Let $p_i(x)$, $p(x)$ be real coefficient polynomials. Suppose that $$\sum_{i=0}^{n-1}x^ip_i(x^{in})=p(x^n), (x-1)\mid p(x).$$ Show that $p_i(x)=0$, $1\leq i\leq n-1$.
I could only show that $p_i(1)=0$, once we take $x$ to be the $n$-th root of $1$...What about the other $x$?
Let $y$ a $n$-th root of unity. Replace $x$ by $yx$. You get $$\sum_{i=0}^{n-1}y^ix^{i}p_{i}(x^{in})-p(x^n)=0$$ Now fix $x$, and let: $$Q(t)=\sum_{i=0}^{n-1}t^ix^{i}p_{i}(x^{in})-p(x^n)$$
This is a polynomial of degree $\leq n-1$ in $t$, and by the above it has as roots the $n$ distincts $n$-th roots of unity. Then $Q$ is the zero polynomial, and all its coefficients (in the variable $t$) are $0$, and we are done (we get also $p_0(1)-p(x^n)=0$).
