Consider solutions to the equation \[x^2-cx+1 = \dfrac{f(x)}{g(x)},\] where $f$ and $g$ are polynomials with nonnegative real coefficients. For each $c>0$, determine the minimum possible degree of $f$, or show that no such $f,g$ exist. 

[i]Proposed by Linus Hamilton and Calvin Deng[/i]
1. **Assume $\gcd(f, g) = 1$ and express $f(x)$ and $g(x)$ as polynomials:**
   \[
   f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0
   \]
   \[
   g(x) = x^{n-2} + b_{n-3}x^{n-3} + \cdots + b_1x + b_0
   \]

2. **Consider the case when $c \ge 2$:**
   - Evaluate the equation at $x = 1$:
     \[
     \frac{f(1)}{g(1)} = 2 - c \le 0
     \]
   - Since $f$ and $g$ have nonnegative coefficients, $f(1) \ge 0$ and $g(1) \ge 0$. Therefore, $f(1) = 0$.
   - This implies:
     \[
     \sum_{i=0}^{n-1} a_i = 0
     \]
   - Given that all $a_i \ge 0$, it follows that $a_i = 0$ for all $i$, leading to $f \equiv 0$, which is a contradiction. Hence, $c \ge 2$ fails.

3. **Consider the case when $c < 2$:**
   - We need to find $f$ and $g$ such that the equation holds. We claim that the minimum possible degree of $f$ is:
     \[
     \boxed{\left\lceil \frac{\pi}{\cos^{-1}(c/2)} \right\rceil}
     \]
   - Let $c = 2 \cos \theta$ where $\theta \in (0, \pi/2)$. We need to show that we can find $f$ with degree $n = \left\lceil \frac{\pi}{\theta} \right\rceil$.

4. **Verify the polynomial identity:**
   - Consider the polynomial identity:
     \[
     (x^2 - (2 \cos \theta)x + 1) \left( x^{n-2} + \frac{\sin(2\theta)}{\sin \theta} x^{n-3} + \cdots + \frac{\sin((n-1)\theta)}{\sin \theta} \right) = x^n - \frac{\sin(n\theta)}{\sin \theta} x + 1
     \]
   - To prove this, we need to check that the coefficients of $x^k$ for $2 \le k \le n-1$ are zero:
     \[
     a_k = \frac{\sin((n-1-k)\theta)}{\sin \theta} - 2 \cos \theta \frac{\sin((n-k)\theta)}{\sin \theta} + \frac{\sin((n-k+1)\theta)}{\sin \theta} = 0
     \]
   - This can be verified using trigonometric identities and sum-to-product formulas.

5. **Show that $n \ge \frac{\pi}{\theta}$ is required:**
   - Since $e^{i\theta}$ is a root of $x^2 - cx + 1$, it must also be a root of $f(x)$. Therefore, the imaginary part of the sum of the roots must be zero:
     \[
     \sum_{j=0}^n a_j \sin(j\theta) = 0
     \]
   - If $n < \frac{\pi}{\theta}$, then all terms except the first are positive, leading to $a_i = 0$ for all $i \ge 0$, implying $f$ is constant, which is not possible. Hence, $n \ge \frac{\pi}{\theta}$ is necessary.