(Kazakstan 2003). Let $\left(a_{n}\right)$ and $\left(b_{n}\right)$ be the sequences defined by $a_{0}=b_{0}=0$ and $a_{n}=a_{n-1}^{2}+3$ and $b_{n}=b_{n-1}^{2}+2^{n}$.

Compare the numbers $a_{2003}$ and $b_{2003}$.
. Since $a_{1}=3$ and $a_{n}>a_{n-1}^{2}$ for all integers $n \geqslant 1$, an obvious recurrence leads to $a_{n}>2^{2^{n-1}}$ for all $n \geqslant 1$.

Another immediate recurrence allows us to assert that $2^{n}>n+3$ for all $n \geqslant 3$. Combining these two inequalities, we deduce that, for all $n \geqslant 3$, we have $a_{n}^{2}>2^{2^{n}}>2^{n+3}$, and thus $\frac{1}{4} a_{n}^{2}>2^{n+1}$. (1)

We now prove by induction that $b_{n}<\frac{1}{2} a_{n}$ for all $n \geqslant 3$:

(B) We have $a_{3}=147$ and $b_{3}=72<\frac{1}{2} a_{3}$.

Let $n \geqslant 3$ be a fixed integer for which we assume that $b_{n}<\frac{1}{2} a_{n}$. Then

$$
\begin{aligned}
b_{n+1} & =b_{n}^{2}+2^{n+1} \\
& <\frac{1}{4} a_{n}^{2}+2^{n+1} \text { from the induction hypothesis }
\end{aligned}
$$

```
\(<\frac{1}{2} a_{n}^{2}\) from (1)
\(<\frac{1}{2} a_{n}^{2}+\frac{3}{2}\)
\(=\frac{1}{2} a_{n+1}\), which proves the desired inequality at rank \(n+1\).
```

It follows obviously that $b_{2003}<a_{2003}$.