# 1970 Canadian MO Problems/Problem 10

## Problem

Given the polynomial with integer coefficients , and given also that there exist four distinct integers and such that , show that there is no integer such that .

## Solution

Set g(x) = f(x) − 5. Since a, b, c, d are all roots of g(x), we must have

g(x) = (x − a) (x − b) (x − c) (x − d) h(x)

for some h(x) ∈ Z[x].

Let k be an integer such that f(k) = 8, giving g(k) = f(k) − 5 = 3. Using the factorization above, we find that

3 = (k − a) (k − b) (k − c) (k − d) h(x).

By the Fundamental Theorem of Arithmetic, we can only express 3 as the product of at most three distinct integers (−3, 1, −1). Since k − a, k − b, k − c, k − d are all distinct integers, we have too many terms in the product, leading to a contradiction.

via Justin Stevens