Olimpiadi di Matematica

Solution:

The answer is $(\mathbf{D})$. Let us first note that if $p(x)$ has degree $d$, then $p(x+1)-p(x)$ has degree exactly $d-1$. For simplicity we denote generically by $\star$ a sum of terms of degree less than $d-1$. Setting $p(x)=a x^{d}+b x^{d+1}+\star$, we have
$$\begin{aligned} p(x+1)-p(x) & =a(x+1)^{d}-a x^{d}+b(x+1)^{d-1}-b x^{d-1}+\star \\ & =\left(a x^{d}+a d x^{d-1}+\star\right)-a x^{d}+\left(b x^{d-1}+\star\right)-b x^{d-1}+\boldsymbol{t} \\ & =a d x^{d-1}+\star . \end{aligned}$$
Thus $p(x+1)-p(x)$ has degree exactly $d-1$. Now, let us denote $r(x):=p(x+1)-p(x)$ and $s(x):=r(x+1)-r(x)$; we have
$$\begin{aligned} s(x) & =(p(x+2)-p(x+1))-(p(x+1)-p(x)) \\ & =p(x+2)-2 p(x+1)+p(x), \\ s(x)-s(x-1) & =(p(x+2)-2 p(x+1)+p(x))-(p(x+1)-2 p(x)+p(x-1)) \\ & =p(x+2)-3 p(x+1)+3 p(x)-p(x-1) . \end{aligned}$$
Applying the result proved above several times, we then have that the degree of $r(x)$ is $2010-1=2009$, the degree of $s(x)$ is $2008$, the degree of $s(x)-s(x-1)$ is $2007$ (to obtain this last one we must set $y:=x-1$ and apply the result to $s(y+1)-s(y)$).

Second Solution

For every positive integer $i$, define the polynomial
$$\binom{x}{i}=\frac{x \cdot(x-1) \cdots(x-i+1)}{i !}$$
Set $\binom{x}{0}=1$ and $\binom{x}{i}=0$ for every integer $i<0$. The facts I need about these polynomials are two:
the first is that, for positive $i$, $\binom{x}{i}$ has degree $i$, and, in particular, the coefficient of $x^{i}$ in $\binom{x}{i}$ is $\frac{1}{i !}$ (it suffices to observe that the coefficient of $x$ in each of the $i$ degree-one factors making up the numerator is 1)
the second is the equation $\binom{x+1}{i+1}-\binom{x}{i+1}=\binom{x}{i}$, which holds for every integer $i$: for positive values of $i$ we indeed have
$$\begin{aligned} \binom{x+1}{i+1}-\binom{x}{i+1}&=\frac{(x+1) \cdot x \cdots(x-i+1)}{(i+1) !}-\frac{x \cdots(x-i+1) \cdot(x-i)}{(i+1) !} \\ &=\frac{x+1}{i+1}\binom{x}{i}-\frac{x-i}{i+1}\binom{x}{i}=\binom{x}{i} \end{aligned}$$
for $i=0$ the equation reduces to $x+1-x=1$, and for $i<0$ it is an immediate consequence of the definition.
Applying this second fact, I obtain
$$\begin{aligned} & \binom{x-1}{i}-3\binom{x}{i}+3\binom{x+1}{i}-\binom{x+2}{i}= \\ & =-\left(\binom{x}{i}-\binom{x-1}{i}\right)+2\left(\binom{x+1}{i}-\binom{x}{i}\right)-\left(\binom{x+2}{i}-\binom{x+1}{i}\right) \\ & =-\binom{x}{i-1}+2\binom{x+1}{i-1}-\binom{x+2}{i-1} \\ & =\left(\binom{x+1}{i-1}-\binom{x}{i-1}\right)-\left(\binom{x+2}{i-1}-\binom{x+1}{i-1}\right) \\ & =\binom{x+1}{i-2}-\binom{x+2}{i-2} \\ & =-\binom{x+2}{i-3} \end{aligned}$$
where in the first and third steps I simply reordered the sum.
Consider now a polynomial $p(x)$ of degree $d$. I claim that it can be written in the form
$$p(x)=a_{d}\binom{x}{d}+\cdots+a_{1}\binom{x}{1}+a_{0}$$
for a suitable choice of real numbers $a_{0}, \ldots, a_{d}$. Indeed, calling $k$ the leading coefficient of $p(x)$ and setting $a_{d}=k d!$, we obtain that $p(x)-a_{d}\binom{x}{d}$ has degree less than $d$, since the degree-$d$ monomials cancel. Repeating the procedure on this new polynomial we obtain an $a_{d-1}$ such that the degree of $p(x)-a_{d}\binom{x}{d}-a_{d-1}\binom{x}{d-1}$ is less than $d-1$. Continuing to lower the degree of the polynomial we arrive at the point where $p(x)-a_{d}\binom{x}{d}-a_{d-1}\binom{x}{d-1}-\cdots-a_{1}\binom{x}{1}$ has degree 0, that is, it is a constant, and we set $a_{0}$ equal to that constant.
Now it is easy to compute the polynomial required by the problem, by collecting the coefficients $a_{i}$
$$\begin{aligned} p(x-1)-3 p(x)+3 p(x+1)-p(x+2) & = \\ & =a_{d}\left(\binom{x-1}{d}-3\binom{x}{d}+3\binom{x+1}{d}-\binom{x+2}{d}\right)+\cdots \\ & =-a_{d}\binom{x+2}{d-3}-a_{d-1}\binom{x+2}{d-4}-\cdots-a_{3} \end{aligned}$$
and this is a polynomial of degree $d-3$, since the degree of $\binom{x+2}{d-3}$, like that of $\binom{x}{d-3}$, is $d-3$, while the degree of the other terms is less. The correct answer is therefore 2007.