Solution for January 2012

The Problem:
	The polynomial $p(x)$ with real coefficients has degree $2011$ and satisfies $$p(n)=\frac{n}{n+1}$$ for all integers $n,\; 0 \le n \le 2011$. Compute $p(2012)$.

Solution: $p(2012) = 1.$

Correct solutions were submitted by

Lamis Alsheikh (Syria)	Bojan Bašić (Serbia)
Lou Cairoli (USA)	Bernard Collignon (France)
Olivier Cyr (France)	Hubert Desprez, (France)
Dan Dima (Romania)	Mei-Hui Fang (Austria)
Philippe Fondanaiche (France)	Gruian Cornel (2 solutions) (Romania)
Benoît Humbert (France)	Ile Ilijevski (Macedonia)
Wolfgang Kais (Germany)	Marc Lichtenberg (France)
Albert Stadler (Switzerland)	Ruben Victor Cohen (Argentina)
Matthew Lim (USA)	Codreanu Ioan Viorel (Romania)
César Días Mijangos (Mexico)	Mathias Schenker (Switzerland)

We also received three incomplete submissions.

Define $q(x):=(x+1)p(x)-x$. Note that $q(x)$ is a polynomial of degree 2012, and we know all 2012 of its zeros, namely $q(0) = q(1) = q(2) \dots = q(2011) =0$. We can therefore write it as a product of its linear factors,
$$q(x) = Cx(x-1)...(x-2011),$$
for some constant $C$. To find $C$ note that by its definition,
$$q(-1) = (-1+1)p(-1) - (-1) = 1,$$
while in factored form,
$$q(-1) = C(-1)(-2)\dots(-2012) = C\cdot 2012!;$$
therefore,
$$C=\frac1{2012!}.$$
Because
$p(x) = \frac1{x+1}(q(x)+x)$, we find that
$$p(2012) = \frac1{2013}\left(\frac1{2012!}\cdot(2012\cdot2011\cdot \dots \cdot 1) + 2012\right) = \frac{2013}{2013} = 1.$$

Comments.
Both Kais and Humbert observed that with the exception of $x=-1$, one can easily find any value of $p(x)$ by direct substitution into the factorized equation for $q(x)$. It is a much greater challenge to show that $p(-1) = 1 - \left(1 + \frac12 + \frac13 + \dots +\frac1{2012}\right)$. Can you prove that they are right?

Many readers solved a more general problem, replacing 2011 by an arbitrary integer $\kappa$: Find $p_\kappa(\kappa+1)$ when $p_\kappa(x)$ is a polynomial of degree $\kappa$ with real coefficients and satisfies $p_\kappa(n) = \frac{n}{n+1}$ for all integers $n$, $0 \le n \le \kappa$. This version was Problem 3 on the 1975 USA Mathematical Olympiad; we thank Codreanu Ioan for this information. The only detail to watch out for in the solution to the generalized problem comes when finding the constant $C$:
$$q(-1) = C(-1)(-2)\dots(-\kappa-1) = C\cdot(-1)^{\kappa+1}\cdot (\kappa+1)!.$$
It follows that
$$p_\kappa(\kappa+1)= \frac{\kappa+1+(-1)^{\kappa+1}}{\kappa+2}.$$
Consequently, when $\kappa$ is odd (as in our problem) $p_\kappa(\kappa+1) = 1$; when $\kappa$ is even, $p_\kappa=\frac{\kappa}{\kappa+2}$.