Evaluate $a+b+c+d$

Question

If $a$, $b$, $c$, and $d$ are distinct integers such that
$$(x-a)(x-b)(x-c)(x-d)=4$$
has an integral root $r$, what is the value of $a+b+c+d$ in terms of $r$?

I tried to analyze graphically by shifting the graph of
$f(x)=(x-a)(x-b)(x-c)(x-d)$ four units downward but couldn't infer anything due to the random nature of $a$, $b$, $c$ and $d$.

Answer 1

The product of the four distinct integers $r-a$, $r-b$, $r-c$, and $r-d$ is $4$. Thus $r-a,r-b,r-c,r-d$ are, in some order, $-1$, $1$, $-2$, and $2$.

We may assume it is in the order we gave. So $r-a=-1$, $r-b=1$, $r-c=-2$, $r+d=2$. Solve for $a,b,c,d$ and add.

Answer 2

Plugging in $x=4$ shows that $4$ can be written as product of four distinct integers. The only divisors of $4$ are $\pm1, \pm2,\pm4$. The only way to obtain $4$ is as product $(-2)\cdot(-1)\cdot 1\cdot 2$. Then $(r-a)+(r-b)+(r-c)+(r-d)=-2+(-1)+1+2=0$, i.e. $a+b+c+d=4r$.