1984 USAMO Problems/Problem 5

Problem

$P(x)$ is a polynomial of degree $3n$ such that

$\begin{eqnarray*} P(0) = P(3) = \cdots &=& P(3n) = 2, \\ P(1) = P(4) = \cdots &=& P(3n-2) = 1, \\ P(2) = P(5) = \cdots &=& P(3n-1) = 0, \quad\text{ and }\\ && P(3n+1) = 730.\end{eqnarray*}$

Determine $n$ .

Solution

By Lagrange Interpolation Formula $f(x) = 2\sum_{p=0}^{n}\left ( \prod_{0\leq r\neq3p\leq 3n}^{{}}\frac{x-r}{3p-r} \right )+ \sum_{p=1}^{n}\left ( \prod_{0\leq r\neq3p-2\leq 3n}^{{}} \frac{x-r}{3p-2-r}\right )$

and hence $f(3n+1) = 2\sum_{p=0}^{n}\left ( \prod_{0\leq r\neq3p\leq 3n}^{{}}\frac{3n+1-r}{3p-r} \right )+ \sum_{p=1}^{n}\left ( \prod_{0\leq r\neq3p-2\leq 3n}^{{}} \frac{3n+1-r}{3p-2-r}\right )$

after some calculations we get $f(3n+1) =\left ( \binom{3n+1}{0}- \binom{3n+1}{3}+\binom{3n+1}{6}- ... \right )\left ( 2.(-1)^{3n}-1 \right )+1$

Given $f(3n+1)= 730$ so we have to find $n$ such that $\left ( \binom{3n+1}{0}- \binom{3n+1}{3}+\binom{3n+1}{6}- ... \right )\left ( 2.(-1)^{3n}-1 \right )= 729$

Lemma: If $p$ is even $\binom{p}{0}- \binom{p}{3}+ \binom{p}{6}- \cdots = \frac{2^{p+1}sin^{p}\left ( \frac{\pi}{3} \right )(i)^{p}\left ( cos\left ( \frac{p\pi}{3} \right ) \right )}{3}$

and if $p$ is odd $\binom{p}{0}- \binom{p}{3}+ \binom{p}{6}- \cdots = \frac{-2^{p+1}sin^{p}\left ( \frac{\pi}{3} \right )(i)^{p+1}\left ( sin\left ( \frac{p\pi}{3} \right ) \right )}{3}$

$i$ is $\sqrt{-1}$ Using above lemmas we do not get any solution when $n$ is odd, but when $n$ is even $3n+1=13$ satisfies the required condition, hence $n=4$ This problem needs a solution. If you have a solution for it, please help us out by adding it.

1984 USAMO (Problems • Resources)
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1984 USAMO Problems/Problem 5

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