_2,$

we can prove $ord_{lcm(m_1,m_2)}a=lcm(d_1,d_2)$

lcm$(2,13^2)=2\cdot13^2=338$ and lcm$(1,12\cdot13)=12\cdot13$

So, $ord_{2\cdot13^2}(2+k\cdot13^2)=12\cdot13=\phi(2\cdot13^2)$ if $k$ is odd

$(2+k\cdot13^2)$ will be odd if $k$ is odd=$2r+1$ where $r$ is any integer.

So, one of the primitive roots of $2\cdot13^2=338$ is $2+(2r+1)169\equiv2+169\pmod{338}\equiv171$

We know from here, $ord_ma=d, ord_m(a^t)=\frac{d}{(d,t)}$ where integer $t>0$

So, $ord_ma=d=ord_m(a^t)$ if $(d,t)=1$

$ord_{338}(171)=\phi(338)=12\cdot13=156$

So, all the primitive roots of $338$ are $171^t$ where integer $0<t<156$ and $(t,156)=1$