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Natural number $n$ is said to be Beautiful if there exists four complex number $a,b,c,d$ such that $a^n=b^n=c^n=d^n=1$ and $a+b+c+d=1$ . Show that there exists beautiful number and explain whether 28 is beautiful or not

I'm more concerned with how I can derive this problem

lio
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    A complex number $z$ is said to be an $n$th root of unity if $z^n = 1$; so really all that line says is $a,b,c,d$ are $n$th roots of unity. There's nothing to derive, really, it's basically just the definition – PrincessEev Aug 10 '20 at 01:25
  • For existence, you can say $n=0$ and $a=-\omega$, $b=-a^2$, $c=a^3$ and $d=-a^4$ where $\omega = \exp{\frac{2i\pi}{5}}$. In fact, the sufficient and necessary conditions for your problem is that $n\equiv 0\mod 5$, if $a,b,c,d$ are distinct. (Otherwise it's just adjusting the norm) Have fun proving this! –  Aug 10 '20 at 01:43
  • If 0 is allowed then any four numbers adding up to 1 like all $1 \over 4$s will do the job so I think the natural number must refer to starting from 1. – cr001 Aug 10 '20 at 01:49
  • Roots of order $5$ of $-1$ – Conrad Aug 10 '20 at 01:57
  • Yes looks like $n=10$ works. To prove that's the only case seems not so easy. – cr001 Aug 10 '20 at 01:59
  • $a=b=1, c,d$ the other roots of unity of order $3$ works if we allow the numbers to repeat – Conrad Aug 10 '20 at 02:07
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    $i,-i,\exp(i\pi/3),\exp(-i\pi/3)$ – Empy2 Aug 10 '20 at 02:34
  • @Empy2 that's interesting! It might not be a necessary condition to have $n\equiv 0\mod 5$ –  Aug 10 '20 at 03:23

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