# Newton's Sums

Newton sums give us a clever and efficient way of finding the sums of roots of a polynomial raised to a power. They can also be used to derive several factoring identities.

## Statement

Consider a polynomial $P(x)$ of degree $n$, $P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$

Let $P(x)=0$ have roots $x_1,x_2,\ldots,x_n$. Define the following sums: $P_1 = x_1 + x_2 + \cdots + x_n$ $P_2 = x_1^2 + x_2^2 + \cdots + x_n^2$ $\vdots$ $P_k = x_1^k + x_2^k + \cdots + x_n^k$ $\vdots$

Newton sums tell us that, $a_nP_1 + a_{n-1} = 0$ $a_nP_2 + a_{n-1}P_1 + 2a_{n-2}=0$ $a_nP_3 + a_{n-1}P_2 + a_{n-2}P_1 + 3a_{n-3}=0$ $\vdots$

(Define $a_j = 0$ for $j<0$.)

We also can write: $P_1 = S_1$ $P_2 = S_1P_1 - 2S_2$

etc., where $S_n$ denotes the $n$-th elementary symmetric sum.

## Proof

Let $\alpha,\beta,\gamma,...,\omega$ be the roots of a given polynomial $P(x)=a_nx^n+a_{n-1}x^{n-1}+..+a_1x+a_0$. Then, we have that $P(\alpha)=P(\beta)=P(\gamma)=...=P(\omega)=0$

Thus, $\begin{cases}a_n\alpha^n+a_{n-1}\alpha^{n-1}+...+a_0=0\\a_n\beta^n+a_{n-1}\beta^{n-1}+...+a_0=0\\~~~~~~~~~~~~~~~~~~\vdots\\a_n\omega^n+a_{n-1}\omega^{n-1}+...+a_0=0\end{cases}$

Multiplying each equation by $\alpha^{k-n},\beta^{k-n},...,\omega^{k-n}$, respectively, $\begin{cases}a_n\alpha^{n+k-n}+a_{n-1}\alpha^{n-1+k-n}+...+a_0\alpha^{k-n}=0\\a_n\beta^{n+k-n}+a_{n-1}\beta^{n-1+k-n}+...+a_0\beta^{k-n}=0\\~~~~~~~~~~~~~~~~~~\vdots\\a_n\omega^{n+k-n}+a_{n-1}\omega^{n-1+k-n}+...+a_0\omega^{k-n}=0\end{cases}$ $\begin{cases}a_n\alpha^{k}+a_{n-1}\alpha^{k-1}+...+a_0\alpha^{k-n}=0\\a_n\beta^{k}+a_{n-1}\beta^{k-1}+...+a_0\beta^{k-n}=0\\~~~~~~~~~~~~~~~~~~\vdots\\a_n\omega^{k}+a_{n-1}\omega^{k-1}+...+a_0\omega^{k-n}=0\end{cases}$

Sum, $a_n\underbrace{(\alpha^k+\beta^k+...+\omega^k)}_{P_k}+a_{n-1}\underbrace{(\alpha^{k-1}+\beta^{k-1}+...+\omega^{k-1})}_{P_{k-1}}+a_{n-2}\underbrace{(\alpha^{k-2}+\beta^{k-2}+...+\omega^{k-2})}_{P_{k-2}}+...+a_0\underbrace{(\alpha^{k-n}+\beta^{k-n}+...+\omega^{k-n})}_{P_{k-n}}=0$

Therefore, $\boxed{a_nP_k+a_{n-1}P_{k-1}+a_{n-2}P_{k-2}+...+a_0P_{k-n}=0}$

## Example

For a more concrete example, consider the polynomial $P(x) = x^3 + 3x^2 + 4x - 8$. Let the roots of $P(x)$ be $r, s$ and $t$. Find $r^2 + s^2 + t^2$ and $r^4 + s^4 + t^4$.

Newton's Sums tell us that: $P_1 + 3 = 0$ $P_2 + 3P_1 + 8 = 0$ $P_3 + 3P_2 + 4P_1 - 24 = 0$ $P_4 + 3P_3 + 4P_2 - 8P_1 = 0$

Solving, first for $P_1$, and then for the other variables, yields, $P_1 = r + s + t = -3$ $P_2 = r^2 + s^2 + t^2 = 1$ $P_3 = r^3 + s^3 + t^3 = 33$ $P_4 = r^4 + s^4 + t^4 = -127$

Which gives us our desired solutions, $\boxed{1}$ and $\boxed{-127}$.