Many Variables

It is given that:

$\begin{cases} x + y + z = 1 &...(1) \\ 13^2 x + 17^2 y + 19^2 z = 169x + 289y + 361z = w^2 &...(2) \\ 13^3 x + 17^3 y + 19^3 z = 2197x + 4913y + 6859z = w^3 &...(3) \\ 13^4 x + 17^4 y + 19^4 z = 28561x + 83521y + 130321z = w^4 &...(4) \end{cases}$

In matrix, (2), (3) and (4) is as follows:

$\begin{aligned} \begin{pmatrix} 169 & 289 & 361 \\ 2197 & 4913 & 6859 \\ 28561 & 83521 & 130321 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} & = \begin{pmatrix} w^2 \\ w^3 \\ w^4 \end{pmatrix} \end{aligned}$

$\begin{aligned} \Rightarrow \begin{pmatrix} x \\ y \\ z \end{pmatrix} & = \begin{pmatrix} 169 & 289 & 361 \\ 2197 & 4913 & 6859 \\ 28561 & 83521 & 130321 \end{pmatrix}^{-1} \begin{pmatrix} w^2 \\ w^3 \\ w^4 \end{pmatrix} \\ & = \frac{1}{423158424} \begin{pmatrix} 33698267 & -3755844 & 104329 \\ -45207669 & 5856864 & -183027 \\ 21587722 & -2930460 & 97682 \end{pmatrix} \begin{pmatrix} w^2 \\ w^3 \\ w^4 \end{pmatrix} \\ \Rightarrow x + y + z & = \frac{10078320w^2 -829440w^3 +18984w^4}{423158424} \\ 1 & = \frac{419930w^2-34560w^3+791w^4}{17631601} \end{aligned}$

$\begin{aligned} \Rightarrow 791w^4-34560w^3 + 419930w^2 -17631601 & = 0 \\ \left(w+\frac{4199}{791} \right)(w-13)(w-17)(w-19) & = 0 \end{aligned}$

We note that $w = - \frac{4199}{791} \approx -5.308470291$ and $|w| \approx \boxed{5.3} < 6$ .

Define $s_n = xa^n+yb^n+zc^n$

The characteristic equation of this recursion has roots $a,b,c$ . Therefore the recursion follows $s_n -ps_{n-1} + qs_{n-2} - rs_{n-3}= 0$ where $(s-a)(s-b)(s-c) = s^3 - ps^2 + qs - r$ .

We know that $s_0 = 1 , s_2 = w^2 , s_3 = w^3 , s_4 = w^4$ $s_4 - ps_3 + qs_2 - rs_1 = 0$ $s_3 - ps_2 + qs_1 - rs_0 = 0$

Plugging in the values and equating the two equalities for $s_1$ , we get that $\frac{w^4 - pw^3 + qw^2}{r} = \frac{-w^3 + pw^2 + r}{q},$ $qw^4 - (pq - r)w^3 + (q^2 - pr)w^2 - r^2 = 0.$ $qw(w^3 - pw^2 + qw - r) + r(w^3 - pw^2 +qw - r) = 0$ $(qw+r)(w - a ) (w - b)( w-c) = 0$ Since $w,a,b,c$ are distinct, you know that $w = \frac{-r}{q} \implies |w| = \frac{abc}{ab+bc+ca}$ Plugging in the values for $a,b,c$ , you get that $|w| \approx \boxed{5.3}$