China Mathematical Olympiad

The answer is YES. From the definition of $\omega$ and $\Omega$, we have $$\omega (ab)\le \omega (a)+\omega (b),\stackrel{◯}{1}$$ $$\Omega (ab)=\Omega (a)+\Omega (b),\stackrel{◯}{2}$$ for any positive integers $a,b$. Given a fixed positive integer $k$ and positive real numbers $\alpha ,\beta$, we take a positive integer $m>(\omega (k)+1)\alpha$. As there are infinitely many prime numbers, we can take a sufficiently large prime $p$ such that $\frac{\Omega (k)+1}{p^m}+{\log }_{p}2<\beta$, and take $m$ pairwise distinct prime numbers $q_1,q_2,\dots ,q_m$ that are all greater than $p$. We will show that $n=2^{q_1{q}_2\cdots q_m}k$ has the desired property.

First, we prove $\frac{\omega (n+k)}{\omega (n)}>\alpha$. Let $n_1=\frac{n+k}{k}=2^{q_1{q}_2\cdots q_m}+1$. As $q_1,q_2,\dots ,q_m$ are all odd prime numbers, $2^{q_i}+1\mid n_1$ when $1\le i\le m$, then $d_i=\frac{2^{q_i}+1}{3}$ is an integer greater than 1.

Note that $(2^r-1,2^s-1)=2^{(r,s)}-1$ for all positive integers $r,s$, ③ and $(q_i,q_j)=1(i\ne j)$, we have $$\begin{array}{rl}(d_i,d_j)& =\frac{1}{3}(2^{q_i}+1,2^{q_j}+1)\le \frac{1}{3}(2^{2q_i}-1,2^{2q_j}-1)\\ & =\frac{2^{(2q_i,2q_j)}-1}{3}=\frac{2^2-1}{3}=1.\end{array}$$ $d_1,d_2,\dots ,d_m$ are the pairwise coprime factors of $n_1$, and each of them is greater than 1. Hence, $\omega (n_1)\ge m$. From ① and the choice of $m$, we have $$\frac{\omega (n+k)}{\omega (n)}\ge \frac{\omega (n_1)}{\omega (n)}\ge \frac{\omega (n_1)}{\omega (k)+1}\ge \frac{m}{\omega (k)+1}>\alpha .$$

Next, we prove $\frac{\Omega (n+k)}{\Omega (n)}<\beta$. As $q_1{q}_2\cdots q_m$ is an odd number and cannot be divided by 3, we have $n_1=2^{q_1{q}_2\cdots q_m}+1\equiv \pm 3(\mathrm{mod}9)$, that is, $3\nmid n_1$. Suppose $q$ is a prime factor of $\frac{n_1}{3}$ and $q\le p$, then $$2^{2q_1{q}_2\cdots q_m}-1=(2^{q_1{q}_2\cdots q_m}-1)\cdot n_1\equiv 0(\mathrm{mod}q).$$ From the Fermat's Little Theorem, $2^{q-1}\equiv 1(\mathrm{mod}q)$. From ③, $q\mid 2^{(2q_1{q}_2\cdots q_m,q-1)}-1$. From $(q-1,2q_1{q}_2\cdots q_m)=(q-1,2)\le 2,q-1<p<q_i(i=1,2,\dots ,m)$, hence $q\mid 2^2-1,q=3$. This contradicts that $\frac{n_1}{3}$ is not a multiple of 3. Therefore, each prime factor of $\frac{n_1}{3}$ is larger than $p$. So $\frac{n_1}{3}>p^{\Omega (n_1/3)}$. From ② and the choice of primes $p$ and $q_1,q_2,\dots ,q_m$, we have $$\begin{array}{rl}\Omega (n+k)& =\Omega (k)+\Omega (3)+\Omega \left(\frac{n_1}{3}\right)<\Omega (k)+1+{\log }_p\left(\frac{n_1}{3}\right)\\ & <\Omega (k)+1+{\log }_p(n_1-1)\\ & =\Omega (k)+1+q_1{q}_2\cdots q_m{\log }_{p}2,\end{array}$$ $$ \frac{\Omega(n+k)}{\Omega(n)} < \frac{\Omega(k) + 1 + q_1 q_2 \cdots q_m \log_p 2}{q_1 q_2 \cdots q_m} < \frac{\Omega(k) + 1}{p^m} + \log_p 2 < \beta. \quad \square