Generalizing the Theorem: Number of Solutions to x^k ≡ 1 (mod n)

[Arian.D]

Homework Statement

Prove that the equation x^k \equiv 1 (mod p) has exactly k solutions if k|p-1.

I'm also curious to know if it's possible to generalize this theorem this way:
Prove that the equation x^k \equiv 1 (mod n) has exactly k solutions if k|\varphi(n) where \varphi(n) indicates the Euler's totient function.

Homework Equations

The Attempt at a Solution

Well, this is what I've done so far, but at some point I couldn't progressed any futher:

Consider the polynomial x^k-1. By a theorem we know that this polynomial would have k solutions at the maximum because its degree is k. Now, since k|p-1, there exists an integer number t such that: kt=p-1. we'll have:

x^{p-1} - 1 = x^{kt} -1 = (x^k - 1)(x^{k(t-1)} + x^{k(t-2)} + ... + x^k + 1)

well, since p is a prime number, by Euler's theorem (or Fermat's little theorem to be more precise) we know that x^{p-1} - 1 \equiv 0 (mod p) would have exactly p-1 solutions (since all the numbers from 1 to p-1 would be prime relative to p).

now we'll have:

x^{p-1} - 1 \equiv 0 (mod p)
x^{kt} -1 = (x^k - 1)(x^{k(t-1)} + x^{k(t-2)} + ... + x^k + 1) \equiv 0 (mod p)
p | (x^k - 1)(x^{k(t-1)} + x^{k(t-2)} + ... + x^k + 1)
since p is prime, p divides x^k -1 or x^{k(t-1)} + x^{k(t-2)} + ... + x^k + 1, therefore:

x^k - 1 \equiv 0 (mod p)
x^{k(t-1)} + x^{k(t-2)} + ... + x^k + 1 \equiv 0 (mod p)

Now this is where I'm stuck. I don't know how to go further. I'm not sure if what I've done is useful to conclude anything either.
Any helps would be appreciated.

 

[Arian.D]

By the way, I found a counter example to that general statement.
x²=1 (mod 8) has 4 solutions even though that 2 divides 4.

 

[Millennial]

I think p is prime in this question.

 

[Arian.D]

Yes, it is.
I'm sorry if having not said that caused confusion because I thought that was obvious from my solution to the problem.

 

[Arian.D]

No ones knows how to prove it?