Prove that for every $k$, there exists a sequence $a_1, a_2, \dots, a_k$, consecutive natural numbers, such that, for every $i \neq j$, $\tau(a_i) \cdot \tau(a_j)$ is not a perfect square. We are denoting $\tau(\ell)$ the number of divisors of a natural number $\ell$.
Approach, ideas:
Consider $\mathcal{A}$ the set of numbers for which is the property is true (they are the begining of a sequence that contains consecutive numbers with that property). And consider $\mathcal{B} = \complement \left(\mathcal{A}\right)$. We should prove that $\mathcal{B}$ is infinite,and that it has arbitrarily large distances between consecutive elements of $\mathcal{B}$, to imply that the property is true for each $k$.
I think $\rho_\mathcal{B}$ should equal $0$ in this case, as it should contain arbitrarily large gaps, so arbitrarily small values of the fraction:
$$\frac{|\mathcal{B}\cap [1, N]|}{N}$$
Therefore, the density:
$$\rho = \lim_{n\to\infty} \frac{|\mathcal{B}\cap [1, N]|}{N}$$
However, I am not able to compute this limit to complete the proof.
Best Answer
Let $2=p_{1}<p_{2}<...$ be the sequence of consecutive primes.
Let $k+1<\beta<r_{1}<r_{2}<...<r_{k}$ be some chosen primes ($\beta$ is some variable, this will be used later.)
We solve the following system;
$w \equiv p_{1}^{r_{1}-1} \mod p_{1}^{r_{1}}$
$w+1 \equiv p_{2}^{r_{2}-1} \mod p_{2}^{r_{1}}$
.
.
.
$w+k-1 \equiv p_{k}^{r_{k}-1} \mod p_{k}^{r_{k}}$
By Chinese remainder theorem the above system has a unique solution mod $\prod_{j=1}^{k}p_{j}^{r_{j}}$, we will call this solution $w_{\beta,r,k} \in [1, \prod_{j=1}^{k}p_{j}^{r_{j}}]$.
$\textbf{Claim}:$ Set $A_{\beta,t,r,k} = (w_{\beta,r,k}+\alpha+t\prod_{j=1}^{k}p_{j}^{r_{j}})_{\alpha =0}^{k-1}$ $\text{ }$ ($A_{\beta,t,r,k}$ is a finite sequence of consecutive natural numbers). For every $k \in \mathbb{N}$ there exists $\beta > 0$, $r$ (viewed as a vector of primes of length $k$ satisfying the above condition) and $t \in \mathbb{N}$ so that if some prime $p$ satisfies $p^{r_{1}-1} | \left(w_{\beta,r,k}+\alpha+t\prod_{j=1}^{k}p_{j}^{r_{j}}\right) $ then $p = p_{\alpha+1}$.
Suppose this claim is not true then for each $t \geq 0$, $\beta>0$, $r$ there is some $\alpha_{t} \in \{0,...,k-1\}$ so that some $p > p_{\alpha_t+1}$ satisfies
$$p^{r_{1}-1}|\left(w_{\beta,r,k}+\alpha_{t}+t\prod_{j=1}^{k}p_{j}^{r_{j}}\right)$$
By infinite pigeon hole principle there exists some $v \in \{0,...,k-1\}$ so that
$$\limsup_{n \rightarrow \infty}\frac{|\{t: \exists \text{ prime } p > p_{k}\text{ so that }p^{r_{1}-1}|\left(w_{\beta,r,k}+v+t\prod_{j=1}^{k}p_{j}^{r_{j}}\right), 1 \leq t \leq n, t \in \mathbb{N}\}|}{n} \geq \frac{1}{k}$$
But by a simple asymptotic estimate
$$\limsup_{n \rightarrow \infty}\frac{|\{t: \exists \text{ prime } p > p_{k}\text{ so that }p^{r_{1}-1}|\left(w_{\beta,r,k}+v+t\prod_{j=1}^{k}p_{j}^{r_{j}}\right), 1 \leq t \leq n, t \in \mathbb{N}\}|}{n} \leq \sum_{s=k+1}^{\infty}\frac{1}{p_{s}^{r_{1}-1}}$$
which becomes smaller than $\frac{1}{k}$ with appropriately sized $r_{1}$; this is a contradiction.
With this claim note that with some $\beta, t$ and $r$ $A_{\beta,t,r,k}$ satisfies your question as for each $\alpha \in \{0,...,k-1\}$ $\tau(w_{\beta,r,k}+\alpha+t\prod_{j=1}^{k}p_{j}^{r_{j}})$ is a multiple of $r_{\alpha+1}$ and prime factors smaller than $r_{1}$. So that when $\mu \neq \alpha \in \{0,...,k-1\}$
$$\tau(w_{\beta,r,k}+\alpha+t\prod_{j=1}^{k}p_{j}^{r_{j}})\tau(w_{\beta,r,k}+\mu+t\prod_{j=1}^{k}p_{j}^{r_{j}}) = r_{\alpha+1}r_{\mu+1}\prod_{\text{ some prime factors smaller than }r_{1}} p$$
which is not a square.