We say two positive integers, $m, k\geq 2$, have a different number of prime factors if in their standard form $m=p_1^{k_1}\cdots p_s^{k_s}$ and $k =q_1^{\ell_1}\cdots q_r^{\ell_r}$ it is the case that $$\sum_{i=1}^{s}{k_i}\neq \sum_{i=1}^{r}{\ell_i}.$$ (Additionally, the number $1$ is said to have zero prime factors). So, for instance, $24=2^3 \cdot 3$ and $25 = 5^2$ have a different number of prime factors because $3 + 1\neq2$; on the other hand, $25 = 5^2$ and $26 = 2 \cdot 13$ have the same number of prime factors since $2 = 1 + 1$.
Exercise. Let $n\in \mathbb{N}_+$, $n\geq 2$, be fixed but arbitrary. Prove that there exist $n$ consecutive positive integers $a + 1, a + 2, \dots, a + n$ such that all these numbers, pairwise speaking, have a different number of prime factors.
(That is, whenever $i\neq j$, then $a+i$ and $a+j$ must have a different number of prime factors; $i, j\in\{1, \dots, n\}$).
Attempt. Whether done constructively or non-constructively, a by-product must be that the sequence $(a+ i)_{i=1}^n$ contains at most one prime number.
- Therefore, one might be led to think toward an algorithm that constructs $n$ consecutive composite numbers, and then prove that it just so happens all of the constructed composite numbers also fulfil the criterium of the exercise. However, the classical option of looking at numbers near a factorial does not necessarily work. A counterexample would be $n = 4$ and $4!$. Maybe one should look at some $f(n)!$ instead?
Another possibility would perhaps be to set up an appropriate equation or system of equations (congruences), and the existance of appropriate $(a+ i)_{i=1}^n$ could instead be derived from some traditional theorem. I can set up a system of congruences to obtain $n$ consecutive composite numbers, but I am unsure what constraints to add that would ensure an outcome entailing a different number of prime factors.
For $n$ big, $a$ should probably be big, and at least somewhat bigger than $n$.
I suppose I am missing some straightforward idea here, so any hints (also as answers) are welcome on how to approach this exercise.
(Problem origin: homework exercise).