I'm dealing with the test of the International Mathematics Competition for University Students, 2011, and I've had a lot of difficulties, so I hope someone could help me to discuss the questions.
I've posted before the questions 2, 3 and 5, these last still open.
The question 4 says:
Let be $f(x)$ a polynomial with real coeficients and degree $n$. Supose that $\dfrac{f(k)-f(m)}{k-m}$ is integer for all the integers $0\leq k\lt m\leq n$. Prove that $a-b$ divides $f(a)-f(b)$ for any couple of distinct integers $a$ and $b$.
The only substantial thing that I've got is that:
Given any $k\in\{1,2,3,\dots,n-1\}$, we have
$\dfrac{f(k)-f(0)}{k}\in\mathbb{Z}$.
So,
$f(k)-f(0)$ is integer by any $k\in\{1,2,3,\dots,n-1\}$.
I also have thinked that derivatives can be some relation, because of the type of fraction…
I thank very much.
Important Edit (October, 04)
I've found a document with these solutions and I'm studying them. These are the documents: http://www.imc-math.org.uk/imc2011/imc2011-day2-solutions.pdf and http://www.imc-math.org.uk/imc2011/imc2011-day1-solutions.pdf.
Best Answer
I don't think the derivative comes anywhere into this. However, the discrete version, i.e., forward differences, does come in in a sense. I'll just sketch a proof.
Let $g(x)=\frac{f(x)-f(0)}{x}$. Then $g(1),g(2),\dots,g(n-1)\in\mathbb{Z}$. Moreover, since $g$ is a polynomial of degree $n-1$, we have $$ 0=\Delta^n[g](x)=\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}g(x+k) $$ giving $g(\mathbb{Z})\subseteq\mathbb{Z}$ inductively. Thus $$ g(x)=\sum_{k=0}^{n-1} c_k\binom{x}{k} $$ with $c_k\in\mathbb{Z}$, i.e., $$ f(x)=f(0)+x\sum_{k=0}^{n-1} c_k\binom{x}{k}=f(0)+\sum_{k=1}^n c'_k\binom{x}{k}. $$ with $c'_k\in\mathbb{Z}$.
Moreover, by examining deeper into the conditon $\dfrac{f(k)-f(m)}{k-m}\in\mathbb{Z}$ for $0\leq k<m\leq n$ and induction on $k$ noting $$ \begin{align*} \binom{a}{k}-\binom{b}{k}&=\sum_{j=1}^k\binom{a-b}{j}\binom{b}{k-j}\\ &=(a-b)\sum_{j=1}^k\frac1j\binom{a-b-1}{j-1}\binom{b}{k-j} \end{align*} $$ one can prove $c'_k$ is an integer multiple of $L_k:=\operatorname{lcm}(1,2,\dots,k)$ for each $k$. Hence the result follows.