If a×b×c×d is 9, what is a+b+c+d ( a b c d are different integers)

Question

If a×b×c×d is 9, what is a+b+c+d ( a b c d are different integers)

Actually, I know the answer to this problem, (which is 1+(-1)+3+(-3)= 0), but I am not sure whether there is a systematic way to solve it.

I thought of separating the a b c and d to

a= 91/b1/c1/d b=91/a1/c1/d And so on...

However, it didn't work. Should I treat a b c d as a whole?

Edit: Sorry I am on my phone so I can't use the formattings. Thanks for your understanding

Edit: Thanks for all your help :).

Answer 1

If $a, b, c, d$ are distinct integers which satisfies $abcd=9$, find $a+b+c+d$.

$a|9, b|9, c|9, d|9$.

The factors of $9$ is $-9, -3, -1, 1, 3, 9$.

WLOG, $a<b<c<d$.

If $a=-9:$ $bcd=-1$ for all distinct integers $b, c, d$, which has a contradiction.

If $b=9:$ $abc=1$ for all distinct integers $a, b, c$, which has a contradiction.

Therefore, $a=-3, b=-1, c=1, d=3.$

So, $a+b+c+d=0$.