$p(x)x^n+q(x)(1-x)^n=1$ for some $p(x), q(x)\in \mathbb{Z}[x]$, what explicitly $p(x), q(x)$ are?

Question

Because $x^n$ and $(x-1)^n$ are relatively prime in $\mathbb{Z}[x]$, so $p(x)x^n+q(x)(1-x)^n=1$ for some $p(x), q(x)\in \mathbb{Z}[x]$. What are the explicit formulas of $p(x), q(x)$?

Answer 1

You have a closed form in the more general case when you have two distinct exponents $n$ and $m$ :

$$ (1-x)^n\Bigg(\sum_{j=0}^{m-1} \binom{n+j-1}{j}x^j\Bigg) -x^m\Bigg(n\binom{n+m-1}{n}\sum_{k=0}^{n-1} \frac{(-1)^{k+1}\binom{n-1}{k}}{m+k}x^k\Bigg)=1. $$

You can also rewrite this more compactly as

$$ (1-x)^n\Bigg(\sum_{j=0}^{m-1} \binom{n+j-1}{j}x^j\Bigg) +x^m\Bigg(\sum_{k=0}^{n-1} \binom{m+k-1}{k}(1-x)^k\Bigg)=1. $$

Answer 2

Hint:

Start with the equality $$x+(1-x)=1$$ raise to a large enough power ( $2n-1$ will do it ), and use Newton binomial formula. Separate the LHS into a multiple of $x^n$ and a multiple of $(1-x)^n$.