Olympiad - Polynomials

This one doesn't need latex and the solution is quick (I literally did it while going to the bathroom and reading Samurai Champloo at the same time) so I'll just type it up now.

(Canada 1970) Let P(x) be a polynomial with integral coefficients. Suppose that there exist four distinct integers a, b, c, d with P(a) = P(b) = P(c) = P(d) = 5. Prove that there is no integer k with P(k) = 8.

Kind of scary looking, but it's really not. We can say that P(x) -5 has 4 distinct integral roots, or:

P(x) - 5 = Q(x)(x-a)(x-b)(x-c)(x-d) = 0

Hopefully I don't need to explain that part to you. Now, suppose that there exists an integer k such that P(k) = 8, then we have the result:

P(x) - 8 = ((P(x) - 5) - 3 = Q(x)(x-a)(x-b)(x-c)(x-d) - 3 = 0

Q(x)(x-a)(x-b)(x-c)(x-d) = 3


Since all of (x-a), (x-b), (x-c), (x-d) are distinct and integers, then our assumed result is impossible. As only one can be 3, and the rest must be either 1 or -1. However, we have 4 roots, and thus, one of these three choices for numbers must occur twice, contradicting that the integers are distinct.

Redefining polynomials is a useful tool.