Problem - 2836
Let $p(x)$ be a polynomial with integer coefficients. Assume that $p(a) = p(b) = p(c) = -1$, where $a, b, c$ are three different integers. Prove that $p(x)$ has no integral zeros.
We have that $p(x)+1$ has zeros at $a, b$, and $c$, hence $p(x)+1 = (x-a)(x-b)(x-c)q(x)$. If $p$ had an integral zero $d$ we would have $(d -a)(d-b)(d-c)q(d) = 1$, where $(d-a)$, $(d-b)$, and $(d -c)$ are distinct integers. But that is impossible, because 1 has only two possible factors, 1 and $-1$.