$1770=2\cdot3\cdot5\cdot59$. Denote $A=4a-b; B=4b-a$. $AB=1770^n\Longrightarrow 3|AB$ and $5|AB$. $3|AB\Longrightarrow 3|A$ or $3|B$. Similarly, $5|A$ or $5|B$. $3|A+B; 3|AB\Longrightarrow 3|A$ and $3|B$. $5|A-B; 5|AB\Longrightarrow 5|A$ and $5|B$. Results: If $15|AB\Longrightarrow 15|A$ and $15|B$. Let $A=15c; B=15d; c,d\in\mathbb{N}$. The system $4a-b=15c; 4b-a=15d; c,d\in\mathbb{Z}$ has the solution $a=4c+d; b=c+4d$. Case 1: $n=1$. The initial equation becomes: $15cd=2\cdot59$, without integer solutions. Case 2: $n\ge2$. The initial equation becomes: $cd=3^{n-2}\cdot5^{n-2}\cdot2^2\cdot59^2$. The number $3^{n-2}\cdot5^{n-2}\cdot2^2\cdot59^2$ has $(n-1)\cdot (n-1)\cdot3\cdot3=9(n-1)^2$ natural divisors. Results: exist $9(n-1)^2$ pairs $(c,d)\in\mathbb{N}\times\mathbb{N}$ with $cd=3^{n-2}\cdot5^{n-2}\cdot2^2\cdot59^2$. The number of the ordered pairs $(a,b)$ is equal to the number of ordered pairs $(c,d)$. A last comment: If we consider $(a,b)$ as ordered pairs, $(a,b)\ne(b,a)$ for $a\ne b$, then the number of the solutions $(a,b)$ is $N_n=9(n-1)^2$. If we consider the pairs $(a,b)$ as sets of two numbers with $(a,b)=(b,a)$, then result the number of the solutions $(a,b)$: $M_n=\begin{cases} \dfrac{9(n-1)^2}{2},\text{ for n odd number};\\\dfrac{9(n-1)^2+1}{2},\text{ for n even number}.\end{cases}$