1970 Canadian MO Problems/Problem 10


Given the polynomial $f(x)=x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{n-1}x+a_n$ with integer coefficients $a_1,a_2,\ldots,a_n$, and given also that there exist four distinct integers $a, b, c$ and $d$ such that $f(a)=f(b)=f(c)=f(d)=5$, show that there is no integer $k$ such that $f(k)=8$.


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