Problems from Ukrainian Authors

$$f(x_1)=n,f(x_2)=n^2,f(x_3)=n^3.$$ Then for a polynomial $g(x)=f(x)-n$ we have $g(x_1)=0$, $g(x_2)=n^2-n$, $g(x_3)=n^3-n$, so $g(x)=(x-x_1)(x-t)$ for some integer $t$. Then we need to find $x_1,x_2,x_3,t$, such that $g(x_2)=(x_2-x_1)(x_2-t)=n^2-n$, $g(x_3)=(x_3-x_1)(x_3-t)=n^3-n$. It suffices to have $$x_2-x_1=1,x_2-t=n^2-n,x_3-x_1=n,x_3-t=(n^2-1).$$ For example, $x_1=0$, $x_2=1$, $x_3=n$ and $t=-n^2+n+1$ satisfy these relations. It's easy to check that the polynomial $f(x)=x(x+n^2-n-1)+n$ attains the desired values at $x=0$, $x=1$, and $x=n$.