Polynomial $P(x)$ such that $P(3k)=2$, $P(3k+1)=1$, $P(3k+2)=0$ for $k=0,1,2,\ldots,n-1$, $P(3n)=2$, and $P(3n+1)=730$

Question

Let $n$ be a positive integer such that there exists a polynomial $P(x)$ over $\mathbb{Q}$ of degree $3n$ satisfying the conditions below: $$P(0) = P(3) = \ldots = P(3n) = 2\,,$$ $$P(1) = P(4) = \ldots= P(3n - 2) = 1\,,$$ $$P(2) = P(5) = \ldots = P(3n - 1) = 0\,,$$ and $$P(3n + 1) = 730\,.$$ Determine the value of $n$.

Answer 1

The OP is equivalent to: given \begin{align} P(x)&=\sum_{i=0}^{3n} a_i x^i \tag{1}\label{1} ,\\ P(3i-2)&=1,\quad i=1,\cdots,n \tag{2}\label{2} ;\\ P(3i-1)&=0,\quad i=1,\cdots,n \tag{3}\label{3} ;\\ P(3i)&=2,\quad i=1,\cdots,n \tag{4}\label{4} ;\\ P(3n+1)&=730 \tag{5}\label{5} ;\\ P(0)&=2 \tag{6}\label{6} , \end{align}

determine $n$.

Conditions \eqref{2}-\eqref{5} define the system of $(3n+1)\times(3n+1)$ linear equations,

\begin{align} Au&=v \tag{7}\label{7} , \end{align}
where $A$ is a special case of the $(3n+1)\times(3n+1)$ Vandermonde matrix

\begin{align} A&=\begin{bmatrix} 1&1&1&\dots &1 \\ 1&2^1&2^2&\dots &2^{3n} \\ 1&3^1&3^2&\dots &3^{3n} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ 1&(3n+1)^1&(3n+1)^2&\dots &(3n+1)^{3n}\\ \end{bmatrix} ,\\ \text{or }\quad A_{ij}&=(i+1)^{j} ,\quad i,j=0,\cdots,3n . \end{align}

$u$ is the vector of coefficients $a_i$ \begin{align} u&=[a_0,\cdots,a_{3n}]^{\mathsf{T}} , \end{align}

and the right-hand side of \eqref{7} is a vector of the form \begin{align} v&=[\underbrace{1,0,2,1,0,2,\cdots,1,0,2}_{3n},730]^{\mathsf{T}} ,\\ \text{or }\quad v_{3n}&=730 ,\\ v_{i}&= 2-(i+1\mod 3) ,\quad i=0,\cdots,3n-1 . \end{align}

It follows from the condition \eqref{6} that \begin{align} a_0&=2 , \end{align}

For any $n$ all the coefficients $a_i$ of $P(x)$, including $a_0$, can be found as the solution of \eqref{7}.

A quick test reveals that $a_0=2$ for $n=4$.

This is a log of Maxima session for the test:

(%i1) a[i,j]:=(i+1)^j$
(%i2) v[i]:=2-mod(i+1,3)$
(%i3) n:4$
(%i4) A:apply('matrix,makelist(makelist(a[i,j],j,0,3*n),i,0,3*n))$
(%i5) V:makelist(v[i],i,0,3*n)$
(%i6) V[3*n+1]:730$
(%i7) V;
(%o7)              [1, 0, 2, 1, 0, 2, 1, 0, 2, 1, 0, 2, 730]
(%i8) a0=invert(A)[1] . V;
(%o8)                               a0 = 2

Answer 2

Let $\omega:=\exp\left(\frac{2\pi\text{i}}{3}\right)=\frac{-1+\sqrt{-3}}{2}$. Define $$\tau(z):=\left(\frac{1-\omega^2}{3}\right)z^2+\left(\frac{1-\omega}{3}\right)z+1\text{ for all }z\in\mathbb{C}\,.$$ Note that, for each $k\in\mathbb{Z}$, we have $$\tau(\omega^k)=\left\{ \begin{array}{ll} 2&\text{if }k\equiv0\pmod{3}\,,\\ 1&\text{if }k\equiv1\pmod{3}\,,\\ 0&\text{if }k\equiv2\pmod{3}\,. \end{array} \right.$$

It is well known (see also here) that, if $f(x)\in \mathbb{K}[x]$ is a polynomial over a field $\mathbb{K}$ of degree $d$, then $$\sum_{r=0}^{d+1}\,(-1)^r\,\binom{d+1}{r}\,f(x+r)\equiv 0\,.$$ From the result above, we have that $$\sum_{r=0}^{3n+1}\,(-1)^r\,\binom{3n+1}{r}\,P(r)=0\,.$$ Recall that $P(r)=\tau(\omega^r)$ for $r=0,1,2,\ldots,3n$, and $P(3n+1)=3^6+1=3^6+\tau(\omega^{3n+1})$. That is, $$\sum_{r=0}^{3n+1}\,(-1)^r\,\binom{3n+1}{r}\,\tau(\omega^r)=(-1)^{3n}\,3^6\,.$$ Ergo, $$\left(\frac{1-\omega^2}{3}\right)\left(1-\omega^2\right)^{3n+1}+\left(\frac{1-\omega}{3}\right)\left(1-\omega\right)^{3n+1}+1(1-1)^{3n+1}=(-1)^{3n}3^6\,.$$ Thus, $$2\,\text{Re}\left((1-\omega)^{3n+2}\right)=\left(1-\omega^2\right)^{3n+2}+\left(1-\omega\right)^{3n+2}=(-1)^{3n}3^7\,.$$ Since $1-\omega=\sqrt{-3}\,\omega^2$, we see that $$(1-\omega)^{3n+2}=\sqrt{-3}^{3n+2}\omega^{6n+4}=\sqrt{-3}^{3n+2}\omega\,.$$ Therefore, $$3^{\frac{3n+2}{2}}\,\text{Re}\left(\sqrt{-1}^{3n+2}\omega\right)=\text{Re}\left(\sqrt{-3}^{3n+2}\omega\right)=(-1)^{3n}\frac{3^7}{2}\,.$$ Since $\frac{1}{2}\leq \Big|\text{Re}\left(\sqrt{-1}^{3n+2}\omega\right)\Big|\leq\frac{\sqrt{3}}{2}$, we have that $$\frac{3^{\frac{3n+2}{2}}}{2}\leq \frac{3^7}{2} \leq \frac{3^{\frac{3n+3}{2}}}{2}\,.$$ Consequently, $$3n+2\leq 14\leq 3n+3\,.$$ This proves that $n=4$ is the only possibility. It is not difficult to see that $n=4$ indeed works.

Answer 3

Define the polynomials $$ L_i(x) = \underset{j \neq i}{\prod_{0 \le j \le n}} \frac{(x-x_j)}{(x_i - x_j)} = \begin{cases} 1 & \text{ if } x = i \\ 0 & \text{ if } x = j \neq i \end{cases} $$ for each $i \in \{0,1,\cdots,3n\}$. It follows that since the $L_i$'s have degree $3n$, we have (I leave the computations up to you) : $$ P(x) = \sum_{i=0}^{3n} P(i) L_i(x) = \sum_{i=0}^n 2 (-1)^{n-i} \binom x{3i} \binom {x-(3i+1)}{3(n-i)} - \sum_{i=0}^{n-1} (-1)^{n-i}\binom x{3i+1}\binom{x-(3i+2)}{3(n-i)-1}. $$ You can evaluate $P(3n+1)$ for a long range using a computer (the notation $\binom xi = x(x-1)\cdots(x-(i-1))/i!$ implies that $\binom xi (n) = \binom ni$ for positive integers). In other words, you are looking for $n$ such that $$ 730 = P(3n+1) = 2 \sum_{i=0}^n (-1)^{n-i} \binom{3n+1}{3i} - \sum_{i=0}^{n-1} (-1)^{n-i} \binom{3n+1}{3i+1} \\ = \sum_{i=0}^n \left( (-1)^{n-i} \left[ 2\binom{3n+1}{3i} - \binom{3n+1}{3i+1} \right] \right) $$

Hope that helps,