2022/054) Number $3^{32}-1$ has exactly two divisors greater than 75 and less than 85. What is their product?

we have $3^{32}-1 =  (3^{16}+ 1)(3^{16}-1) =   (3^{16}+ 1)(3^{8}+1) (3^{8}- 1)$

$= (3^{16}+1)(3^8+1)(3^4+1)(3^4-1)$ 

out of the above $3^4+1 = 82$ and %3^4-1=80$ are between 75 and 85 and product is 80 * 82 = 6560