Proving that : $(\forall n: \sum_{i=1}^{m} a_i x_i^{n} =\sum_{i=1}^{m} a_i y_i^{n} )\rightarrow (\sum_{i=1}^{m}x_i =\sum_{i=1}^{m}y_i )$ Let $(x_i)^{m}_{i=1}$,$(y_i)^{m}_{i=1}$ and $(a_i)^{m}_{i=1}$ be  tuples of positive non-zero reals. If for all positive integers $n$:
\begin{equation}
   \sum_{i=1}^{m}a_ix_i^{n} = \sum_{i=1}^{m}a_iy_i^{n} \tag{1}
  \end{equation}
Then
\begin{equation*}
  \sum_{i=1}^{m}x_i = \sum_{i=1}^{m}y_i
\end{equation*}
NOTE: This is not a textbook problem statement. It could potentially be false, so I would accept a counter example as an answer too.
MY WORK
I have been trying to prove this statement. Through help of MSE, I have been able to prove that if $\{u_{i}\}^{m'}_{i=1}$ and  $\{v_{i}\}^{m''}_{i=1}$ are the set of unique entries in $(x_i)^{m}_{i=1}$ and $(y_i)^{m}_{i=1}$ respectively. Then $\{u_{i}\}^{m'}_{i=1}$ and  $\{v_{i}\}^{m''}_{i=1}$ are the same set, the proof goes as follows:
Let $\{u_{i}\}^{m'}_{i=1}$ and  $\{v_{i}\}^{m''}_{i=1}$ be the set of unique entries in $(x_i)^{m}_{i=1}$ and $(y_i)^{m}_{i=1}$ respectively. Also, without loss of generality, we may assume an ordering such that $u_1 > u_2> ... >u_{m'} $ and $v_1 > v_2> ...>v_{m''} $ and also that $m''\geq m'$. We can rewrite $(1)$ as:
\begin{equation}
  \label{eq: lemma_1_equivalence}
 \forall n \in \mathbb{Z^{+}}:  \sum_{i=1}^{m'}c_iu_i^{n} = \sum_{i=1}^{m''}d_iv_i^{n} \tag{2} 
\end{equation}
As $n$ grows the leading term on LHS is $c_1u_{1}^{n}$ and on the RHS is $d_1v_{1}^{n}$. Hence, it must be :
\begin{equation*}
  \forall n \in \mathbb{Z^{+}}: c_1 u_{1}^{n}  = d_1 v_{1}^{n} 
\end{equation*}
Since, $u_1,v_1,c_1$ and $d_1$ are non-zero positive reals, we can conclude that $u_1=v_1$ and $c_1 = d_1$. Hence, we may subtract $c_1 u_{1}^{n}$ from both sides in $(2)$ to get :
\begin{equation}
  \label{eq: lemma_1_equivalence_1}
 \forall n \in \mathbb{Z^{+}}:  \sum_{i=2}^{m'}c_iu_i^{n} = \sum_{i=2}^{m''}d_iv_i^{n} 
\end{equation}
We may now repeat the aforementioned argument and infer that $u_2=v_2$ and $c_2 = d_2$. Furthermore, repeating this argument $m'$ times, we can infer that $\{u_i\}^{m'}_{i=1} = \{v_i\}^{m'}_{i=1}$, leaving us with $0 =\sum_{i=m''-m'+1}^{m''}d_iv_i^{n}$, which is a contradiction, hence, $m'=m''$. Hence, we have that $\{u_{i}\}^{m'}_{i=1}$ = $\{v_{i}\}^{m''}_{i=1}$.
 A: The statement is not true. Consider $m=3$, and
$$(x_0,x_1,x_2)=(1,2,2), \quad \quad (y_0,y_1,y_2)=(2,1,1), \quad \quad \text{and } \quad (a_0,a_1,a_2)=(2,1,1)$$
Then for every $n \geq 0$, one has
$$\sum_{i=1}^m a_ix_i^n = 2 \times 1^n + 1 \times 2^n + 1 \times 2^n = 2 + 2^{n+1}$$
and
$$\sum_{i=1}^m a_iy_i^n = 2 \times 2^n + 1 \times 1^n + 1 \times 1^n = 2 + 2^{n+1}$$
but obviously,
$$\sum_{i=1}^3 x_i = 5 \neq 4 = \sum_{i=1}^3 y_i$$
A: Though the statement is not true, we can get some simpler conclusions on $a,x,y$:
$$
\forall r,\sum_{i=1}^m[x_i=r]a_i=\sum_{i=1}^m[y_i=r]a_i
$$
where $[P]$ is $1$ if $P$ if true, and $0$ otherwise.
Considering the generating function using the placeholder $z$. We have
$$
\sum_{i=1}^m\frac{a_i}{1-x_iz}=\sum_{i=1}^m\sum_{n=0}^{+\infty}a_iz^nx_i^n=\sum_{n=0}^{+\infty}z^n\sum_{i=1}^ma_ix_i^n=\sum_{n=0}^{+\infty}z^n\sum_{i=1}^ma_iy_i^n=\sum_{i=1}^m\frac{a_i}{1-y_iz}
$$
Lemma: If the g.f. of $z$ satisfies
$$
\sum_{i=1}^t\frac{a_i}{1-x_iz}=0
$$
and $x_i$ has no repeating terms, then $\forall i,a_i=0$.
Proof: For a fixed $i$, multiply both sides with $(1-x_iz)$, and then let $z$ take the value $x_i^{-1}$, we get $a_i=0$. $\blacksquare$
Finally, it is not hard to get the conclusion considering coefficient of $1/(1-rz)$ on both sides.
