How is it possible to prove that there are no solutions for n-odd?

That is good but what about the solution set?

Any further ideas?

Anar24 wrote: How is it possible to prove that there are no solutions for n-odd? Here is a counterexample: Consider the following complex numbers: $z_{1}=\frac{1}{3}(1+i)$ , $z_{2}=\frac{1}{3}(1-i)$, $z_{3}=\frac{1}{3}$ so $Re(z_{1}+z_{2}+z_{3})=1$ and $Im(z_{1}+z_{2}+z_{3})=0$ hence there exist $x,y,z$ real number with $cosx+cosy+cosz=1$ and $sinx+siny+sinz=0$.

I think the length of complex numbers should be 1.

I deleted the generalization, one should think about!

I think that this is true: Generalization: let be $\omega_{1},$$\omega_{2},$...,$\omega_{n-1}$ the roots of the equation $\omega^{n-1}=-1$, and we define the following complex numbers: $z_{1}=\omega_{1},$$z_{2}=\omega_{2},$...,$z_{n-1}=\omega_{n-1},$ and $z_{n}=1$, so $z_{1}+z_{2}+...+z_{n}=1$ (using Viéte), thus $Re(z_{1}+z_{2}+...+z_{n})=1$ and $Im(z_{1}+z_{2}+...+z_{n})=0$

Let $n=2m+1$ $x_i=1$, if $1\le i\le m$. $x_j=-1$, if $m+1\le j\le 2m$ and $x_{2m+1}=0$. $ x_1 + x_2 + ... + x_n = 0$, and $\sqrt{1-x_i^2}=0$ if $i\neq 2m+1$. So $(iii)$ is also true.

socrates wrote: Find all POSITIVE integers ..... But $x_{2n+1}=0$

...$n$, but $-1\le x_i\le 1$ oops it's $-1<x_i<1$

But $x_{k}\neq0$

TuZo wrote: But $x_{k}\neq0$ That is not a condition of the problem.

Let $H_m^+$ be a set of complex numbers, which can be written as sum of $m$ $z$ complex numbers, where $|z|=1, Im(z),Re(z)> 0$. And $H_m^-$ same as $H_m^+$, but $Im(z)<0$. It's easy to see $H_2^+\cap (-H_1^-+1)=\{\}$, but $H_{m+1}^+\cap (-H_m^-+1)\neq \{\}$, if $m\ge 2$ So every positvie integer $n$ are good, expcept $n=3$ (I dont have time to write down every details.)

OK x$_{k}=0$ it is possible!

@Ayers For n=3 see the this: TuZo wrote: Anar24 wrote: How is it possible to prove that there are no solutions for n-odd? Here is a counterexample: Consider the following complex numbers: $z_{1}=\frac{1}{3}(1+i)$ , $z_{2}=\frac{1}{3}(1-i)$, $z_{3}=\frac{1}{3}$ so $Re(z_{1}+z_{2}+z_{3})=1$ and $Im(z_{1}+z_{2}+z_{3})=0$ hence there exist $x,y,z$ real number with $cosx+cosy+cosz=1$ and $sinx+siny+sinz=0$.

blawho12 wrote: TuZo wrote: But $x_{k}\neq0$ That is not a condition of the problem. Yes now I see, you have right, but this is a trivial case! I gave example for all x$_{k}\neq0$

$|z_1|=\frac13 \sqrt{2}\neq1$ TuZo wrote: I think that this is true: Generalization: let be $\omega_{1},$$\omega_{2},$...,$\omega_{n-1}$ the roots of the equation $\omega^{n-1}=-1$, and we define the following complex numbers: $z_{1}=\omega_{1},$$z_{2}=\omega_{2},$...,$z_{n-1}=\omega_{n-1},$ and $z_{n}=1$, so $z_{1}+z_{2}+...+z_{n}=1$ (using Viéte), thus $Re(z_{1}+z_{2}+...+z_{n})=1$ and $Im(z_{1}+z_{2}+...+z_{n})=0$ There are some $z_i$, which dont satisfy $Re(z_i)<0$.

Ayers wrote: $|z_1|=\frac13 \sqrt{2}\neq1$...There are some $z_i$, which dont satisfy $Re(z_i)<0$. There is no need! See this: Put $x_{k}=sin(a_{k})$ and you got that $\sum_{k=1}^{n}sin(a_{k})=0$ and $\sum_{k=1}^{n}cos(a_{k})=1$

But $\sqrt{1-sin^2(x)}=|cos(x)|$

Guys,I understand all your ideas.But let me remind:It is first question of Peru TST and i don't think there will be so complicated solution!Certainly,finding solution for n-even is trivial:just put number and its opposite and you are going to find that real number!But what are ideas behind n-odd?Please,if possible,without complex roots!

The case when $n=3$ has no solution.

n=3 has no solution but what about n=5?

Anar24 wrote: Guys,I understand all your ideas.But let me remind:It is first question of Peru TST and i don't think there will be so complicated solution!Certainly,finding solution for n-even is trivial:just put number and its opposite and you are going to find that real number!But what are ideas behind n-odd?Please,if possible,without complex roots! Well, usually IMO pre-selection in Peru consists of 4 problems (one algebra, one combinatorics, one geometry and one number theory in THIS order). Students were aware that the problems are not necessarily in ascending order of difficulty.