Suppose that all of $\displaystyle a$, $\displaystyle b$, and $\displaystyle c$ are distinct. $\displaystyle F(a) -F(b) = b - c$, $\displaystyle F(b) - F(c) = c - a$, and $\displaystyle F(c) - F(a) = a - b$. Multiplying these all together yields $\displaystyle (F(a) - F(b))(F(b) - F(c))(F(c) - F(a)) = (b-c)(c-a)(a-b)$.
Rearrange to get $\displaystyle \displaystyle \left(\frac{F(a) - F(b)}{a-b}\right) \left(\frac{F(b) - F(c)}{b - c}\right) \left(\frac{F(c) - F(a)}{c-a}\right) = 1$ (possible since $\displaystyle a-b, b-c, c-a \ne 0)$. Since $\displaystyle f(x) $ is a polynomial with integer coefficients, $\displaystyle \displaystyle \frac{F(a) - F(b)}{a-b}, \frac{F(b) - F(c)}{b - c}, \frac{F(c) - F(a)}{c-a} \in\mathbb{Z}$.
There's my only doubt: why is this true? Couldn't it be, for example, that one of these quotients is one whereas the other two are rationals inverse to each other?
Tonio
Since their product is one, $\displaystyle \displaystyle \left|\frac{F(a) - F(b)}{a-b}\right| = \left|\frac{F(b) - F(c)}{b - c}\right| = \left|\frac{F(c) - F(a)}{c-a}\right| = 1.$
If one of $\displaystyle \displaystyle \frac{F(a) - F(b)}{a-b}$, $\displaystyle \displaystyle \frac{F(b) - F(c)}{b - c}$, and $\displaystyle \displaystyle \frac{F(c) - F(a)}{c-a}$ is -1 (without loss of generality, let us suppose that $\displaystyle \dfrac{F(a) - F(b)}{a-b} = -1)\right)$ then $\displaystyle F(a) - F(b) = b - a$. But $\displaystyle F(a) = b$ and $\displaystyle F(c) = c$, so $\displaystyle b - c = b - a$, yielding $\displaystyle a = c$, contradicting our assumption that all the variables were distinct.
It follows that each of $\displaystyle \displaystyle \frac{F(a) - F(b)}{a-b}$, $\displaystyle \displaystyle \frac{F(b) - F(c)}{b - c}$, and $\displaystyle \displaystyle\frac{F(c) - F(a)}{c-a}$ is equal to 1, so we have $\displaystyle F(a) - F(b) = a - b$, $\displaystyle F(b) - F(c) = b - c$, and $\displaystyle F(c) - F(a) = c - a$. Substituting, $\displaystyle b - c = a - b$, $\displaystyle c-a = b - c$, and $\displaystyle a-b = c - a$. Rearrange to get $\displaystyle 2b = a + c$, $\displaystyle 2a = b + c$, and $\displaystyle 2c = a + b$. These equalities quickly yield $\displaystyle a=b=c$, which again contradicts our assumption that $\displaystyle a, b, c$ are distinct. $\displaystyle \blacksquare$