1970
Problem - 2853
Let $P(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ 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 $k$ satisfying $P(k) = 8$.
We have that $a, b, c, d$ are distinct roots of $P(x)-5$, hence $$P(x)-5 = g(x)(x-a)(x-b)(x-c)(x -d)$$ where $g(x)$ is a polynomial with integral coefficients. If $P(k) = 8$ then $g(x)(x-a)(x-b)(x-c)(x-d) = 3$. but 3 is a prime number, so all the factors on the left but one must be $\pm 1$. So among the numbers $(x - a), (x-b), (x -c), (x-b)$, there are either two $1$'s or two $-1$'s, which implies that $a, b, c, d$ cannot be all distinct. This is a contradiction.