Given a monic quadratic polynomial $Q(x)$, define \[ Q_n (x) = \underbrace{Q(Q(\cdots(Q(x))\cdots))}_{\text{compose $n$ times}} \]
for every natural number $n$. Let $a_n$ be the minimum value of the polynomial $Q_n(x)$ for every natural number $n$. It is known that $a_n > 0$ for every natural number $n$ and there exists some natural number $k$ such that $a_k \neq a_{k+1}$.
(a) Prove that $a_n < a_{n+1}$ for every natural number $n$.
(b) Is it possible to satisfy $a_n < 2021$ for every natural number $n$?

[i]Proposed by Fajar Yuliawan[/i]
### Part (a)
1. **Define the polynomial and initial conditions:**
   Let \( Q(x) = (x - a)^2 + a_1 \) where \( Q(a) = a_1 \). We are given that \( a_n > 0 \) for every natural number \( n \) and there exists some \( k \) such that \( a_k \neq a_{k+1} \).

2. **Claim 1: \( a_1 > a \)**
   - **Proof by contradiction:**
     Suppose \( a \geq a_1 \). Notice that \( Q(a) = a_1 \). Assume \( Q_n(b_n) = a_1 \) where \( b_n \geq a \) for some \( n \geq 1 \). Let \( b_{n+1} = a + \sqrt{b_n - a_1} \geq a \), hence \( Q(b_{n+1}) = b_n \) and
     \[
     Q_{n+1}(b_{n+1}) = Q_n(Q(b_{n+1})) = Q_n(b_n) = a_1.
     \]
     Since \( Q_n(x) \geq a_1 \) for every natural number \( n \), we obtain that \( a_n = a_1 \) for all \( n \), a contradiction! Therefore, \( a_1 > a \).

3. **Claim 2: \( a_n = Q_n(a) \)**
   - **Proof by induction:**
     For \( n=1 \), it's true since \( a_1 = Q_1(a) \). Assume \( a_n = Q_n(a) \) for some \( n \geq 1 \). Thus, \( Q_n(x) \geq Q_n(a) \geq a_1 > a \), so
     \[
     Q_{n+1}(x) = Q(Q_n(x)) = (Q_n(x) - a)^2 + a_1 \geq (Q_n(a) - a)^2 + a_1 = Q_{n+1}(a).
     \]
     So \( a_{n+1} = Q_{n+1}(a) \). Claim 2 is proven by induction.

4. **Claim 3: \( a_{n+1} > a_n \) for every natural number \( n \)**
   - **Proof by induction:**
     Since \( a_1 > a \), then
     \[
     a_2 = Q_2(a) = Q(a_1) = (a_1 - a)^2 + a_1 > a_1 > a.
     \]
     Suppose it holds that \( a_{n+1} > a_n > a \) for some natural number \( n \), then
     \[
     a_{n+2} = Q(a_{n+1}) = (a_{n+1} - a)^2 + a_1 > (a_n - a)^2 + a_1 = a_{n+1} > a.
     \]
     So claim 3 is proven by induction. \(\blacksquare\)

### Part (b)
1. **Construct a specific polynomial:**
   Take \( Q(x) = x^2 + \frac{1}{4} \). Notice that \( a_1 = \frac{1}{4} \) and
   \[
   a_{n+1} = Q_{n+1}(0) = Q(Q_n(0)) = Q(a_n) = a_n^2 + \frac{1}{4}.
   \]

2. **Claim: \( a_n < \frac{1}{2} \) for every \( n \geq 1 \)**
   - **Proof by induction:**
     It obviously holds for \( n = 1 \). If \( a_n < \frac{1}{2} \), then
     \[
     a_{n+1} = a_n^2 + \frac{1}{4} < \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.
     \]
     So the claim is proven by induction. \(\blacksquare\)

The final answer is \( \boxed{ a_n < 2021 } \) for every natural number \( n \).