cube roots of unity

Question

Let $\omega \ne 1$ be a complex cube root of unity.

If $$(4 + 5\omega + 6\omega ^2)^{n^2 + 2} + (6 + 5\omega ^2 + 4\omega)^{n^2 + 2} + (5 + 6\omega + 4\omega ^2)^{n^2 + 2} = 0$$ then $n$ can be...

Answer 1

The LHS of the given expression can be written thus:

$$\left( 4 + 5\omega + 6\omega ^2 \right)^{n^2 + 2}+ \left( (4 + 5\omega + 6\omega ^2)w \right)^{n^2 + 2}+\left( (4 + 5\omega + 6\omega ^2)w^2 \right)^{n^2 + 2}$$

$$=\left( 4 + 5\omega + 6\omega ^2 \right)^{n^2 + 2}.F \ \ \text{with} \ \ F=1+w^{n^2 + 2}+(w^2)^{n^2 + 2}$$

The first factor is never zero, being a power of a non-zero element.

Thus, it suffices to examine the cases where $F=0$.

As $n^2+2\equiv \begin{cases}2 \ (mod. 3) & when \ n \equiv 0 \ (mod. 3)\\ 0 \ (mod. 3)& when \ n \equiv1 \ (mod. 3)\\ 0 \ (mod. 3)& when \ n \equiv 2 \ (mod. 3)\end{cases}\ \ $, thus

• if $n \ \equiv 0 \ $ mod. 3: $F=1+w^{n^2 + 2}+(w^{n^2 + 2})^2=\dfrac{1-(w^{n^2 + 2})^3}{1-w^{n^2 + 2}}=\dfrac{1-(w^3)^{n^2 + 2}}{1-w^{n^2 + 2}}=0$

• otherwise $F=1+1+1=3\neq 0$ iff $n=1,2 \ mod. 3$.

Thus the result is that the LHS of your formula is 0 iff $n=3k$ for some integer $k$.