Discrete Logarithm Problem Question: Discrete Logarithm Problem: Let $g$ be a primitive root for $F_{p}$. Suppose that $x = a$ and $x = b$ are both integer solutions to the congruence $g^{x} \equiv h \pmod{p}$. Prove that $a \equiv b\pmod{p-1}$. 
So far, I have the following as my proof:
Proof: Let $g$ be a primitive root for $F_{p}$. Suppose that $x = a$ and $x = b$ are both integer solutions to the congruence $g^{x} \equiv h\pmod{p}$. Then, $g^{a} \equiv h\pmod{p}$ and $g^{b} \equiv h\pmod{p} \rightarrow g^{a} - h = kp$ and $g^{b} - h = kp$, for some $k \in Z$. By substitution, we have $g^{a} = g^{b} \rightarrow a \log (g) = b \log (g) \rightarrow a = b$.
It doesn't seem as if I'm going in the right direction, but I'm stumped on what other routine I can take. Any ideas?
 A: The definition of a  primitive root (that I learned first anyway) is that $g$ is a generator in $\pmod p$ if and only if the function $f$ :
$$f(x) \equiv g^x \pmod p$$
is a bijection over the domain $D$, for $D=\{1, 2, \dots p-1\}$.  A bijection is a function that is both 


*

*onto, that is it has every value of the domain in its output, and

*invertible, that is it doesn't output a value into the domain more than once


Your question comes down to proving that
$$g^{p - 1} \equiv 1 \pmod p \tag{B}$$
To prove it, assume the opposite for the sake of contradiction.  Then $f(p-1) \ne 1$.  Then because $f$ is onto (by the definition of a generator), there must be some $y$ such that 
$$f(y) = 1$$
So $f(y+1) = g$.  But we know that $f(1) = g$ from the definition of $f$.  But this contradicts the assumption that $f$ is invertible since it outputs $g$ twice.  So by contradiction
$$f(p-1) \equiv 1$$
$$g^{p-1} \equiv 1 \pmod p$$
Can you now use (B) to fill in the rest of the details to establish that $g^a \equiv g^b \pmod p \quad \rightarrow \quad a \equiv b \pmod {p-1}$ ?
Hint 1:

 This is the same as $\forall k ~~ g^{a + (p-1)k} \equiv g^{a} \pmod p$ 

Hint 2:

 Which is the same as $\forall k ~~ g^{a}(g^{p-1})^k \equiv g^{a} \pmod p$

