Determine the integers $n$ such that $\frac{1}{3} n^{4}-\frac{1}{21} n^{3}-n^{2}-\frac{11}{21} n+\frac{4}{42}=0$.
By multiplying both sides of the equality by 21, we end up with the equation $7 n^{4}-n^{3}-n^{2}-$ $11 n+2=0$. The same reasoning as in the previous exercises indicates that a solution $n$ satisfies $n \mid 2$. The solutions of this equation are then -1 and 2.