Newton's sums in reverse

Well, this was my attempt to solve this:

Let the polynomial $P(x)=x^3+mx^2+nx+p$ have roots $a$ , $b$ and $c$ . Now, if we apply Newton's sums recursively we obtain:

$a+b+c=-m$

$a^2+b^2+c^2=m^2-2n$

$a^3+b^3+c^3=-m^3+3mn-3p$

$a^4+b^4+c^4=m^4-4m^2n+4mp+2n^2$

$a^5+b^5+c^5=-m^5+5m^3n-5m^2p-5mn^2+5np$

$a^6+b^6+c^6=m^6-6m^4n+6m^3p+9m^2n^2-12mnp-2n^3+3p^2$

With the known values, the following system is formed:

$m^4-4m^2n+4mp+2n^2=32$

$-m^5+5m^3n-5m^2p-5mn^2+5np=186$

$m^6-6m^4n+6m^3p+9m^2n^2-12mnp-2n^3+3p^2=803$

It's easy to solve for $p$ from the first equation:

$p=\dfrac{32+4m^2n-2n^2-m^4}{4m}$

But the things get worse when we substitute this value, we get the following two equations after simplifying:

$m^6+10m^2n^2-10n^3-160m^2-5m^4n-744m+160n=0$

$14m^4n^2+32m^2n-34m^2n^3-m^6n+12n^4-384n^2-224m^4-3720m^3-12848m^2+3072=0$

I got stuck there, so the only thing I could do was solving for $n$ from the first equation of these two and substitute on the second equation. But the expression for $n$ is horrible because it's from a 3rd degree polynomial. So, after hours and hours, I got an equation in $m$ with the help of Excel and Wolfram Alpha to do the sums with very large numbers. This is the equation in $m$ of degree 22:

$m^{22}-4320m^{18}-84816m^{17}-401500m^{16}+1228800m^{14}+29283840m^{13}+329484096m^{12}\\+1792296000m^{11}+3273475750m^{10}-9256550400m^9-83841807360m^8-434166425856m^7\\-1696694220000m^6-5555724768000m^5-9731157199200m^4+296518164480m^3+\\1071516905856m^2=0$

Since the coefficients of $m$ and $m^0$ are $0$ and $m \neq 0$ , the degree of the equation is now 20, it's a lot easier now, don't you think? :D

Finally, using Newton-Raphson method, we see that the equation has 4 real roots and 16 complex roots. Since $m$ , $n$ and $p$ are real, these are the solutions for $m$ :

$m=-6$ , $m\approx -0.3482217748938019$ , $m\approx 0.3159770443715433$ or $m\approx 11.02483070176856$

Hence, $a+b+c=6$ , $a+b+c \approx 0.3482217748938019$ , $a+b+c \approx -0.3159770443715433$ or $a+b+c \approx -11.02483070176856$ .

Assume a third grade monic polynomial with roots a, b and c. With notations from the wiki on Newton's Identities, in excel create a table with the columns: $p_1, p_2, p_3, e_2, e_3, p_4, p_5, p_6$ Use the following values/formula's:

$p_1=1$ a positive integer we start with

Skip $p_2$ for a moment

$p_3=(p_1^4/24-p_1^2*p_2/4+p_2^2/8-32/4)/(-p_1/3)$

$e_2=(p_1^2-p_2)/2$

$e_3=(p_1^3-3*p_1*p_2+2*p_3)/6$

$p_4=p_1*p_3-e_2*p_2+e_3*p_1$

$p_5=p_1*p_4-e_2*p_3+e_3*p_2$

$p_6=p_1*p_5-e_2*p_4+e_3*p_3$

Now use the 'What-If Analysis' and then 'Goal seek...' tool in Excel (see tab 'Data') to

'Set cell' = $p_6$

'To value' = 803

'By changing' = $p_2$

You will then get $p_5=-718.72$ , clearly wrong. Changing $p_1$ to 2 and 'Goal seek...' again, gives $p_5=529.95$ . Repeat this until the values of $p_4, p_5, p_6$ are 32, 186 and 803.

This happens at $p_1=a+b+c=6$ .

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Now it appears that the monic polynomial is

$x^3-6x^2+14x-15$

and thus $a, b, c$ equal $3, \frac{3+\sqrt{11}i}2, \frac{3-\sqrt{11}i}2$ or any other permutation.

{6, 14, 15}

S = 6

P = 15

I = 14/ 15

This is simple with computer. The answer is 6.