Let $a$ and $b$ be real numbers. Define $f_{a,b}\colon R^2\to R^2$ by $f_{a,b}(x;y)=(a-by-x^2;x)$. If $P=(x;y)\in R^2$, define $f^0_{a,b}(P) = P$ and $f^{k+1}_{a,b}(P)=f_{a,b}(f_{a,b}^k(P))$ for all nonnegative integers $k$.

The set $per(a;b)$ of the [i]periodic points[/i] of $f_{a,b}$ is the set of points $P\in R^2$ such that $f_{a,b}^n(P) = P$ for some positive integer $n$.

Fix $b$. Prove that the set $A_b=\{a\in R \mid per(a;b)\neq \emptyset\}$ admits a minimum. Find this minimum.
1. **Define the function and periodic points:**
   Given the function \( f_{a,b} \colon \mathbb{R}^2 \to \mathbb{R}^2 \) defined by \( f_{a,b}(x, y) = (a - by - x^2, x) \), we need to find the set \( A_b = \{ a \in \mathbb{R} \mid \text{per}(a, b) \neq \emptyset \} \) and its minimum.

2. **Fixed points of \( f_{a,b} \):**
   A point \( P = (x, y) \) is a fixed point of \( f_{a,b} \) if \( f_{a,b}(x, y) = (x, y) \). This gives us the system of equations:
   \[
   \begin{cases}
   a - by - x^2 = x \\
   x = y
   \end{cases}
   \]
   Substituting \( y = x \) into the first equation, we get:
   \[
   a - bx - x^2 = x
   \]
   Simplifying, we obtain:
   \[
   x^2 + (b+1)x + a = 0
   \]

3. **Discriminant condition for real solutions:**
   For \( x \) to be real, the discriminant of the quadratic equation must be non-negative:
   \[
   (b+1)^2 - 4a \geq 0
   \]
   This simplifies to:
   \[
   a \leq \frac{(b+1)^2}{4}
   \]

4. **Existence of periodic points:**
   If \( a \geq \frac{-(b+1)^2}{4} \), then the quadratic equation \( x^2 + (b+1)x + a = 0 \) has real solutions, and thus there exist fixed points of \( f_{a,b} \). Therefore, \( \text{per}(a, b) \neq \emptyset \).

5. **Behavior for \( a < \frac{-(b+1)^2}{4} \):**
   Define \( g \colon \mathbb{R}^2 \to \mathbb{R} \) by \( g(x, y) = x - by \). For \( a < \frac{-(b+1)^2}{4} \), we have:
   \[
   x^2 + (b+1)x \geq \frac{-(b+1)^2}{4} > a
   \]
   This implies:
   \[
   g(f_{a,b}(x, y)) < g(x, y)
   \]
   for every \( (x, y) \in \mathbb{R}^2 \). Therefore, \( g(f_{a,b}^n(P)) < g(f_{a,b}^{n-1}(P)) < \cdots < g(P) \), which means \( f_{a,b}^n(P) \neq P \) for any positive integer \( n \). Hence, there are no periodic points for \( a < \frac{-(b+1)^2}{4} \).

6. **Conclusion:**
   The set \( A_b = \left[ \frac{-(b+1)^2}{4}, +\infty \right) \) admits a minimum value of \( \frac{-(b+1)^2}{4} \).

The final answer is \( \boxed{\frac{-(b+1)^2}{4}} \).