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> Let $x\_1,\cdots, x\_n$ be different real numbers. Prove that $$\sum\_{1\leq i\leq n} \prod\_{j\neq i} \frac{1-x\_ix\_j}{x\_i-x\_j} = \begin{cases} 0,&\text{if }n\text{ is even}\\
> 1,&\text{if }n\text{ is odd}\end{cases}.$$
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Let $f(t) = \prod\_{i=1}^n (1-x\_i t)$. Note that $f(x\_i) = (1-x\_i^2)\prod\_{j\neq i} (1-x\_ix\_j)$.

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> First, I think one can ignore the case where $\{x\_1,\dots, x\_n\}\cap \{-1,1\}\neq \emptyset$, but I'm not sure how to justify this, which is the point of this question.
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I think I should use a similar approach to below. For instance, if one assumes that $x\_i = 1$ and $x\_j=-1$ for some $i,j$, Lagrange interpolation gives the following expression: $\sum\_{1\leq k\leq n, k\neq i,j} f(x\_k) \dfrac{(x-1)(x+1)}{(x\_k - 1)(x\_k + 1)} \prod\_{a\neq j,i,k} \dfrac{x-x\_a}{x\_k - x\_a} + f(1) \dfrac{x+1}{1+1}\prod\_{1\leq k\leq n, k\neq j,i} \dfrac{x-x\_k}{1-x\_k} + f(-1)\dfrac{x-1}{-1-1}\prod\_{1\leq k\leq n,k\neq i,j}\dfrac{x-x\_k}{-1-x\_k}$.

The issue is that we need one more interpolation point to uniquely determine $f$ in this case; the above expression clearly does not have to equal $f$.

Now assuming that we can ignore the above case, one can use Lagrange interpolation to get the following expression for $f$: $\sum\_{i=1}^n f(x\_i) \dfrac{(x-1)(x+1)}{(x\_i - 1)(x\_i + 1)} \prod\_{j\neq i} \dfrac{x-x\_j}{x\_i - x\_j} + f(1) \dfrac{x+1}{1+1}\prod\_{1\leq i\leq n} \dfrac{x-x\_i}{1-x\_i} + f(-1)\dfrac{x-1}{-1-1}\prod\_{1\leq i\leq n}\dfrac{x-x\_i}{-1-x\_i}$. Setting the coefficient of $x^{n+1}$ on both sides equal to zero, we eventually get that if $H(x\_1,\cdots, x\_n)$ denotes the desired rational function, then $H(x\_1,\cdots, x\_n) = \dfrac{1}2 (1 +(-1)^{n+1})$. Also there may be other methods involving multivariate polynomials.