Suppose $A$ is a complex number and $n\in \mathbb{N}$ such that $A^n = (A+1)^n= 1$, then the least value of $n$ is?
I really doubt this question is correct because if $A= A+1$ then $1=0$, which is obviously false. If it correct, I'd like to know the reason(s) and how do I begin solving the problem (full solution not needed).
If $A^n=1$ and $(A+1)^n=1$ then both $A$ and $A+1$ lie on the unit circle. Moreover $0$, $A$ and $A+1$ form the vertices of an equilateral triangle $T$. The side of $T$ opposite $0$ is horizontal, so the only possible $A$ are $A=\frac12(-1+i\sqrt 3)$ and $A=\frac12(-1-i\sqrt 3)$, as $T$ must be invariant under reflection in the imaginary axis. Both values of $A$ are primitive cube roots of unity and both values of $A+1$ are primitive sixth roots of unity. Therefore both $A^n=1$ and $(A+1)^n=1$ iff $n$ is a multiple of $6$.
