Problem - 2134
Find all pairs $(a,b)$ of nonnegative reals such that $(a-bi)^n = a^n - b^n i$ for some positive integer $n>1$.
Taking absolute value of both sides of $(a-bi)^n=a^n - b^ni$ yields $(a^2+b^2)^n=a^{2n}+b^{2n}$. Expanding the left side will lead to the conclusion that at least one of $a$ and $b$ equals zero.
If $b=0$, then the give relation always holds. Otherwise, if $a=0$, then it works when $n \equiv 1\pmod 4$.