How many solutions of $x^{p+1} \equiv 1 \mod p^{2017}$ How many solutions does $x^{p+1} \equiv 1 \mod p^{2017}$ have in set $\left\{0,1,...,p^{2017}-1 \right\}$?
$p$ is prime > 2. 
My observations 
$1$ is one of solutions of given equation. 
$p$ is prime so:
$$x^{p-1} \equiv 1 \mod p $$
$$x^{p+1} \equiv x^2 \mod p $$
but I have problem with increase power of $p$ to $2017$ without breaking the rest... I know that I can do:
$$x^{p+1} p^{2016} \equiv x^2 p^{2016} \mod p^{2017} $$
and due to this is group I can say that there is one element such that 
$$p^{2016} \cdot t \equiv 1 \mod p^{2017} $$
but after multiplication from both side I get starting equation...
 A: Using Euler's Totient Theorem, $$x^{\phi(p^n)}\equiv1\pmod{p^n}, \ge1$$
If modulo order org$\displaystyle_{(p^n)}x=d,$
$d$ must divide $\phi(p^n)\ \ \ \  (1)$
and again $x^{p+1}\equiv1\pmod{p^n}\implies d$ must divide $p+1\ \ \ \  (2)$
$(1),(2)\implies d$ must divide $(p+1,\phi(p^n))=(p+1,p-1)=2$
So, if $d=2,x\equiv\pm1$ there two solutions namely $x\equiv\pm1$
What if $d=1?$
A: Use http://mathworld.wolfram.com/DiscreteLogarithm.html
$(p+1)\cdot$ind$_gx\equiv\phi(p^n)$
where $g$ is one of the primitive roots of $p^n,n\ge1$
Use https://proofwiki.org/wiki/Solution_of_Linear_Congruence to find the number of solutions to be
$(p+1,p^{n-1}(p-1))=(p+1,p-1)=2$ for odd $p$
A: Clearly $x$ must be coprime to $p^k$, so we're looking exactly for how many elements of the multiplicative group modulo $p^{2017}$ satisfy $x^{p+1}=1$.
The central fact we need to use is that the multiplicative group is cyclic when the modulus is a power of an odd prime. I don't recall offhand how to prove this, and you say you don't know about primitive roots, so the following is probably not the answer you need, but it would at least tell you the answer.
The multiplicative group has order $(p-1)p^{2016}$, so equivalently we're looking for solutions to
$$ y(p+1) \equiv 0 \pmod{(p-1)p^{2016}} $$
where we fix a generator $g$ and set $x=g^y$.
By the Chinese Remainder Theorem this means we have to solve
$$ yp+y \equiv 0 \pmod{p-1} \qquad yp+y \equiv 0 \pmod{p^{2016}} $$
The first of these is the same as
$ 2y \equiv 0 \pmod{p-1} $
and since $p-1$ is even, this has exactly two solutions, namely $y=0$ and $y=\frac{p-1}{2}$.
For modulus $p^{2016}$ it is easy to prove by induction on $k$ that
$$ yp+p\equiv 0 \pmod{p^k} \implies p^k\mid y $$
so this equation has the single solution $y=0\pmod{p^{2016}}$.
The number of solutions to the combined equation is therefore $2\cdot 1=2$.
