Let $$p(x)$$ be a monic polynomial of degree four with distinct integer roots $$a, b, c$$ and $$d$$. If $$p(r)=4$$ for some integer $$r$$, prove that $$r=\frac{1}{4}(a+b+c+d)$$
My only idea was to let $$p(x)=(x-a)(x-b)(x-c)(x-d)$$, so that: $$4=(r-a)(r-b)(r-c)(r-d)$$. But the casework here, looking for $$4$$ factors of $$4$$, seems too tedious
As you said, we have $$4=(r-a)(r-b)(r-c)(r-d)$$. Therefore, as $$a,b,c,d$$ are integers. the absolute values of any two of the factors of $$4$$ above should be $$2$$ and two of them should be $$1$$. Now, the numbers which have the same absolute values must have different sign, as otherwise, the numbers would become equal. Thus, let us assume, without loss of generality $$r-a=2, r-b=-2,r-c=1,r-d=-1$$, which gives us $$a+b+c+d=4r$$, from which the desired conclusion follows.
Observe that $$\{r-a, r-b, r-c, r-d \}=\{\pm 1, \pm 2\}$$. Thus $$r-a + r-b + r-c + r-d=4r-(a+b+c+d)=0$$
Since a,b,c and d are distinct, we can suppose that $$a. The only possibility to have $$4=(r−a)(r−b)(r−c)(r−d)$$ is when $$a and when $$r-a=2,r-b=1,r-c=-1,r-d=-2.$$ In this configuration, one has $$r-a+r-b+r-c+r-d=4r-(a+b+c+d)=0$$ and then $$r=\frac{a+b+c+d}{4}.$$