Let a, b, c, d be distinct integers such that
(x-a)(x-b)(x-c)(x-d) - 4 = 0 has an integer root r.

Show that 4r=a+b+c+d.

a, b, c, d, and r are all integers.

r-a, r-b, r-c, r-d must be distinct, following from a, b, c, d being distinct.

(r-a)*(r-b)*(r-c)*(r-d) = 4 from the given polynomial.

Then r-a, r-b, r-c, r-d form a four term factorization of distinct integer factors of 4.

The only way for this to work is if those factors are -2, -1, 1, and 2.

Then without loss of generality r-a=-2, r-b=-1, r-c=1, r-d=2.

Then a=r+2, b=r+1, c=r-1, and d=r-2.

Then a+b+c+d = (r+2)+(r+1)+(r-1)+(r-2) = 4r. QED