Let $p>13$ be a prime of the form $2q+1$, where $q$ is prime. Find the number of ordered pairs of integers $(m,n)$ such that $0\le m<n<p-1$ and
\[3^m+(-12)^m\equiv 3^n+(-12)^n\pmod{p}.\]

[i]Alex Zhu.[/i]

[hide="Note"]The original version asked for the number of solutions to $2^m+3^n\equiv 2^n+3^n\pmod{p}$ (still $0\le m<n<p-1$), where $p$ is a Fermat prime.[/hide]
1. **Given Conditions and Initial Setup:**
   - Let \( p > 13 \) be a prime of the form \( p = 2q + 1 \), where \( q \) is also a prime.
   - We need to find the number of ordered pairs of integers \((m, n)\) such that \( 0 \leq m < n < p-1 \) and
     \[
     3^m + (-12)^m \equiv 3^n + (-12)^n \pmod{p}.
     \]

2. **Primitive Root Analysis:**
   - We claim that exactly one of \( 3 \) and \( -12 \) is a primitive root modulo \( p \).
   - The order of any residue \( x \) modulo \( p \) must divide \( p-1 = 2q \), so it must be one of \( 1, 2, q, 2q \).
   - If the order is \( 1 \) or \( 2 \), then the residue is \( \pm 1 \), which is not the case here.
   - Thus, the order of \( x \) is either \( q \) or \( 2q \).

3. **Legendre Symbol and Primitive Root:**
   - Using the Legendre symbol, we have:
     \[
     \left( \frac{x}{p} \right) \equiv x^{\frac{p-1}{2}} \pmod{p}.
     \]
   - A residue \( x \neq \pm 1 \) is a primitive root if and only if \( \left( \frac{x}{p} \right) = -1 \).
   - Since \( p \equiv 3 \pmod{4} \), we have:
     \[
     \left( \frac{3}{p} \right) \left( \frac{-12}{p} \right) = \left( \frac{-36}{p} \right) = \left( \frac{-1}{p} \right) = -1.
     \]
   - Therefore, one of the Legendre symbols is \( -1 \) and the other is \( 1 \).

4. **Identifying the Primitive Root:**
   - Let \( g \in \{3, -12\} \) be the primitive root, and let \( h \) be the other one.
   - Since \( h \) is a square, we must have \( h = g^{2k} \) for some integer \( 1 \leq k < q \) and \( 2k - 1 \neq q \) (since \( h \not\equiv -g \pmod{p} \)).

5. **Restating the Problem:**
   - We need to find the number of ordered pairs \((m, n)\) such that:
     \[
     g^m + (g^m)^{2k} \equiv g^n + (g^n)^{2k} \pmod{p}.
     \]
   - Since \( g^m \) uniquely attains each value in the range \([1, p-1]\) modulo \( p \), we are looking for pairs \((x, y)\) such that \( 1 \leq x, y \leq p-1 \) and:
     \[
     x + x^{2k} \equiv y + y^{2k} \pmod{p}.
     \]

6. **Solving the Restated Problem:**
   - Rearrange the equation:
     \[
     x^{2k-1} \equiv \frac{(y/x)^{2k} - 1}{1 - (y/x)} \pmod{p}.
     \]
   - The right side is defined since \( x \neq y \).
   - If \( y/x \equiv -1 \pmod{p} \), then \( x = 0 \), which is not allowed.
   - Thus, \( y/x \in \{2, \ldots, p-2\} \pmod{p} \).

7. **Counting the Solutions:**
   - For each value of \( y/x \), \( x \) is uniquely determined since we may raise both sides to the power \( (2k-1)^{-1} \pmod{2q} \).
   - There are \( p-3 \) values of \( y/x \), so there are \( p-3 \) solutions.

\(\blacksquare\)

The final answer is \( \boxed{ q-1 } \).