China National Olympiad

Let $n = p_1^{a_1} p_2^{a_2} \cdots p_t^{a_t}$ be the prime factorization of $n$. Define $\omega(n) = t$ and $\Omega(n) = a_1 + a_2 + \ldots + a_t$. We aim to prove or disprove the following statements for any fixed positive integer $k$ and positive reals $\alpha$ and $\beta$:

i) $\frac{\omega(n+k)}{\omega(n)} > \alpha$

ii) $\frac{\Omega(n+k)}{\Omega(n)} < \beta$.

To address statement (i), we need to show that $\limsup_{n \to \infty} \frac{\omega(n+k)}{\omega(n)} = \infty$. This can be demonstrated by considering the behavior of $\omega(p+k)$ for prime $p$. Specifically, we show that $\limsup_{p \to \infty} \omega(p+k) = \infty$.

We start by examining the sum $\sum_{p \le x} \omega(p+k)$. For some $y = x^{\delta}$ with $0 < \delta < \frac{1}{2}$, let $\omega_y(n)$ denote the number of prime factors of $n$ that are $\le y$. It can be shown that $\omega(n) = \omega_y(n) + O(1)$. Thus,
$$\sum_{p \le x} \omega(p+k) = \sum_{p \le x} \omega_y(p+k) + O\left(\frac{x}{\log x}\right).$$

Using the Bombieri-Vinogradov Theorem, we obtain:
$$\sum_{\ell \le y} \pi(x; \ell, -k) = \sum_{\ell \le y} \frac{\text{li} x}{\ell - 1} + O\left(\frac{x}{\log^A x}\right),$$
where $\pi(x; \ell, -k)$ counts primes $p \le x$ such that $p \equiv -k \pmod{\ell}$.

Summing over primes $\ell \le y$, we get:
$$\sum_{\ell \le y} \frac{1}{\ell - 1} = \log \log y + O(1).$$

Therefore,
$$\sum_{p \le x} \omega(p+k) = \pi(x) \log \log x + O\left(\frac{x}{\log x}\right).$$

Assuming $\omega(p+k) = O(1)$ leads to a contradiction, implying $\limsup_{p \to \infty} \omega(p+k) = \infty$. Hence, $\limsup_{n \to \infty} \frac{\omega(n+k)}{\omega(n)} = \infty$, proving statement (i).

For statement (ii), note that $\omega(n) \le \Omega(n)$. Therefore, $\frac{\Omega(n+k)}{\Omega(n)} < \beta$ follows from $\frac{\omega(n+k)}{\omega(n)} > \alpha$ by choosing appropriate $\alpha$ and $\beta$.

Thus, both statements are proven to be true.

The answer is: $\boxed{\text{True}}.$