An application of newtons sum

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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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, $p_1 = p_2 = \cdots = p_k = 0$ .

Pi Han Goh - 3 years, 11 months ago

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