I want to show that if I have $2n$ numbers $a_1,a_2,\dots ,a_n,b_1,b_2,\dots b_n$ such that $\displaystyle \sum_{i=1}^{n}(a_i)^j=\sum_{i=1}^{n}(b_i)^j$ for every $j=1,2,\dots ,n$, then $\{a_1,a_2,\dots ,a_n\}=\{b_1,b_2,\dots ,b_n\}$. The only idea I've had so far is defining in some way two polynomials that are equal and whose roots are, respectively, $a_1,a_2,\dots ,a_n$ and $b_1,b_2,\dots b_n$. However, I have no idea if I'm on the right track. I would appreciate any kind of help.


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    $\begingroup$ Newton's identities. $\endgroup$ Oct 13 '14 at 3:23

Here I will give a proof for $n=2,3.$ Similarly using Elementary symmetric polynomials you can prove the result.

For $n=1$ assertion is clearly true.

For $n=2,$
We have $a_1+a_2=b_1+b_2$ and $a_1^2+a_2^2=b_1^2+b_2^2.$
Using the above expressions, we can obtain $a_1a_2=b_1b_2.$
Since $a_1+a_2=b_1+b_2$ and $a_1a_2=b_1b_2,$ it is the case $\{a_1, a_2\}=\{b_1, b_2\}.$

For $n=3,$
We have $a_1+a_2+a_3=b_1+b_2+b_3,\,\,a_1^2+a_2^2+a_3^2=b_1^2+b_2^2+b_3^2$ and $a_1^3+a_2^3+a_3^3=b_1^2+b_2^3+b_3^3.$
Using these identities we can obtain $a_1a_2+a_2a_3+a_3a_1=b_1b_2+b_2b_3+b_3b_1$ and
Hence $\{a_1, a_2, a_3\}=\{b_1, b_2, b_3\}.$



Here is one line of attack. First try to first show that the condition $\sum_{i=1}^n a_i^j = \sum_{i=1}^n b_i^j$ implies

$$e_k(a_1,a_2,\ldots,a_n) = e_k(b_1,b_2,\ldots,b_n)$$

for $k=0,1,2,\ldots,n$ where $e_k$ are the elementary symmetric polynomials. This result follows from the Newton identities using induction.

Then you can use the identity

$$\prod_{i=1}^n(x+a_i) = x^n + e_1(a_1,a_2,\ldots,a_n)x^{n-1} + e_2(a_1,a_2,\ldots,a_n)x^{n-2} + \ldots + e_n(a_1,a_2,\ldots,a_n)$$

to show that

$$\prod_{i=1}^n(x+a_i) = \prod_{i=1}^n(x+b_i)$$

and conclude from there.


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