IMC 2007 · Day 1 · P1

Let $f(x) = ax^2 + bx + c$. Substituting $x = 0$, $x = 1$ and $x = -1$, we obtain that $5 \mid f(0) = c$, $5 \mid f(1) = (a + b + c)$ and $5 \mid f(-1) = (a - b + c)$. Then $5 \mid f(1) + f(-1) - 2 f(0) = 2a$ and $5 \mid f(1) - f(-1) = 2b$. Therefore 5 divides $2a$, $2b$ and $c$ and the statement follows.

Consider $f(x)$ as a polynomial over the 5-element field (i.e. modulo 5). The polynomial has 5 roots while its degree is at most 2. Therefore $f \equiv 0 \pmod 5$ and all of its coefficients are divisible by 5.