Let $P(x)$ be a polynomial of degree $3n$ so that $P(3i)=2, P(3j-2) = 1, P(3j-1) = 0$ for $0\leq i, j\leq n, j\ge 1$ and $P(3n+1)=730$. Determine the value of $n$ and prove that it is unique.
One value of $n$ that works is $n=4$. To see this, one can use the fact that $\sum_{k=0}^{3n+1} {3n+1\choose k}(-1)^k p(3n+1-k)=0$ and then solve for $p(3n+1)$ using the given values of $p(x)$ when $n=4$. However, I'm not sure how to prove this fact and it seems this fact generalizes to arbitrary polynomials with integer coefficients. That is, for a polynomial with integer coefficients $p$ of degree $n, \sum_{k=0}^{n+1}{n+1\choose k}(-1)^k p(n+1-k) = 0$, but I'm not sure how to prove this. I know Vieta's equations, which state that for a polynomial $p(x) = \sum_{k=0}^n a_k x^k$ with $n$ complex roots $r_1,\cdots, r_n$, we have $\frac{a_k}{a_n} = (-1)^{n-k} e_{n-k}(r_1,\cdots, r_n)$, where $e_i$ denotes the symmetric polynomial in $n$ variables so that $e_1(r_1,\cdots, r_n) = \sum_{i=1}^n r_i$, etc.
Also for a polynomial with integer coefficients $p$ such that $p(p(p(x)))=x$ for three distinct integer values of $x,p(x)=x$, though this probably isn't useful.
Alternatively, I think one can apply the Lagrange interpolation formula, which states that $P(x) = 2\sum_{p=0}^{n} \prod_{r\neq 3p} \frac{x-r}{3p-r} + \sum_{p=1}^n \prod_{r\neq 3p-2} \frac{x-r}{3p-2-r}$, but I'm not sure how to simplify this formula. Would the binomial theorem/multinomial theorem be useful?
