find possible values

Question

$abcd=-2020$ Find all possible 4-tuples of $a,b,c,d$ So I tried to count this one by one but I think the cases were just too many to consider, I was trying to find a way of doing this quicker. Is there any insight toward how to do this fast? I know this question may seem dumb to ask but help will really be appreciated!

Answer 1

$$2020=101*2^2*5$$ Now use star and bars for the powers of prime factors. For example for $101$ $$x_1+x_2+x_3+x_4=1$$

For this we get number of solutions as $\binom {4}{3}$

Similarly for $2$ we have $$y_1+y_2+y_3+y_4=2$$ and get number of solutions as $\binom {5}{3}$ and for $5$ we have $$z_1+z_2+z_3+z_4=1$$ and get number of solutions as $\binom {4}{3}$.

We need to also consider which of the numbers is negative.

Hence an odd number of the factors must be negative. For this we have the ways to select negative integers as $\binom {4}{3}+ \binom {4}{1}$

Hence the number of solutions would be $$\binom {4}{3}.\binom {5}{3}.\binom {4}{3}. \left[\binom {4}{3}+ \binom {4}{1}\right]=1280 $$