We have that $x^2-y^2=(x+y)(x-y)$ Note that for every positive integer $n$ and any two divisors $d_1, d_2$ such that $d_1\cdot d_2=n$, we need to find the solutions of the system of equations: \begin{align*} x+y&=d_1\\ x-y&=d_2\\ \end{align*}Hence without loss of generality we may assume that $d_1\neq d_2$ and $d_1 > d_2$ since otherwise the solutions would contain $0$ or a negative integer. Solving for y in the system we get that $y=\frac{d_1-d_2}{2}$ Hence in order for y to be an integer $d_1$ and $d_2$ must have the same parity. 1) The numbers of unique solutions for an odd number is given by $\frac{\tau(n)}{2}$ furthermore we need to find the solutions to $\frac{\tau(n)}{2}=2021$. were $2021 = 47\cdot 43$. Because to each divisor there is a unique different divisor (if the number is not a perfect square) such that $d_id_{\tau(n)-i+1}$. the possible solutions are $3^{46}* 5^{42}*7$ or $3^{4041}$ or $3^{84}*5^{46}$ or $3^{92}*5^{42}$ Calculation\begin{align*} 3^{4041}=3^{46}*3^{995} &> 3^{46}*5^{42}*7\\ 3^{84}*5^{46}=3^{46}*3^{36}*5^{46} > 3^{46}*5^{42}*7\\ 3^{46}*3^{46}*5^{42}>3^{46}*5^{42}*7\\ \end{align*}. From all the possible solutions the least is the first one s $\boxed{3^{46} \cdot 5^{42} \cdot 7}$ 2) The least positive integer that satisfies this conditions must be divisible by 4, otherwise one of the divisors would be odd and hence would not be a solution. let that number be $n = 2^{\alpha}p$ were p is odd. Hence the number of solutions is given by \[\frac{\tau(2^{\alpha}p)-\tau(p)}{2}=2021\]since $\tau$ is a multiplicative function we have that $\tau(2^{\alpha}p)=\tau(2^{\alpha})\tau(p)$ this result enables us to factor the equation as follows \[\tau(p)\tau(2^{\alpha})-\tau(p) = \tau(p)((\alpha +1)-1)=\tau(p)(\alpha)=4042\]From the prime factorization of $2021$ we get that the least number is $\boxed{2^{48}\cdot 3^{42} \cdot 5}$ $\blacksquare$

We have that $x^2-y^2=(x+y)(x-y)$ Note that for every positive integer $n$ and any two divisors $d_1, d_2$ such that $d_1\cdot d_2=n$, we need to find the solutions of the system of equations: \begin{align*} x+y&=d_1\\ x-y&=d_2\\ \end{align*}Hence without loss of generality we may assume that $d_1\neq d_2$ and $d_1 > d_2$ since otherwise the solutions would contain $0$ or a negative integer. Solving for y in the system we get that $y=\frac{d_1-d_2}{2}$ Hence in order for y to be an integer $d_1$ and $d_2$ must have the same parity. 1) The numbers of unique solutions for an odd number is given by $\frac{\tau(n)}{2}$ furthermore we need to find the solutions to $\frac{\tau(n)}{2}=2021$. were $2021 = 47\cdot 43$. Because to each divisor there is a unique different divisor (if the number is not a perfect square) such that $d_id_{\tau(n)-i+1}$. the possible solutions are $3^{46}* 5^{42}*7$ or $3^{4041}$ or $3^{84}*5^{46}$ or $3^{92}*5^{42}$ Calculation\begin{align*} 3^{4041}=3^{46}*3^{995} &> 3^{46}*5^{42}*7\\ 3^{84}*5^{46}=3^{46}*3^{36}*5^{46} > 3^{46}*5^{42}*7\\ 3^{46}*3^{46}*5^{42}>3^{46}*5^{42}*7\\ \end{align*}. From all the possible solutions the least is the first one s $\boxed{3^{46} \cdot 5^{42} \cdot 7}$ 2) The least positive integer that satisfies this conditions must be divisible by 4, otherwise one of the divisors would be odd and hence would not be a solution. let that number be $n = 2^{\alpha}p$ were p is odd. Hence the number of solutions is given by \[\frac{\tau(2^{\alpha}p)-\tau(p)}{2}=2021\]since $\tau$ is a multiplicative function we have that $\tau(2^{\alpha}p)=\tau(2^{\alpha})\tau(p)$ this result enables us to factor the equation as follows \[\tau(p)\tau(2^{\alpha})-\tau(p) = \tau(p)((\alpha +1)-1)=\tau(p)(\alpha)=4042\]From the prime factorization of $2021$ we get that the least number is $\boxed{2^{48}\cdot 3^{42} \cdot 5}$ $\blacksquare$