Deceiving powers #1

Let $P^n = x^n+y^n+z^n$ , where $n$ is a positive integer; $S_1 = x+y+z = -1$ ; $S_2 = xy+yz+zx$ , and $S_3 = xyz$ . By Newton's sums or identities , we have:

$\begin{aligned} P_1 & = S_1 = -1 \\ P_2 & = S_1 P_1 - 2S_2 = -1(-1) - 2S_2 = -2 & \small \blue{\implies S_2 = \frac 32} \\ P_3 & = S_1 P_2 - S_2P_1 + 3S_3 = -1(-2) - \frac 32(-1) + 3S_3 = -3 & \small \blue{\implies S_3 = - \frac {13}2} \\ P_4 & = S_1 P_3 - S_2P_2 + S_3P_1 = -1(-3) - \frac 32(-2) - \frac {13}6(-1) = \frac {49}6 \end{aligned}$

Therefore, $|a|+|b| = 49+6 = \boxed{55}$ .

We can start by squaring equation #1:

$(x+y+z)^2=(-1)^2$

$\color{#3D99F6}{x^2+y^2+z^2}\color{#69047E}{+2xy+2xz+2yz } \color{#333333}{=1}$

We recognize $\color{#3D99F6}{x^2+y^2+z^2}=-2$ so:

$\implies\color{#69047E}{2xy+2xz+2yz=3}$

$\color{#69047E}{xy+xz+yz=\dfrac{3}{2}}\text{ (a)}$

Now, if we cube equation #1:

$(x+y+z)^3=(-1)^3$

$\color{#3D99F6}{x^3+y^3+z^3}\color{#69047E}{+3x^2y+3x^2z+3xy^2+3xz^2+3y^2z+3yz^2+6xyz}\color{#333333}{=-1}$

$\implies\color{#69047E}{x^2y+x^2z+xy^2+xz^2+y^2z+yz^2+2xyz=\dfrac{2}{3}}\text{ (b)}$

Observe that if we multiply $\text{(a)}$ by $x$ , $y$ and $z$ :

$x^2y+x^2z+xyz=\dfrac{3x}{2}$

$xy^2+xyz+y^2z=\dfrac{3y}{2}$

$xyz+xz^2+yz^2=\dfrac{3z}{2}$

And add them we get $\text{(b)}$ with an extra $xyz$ term, so:

$\implies\color{#69047E}{x^2y+x^2z+xy^2+xz^2+y^2z+yz^2+2xyz}\color{#333333}{=\dfrac{3}{2}(x+y+z)-xyz}$

$\color{#69047E}{\dfrac{2}{3}}\color{#333333}{=-\dfrac{3}{2}}-xyz$

$\implies xyz=-\dfrac{13}{6}$

We can also square equation #2 to get:

$(x^2+y^2+z^2)^2=\color{#EC7300}{x^4+y^4+z^4}\color{#D61F06}{+2x^2y^2+2x^2z^2+2y^2z^2}\color{#333333}{ \equiv 4}$

To find out what the red terms are let us square equation $\text{(a)}$ :

$(xy+xz+yz)^2=\color{#D61F06}{x^2y^2+x^2z^2+y^2z^2}\color{#20A900}{+2x^2yz+2xy^2z+2xyz^2}\color{#333333}{\equiv \dfrac{9}{4}}$

We can factor $2xyz=-\dfrac{13}{3}$ from the green terms:

$\implies \color{#20A900}{-\dfrac{13}{3}(x+y+z)=\dfrac{13}{3}}$

And we find out that $\color{#D61F06}{x^2y^2+x^2z^2+y^2z^2}\color{#333333}{=-\dfrac{25}{12}}$ . Finally:

$\color{#EC7300}{x^4+y^4+z^4}\color{#D61F06}{+2x^2y^2+2x^2z^2+2y^2z^2}\color{#333333}{=4}$

$\color{#EC7300}{x^4+y^4+z^4}\color{#D61F06}{-\dfrac{25}{6}}\color{#333333}{=4}$

$\implies\color{#EC7300}{x^4+y^4+z^4}\color{#333333}{=\dfrac{49}{6}}$

The final answer is $49+6=\boxed{55}$