How many cube roots does 1 have modulo 162? How many cube roots does $1$ have modulo $162$
this is equivalent to saying how many solutions to $x^3 \equiv 1 $ mod$162$
all my attempts are leading a dead end any help appreciated 
the fact that 162 is Non-Carmichael doesn't seem to be an use 
also this is at elementary number theory level so no group theory please 
thanks
 A: You have $162 = 2 \times 3^4$, so
$$(\mathbb Z/162 \mathbb Z)^\times = (\mathbb Z/2 \mathbb Z)^\times \times (\mathbb Z/3^4 \mathbb Z)^\times = (\mathbb Z/3^4 \mathbb Z)^\times.$$
Now, $(\mathbb Z/3^4 \mathbb Z)^\times = (\mathbb Z/3 \mathbb Z)^\times \times \mathbb Z/3^3 \mathbb Z$ is cyclic of order $\varphi(3^4)=54$, being a product of two cyclic groups of relatively prime orders $2$ and $3^3$. Since $3 \mid 54$, there are exactly $3$ elements in $(\mathbb Z/3^4 \mathbb Z)^\times$ of order $3$. 
A: $162=3^4*2$
So you want
$x^3\equiv 1 \bmod 3^4$ and $x^3\equiv1 \bmod 2$, by the Chinese remainder theorem.
So $x^3\equiv1 \bmod 2.$
For the other part, $3^4$ has a cyclic multiplicative subgroup since $\lambda(3^4)=\varphi(3^4)$; by Carmichael's theorem, we have $\varphi(3^4)=54$. Find a primitive root $p$ (we know one exists); your solutions will be $1,p^{18},p^{36}.$
$2$ is a primitive root: So we want $1,2^{18}$ and $2^{36}$. We can calculate both of these via exponentiation by squaring.
We get $2^{18}\equiv28$ and $2^{36}\equiv 55$.
So the solutions are $1,55$ and $109$.
A: As $162$ is of the form $2p^m$ where prime $p=3,m=4$
it has at least one primitive root $g$(say).
Applying Discrete Logarithm on $x^3\equiv1\pmod{162}\ \ \ \ (1)$, 
$3$indx$_gx\equiv0\pmod{54}\ \ \ \ (2)$ 
as $\phi(162)=\phi(2)\phi(81)=1\cdot3^{4-1}(3-1)=54$
Now using  Linear Congruence Theorem, as $(3,54)=3$ divides $0,$ equation$(2),$ has consequently $(1)$ will have exactly three solutions.
A: I believe the question is equivalent to asking what $x$ satisfies $x^3 \equiv 1 \mod 162$
I wrote the simple script:
for x = 1 to 161
    if ((x*x*x) mod 162 == 1) print x

This gave me 1, 55, and 109.
