Let $X=\{x\_1,x\_2,...,x\_n\}$ and $Y=\{y\_1,y\_2,...,y\_n\}$ be sets over a finite field $F$ with $p=char(F)>2$. Assume 


$$x\_1^k+x\_2^k+...+x\_n^k=y\_1^k+y\_2^k+...+y\_n^k,\ 1\leq k\leq n$$ 


I wanna show that $X=Y$. 



Show that sets are equal
Use the standard notations $e\_k=\sum\_{A\subset \{1,\dots,n\}, |A|=k} \prod\_{i\in A} x\_i$, with the conventions $e\_0=1$ and $e\_m=0$ for $m>n$; $p\_k=\sum\_{i=1}^n x\_i^k$.


If $n=p$, the statement is true if you require your conditions for all $k$, not just $k\le n$. 


Indeed, Newton's identities say that
 $$
ke\_k=\sum\_{i=1}^k (-1)^{i-1} p\_ie\_{k-i} 
 $$
for all $k$
Of course for $k=p$ we shall have an issue with finding $e\_p$ from the $p$-th equation. But, if you look at the $p+1$-st equation, you can recover $e\_p$ from it unless $p\_1=0$, or if $p\_1=p\_2=\cdots=p\_{m-1}=0$ and $p\_m\ne0$, then the $m+p$-th equation will give you $e\_p$, the $m+p+1$st equation will give you $e\_{p+1}$ etc. Thus, you recover all the elementary symmetric polynomials unless $p\_k=0$ for all $k$. In this case, Newton's identities tell us that $e\_k=0$ for $k$ not divisible by $p$, so the polynomial 
 $$
(x-x\_1)(x-x\_2)\cdots(x-x\_n)
 $$
is a polynomial in $x^p$. Over a finite field of characteristic $p$, we have $g(x^p)=(h(x))^p$ for some $h(x)$, since every element is a $p$-th power. Thus, there will be repetitions among $x\_i$, which is what you do not allow.


Originally, I claimed this for all $n$, but as noted in comments, for $n>p$ one has to be more careful.


In addition, for $n=3=p$, the formula given by @zibadawatimmy [here](https://chat.stackexchange.com/rooms/75017/discussion-between-zibadawa-timmy-and-randomuser) can be made explicit: $(0,1,-1)$ and $(-1+i,i,1+i)$ have the same first three power sums in $\mathbb{F}\_9=\mathbb{F}\_3[i]/(i^2+1)$. Indeed: 
\begin{gather}
0+1+(-1)=0=(-1+i)+i+(1+i),\\
0^2+1^2+(-1)^2=-1=(-1+i)^2+i^2+(1+i)^2,\\
0^3+1^3+(-1)^3=0=(-1+i)^3+i^3+(1+i)^3.
\end{gather}
Thus, it is not sufficient to ask for these formulas for $1\le k\le n$.


