For a given polynomial of degree \(n\) with roots \( r_1, r_2, \ldots r_n \), let \(s_k \) be the kth symmetric sum and \(p_k = \sum_{i=1}^n r_i ^k \) is the kth power sum

**Claim:** If \(s_1=s_2=s_3=...=s_k=0\),then \(p_1=p_2=p_3=.,..=p_k=0\). where

**Proof:** By induction on \(k \).
Base case: Since \( s_1 = p_1 \), hence if \( s_1 = 0 \) then \( p_1 = 0 \).

Induction step: Suppose that for some \(j\), we know that \( p_1, p_2, \ldots p_j = 0 \). Then, by newtons sum, we have

\[p_k=s_1p_{k-1}-s_2p_{k-2}+....+(-1)^{k-1}s_{k-2}p_2+(-1)^{k}s_{k-1}p_1+(-1)^{k+1}ks_k = 0 + 0 + \ldots + 0. \]

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## Comments

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TopNewest@Aareyan Manzoor Great note! I've cleaned it up so that it's much easier to understand what you are saying.

Presenting it via a proof by induction isn't necessary, but helps build the idea that all we need to show \( p_j = 0 \) is up to \( s_j = 0 \).

By improving the format of your presentation, you make it immediately clear what you want to show, and why it's interesting.

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thank you sir!

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There's no need to apply Newton's Sum.

If \(s_1 = s_2 = \cdots = s_k = 0 \), then the roots satisfy the equation, \(x^k = 0 \). So all its roots are 0, then by definition, \(p_1 = p_2 = \cdots = p_k = 0 \).

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what about \[x^{100}-x^2+x+1=0\] i meant for that

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I don't follow. What about that polynomial?

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this

look atLog in to reply

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nice way....

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