Vieta's Job

The problem can be solved using Newton's sums . Let $P^n = a^n + b^n + c^n + d^n$ , where $n$ is an positive integer, $S_1 = a + b + c + d = 2$ , $S_2 = ab + ac+ ad + bc + bd + cd$ , $S_3 = abc+ abd + acd + bcd$ , and $S_4 = abcd$ . Then

$\begin{aligned} P_1 & = S_1 = 2 & \implies S_1 = 2 \\ P_2 & = S_1 P_1 - 2S_2 = 4 - 2S_2 = 4 & \implies S_2 = 0 \\ P_3 & = S_1 P_2 - S_2P_1 + 3S_3 = 8 - 0 + 3S_3 = 5 & \implies S_3 = -1 \\ P_4 & = S_1 P_3 - S_2P_2 + S_3P_1 - 4S_4 = 10 - 0 - 2 - 4S_4 = 8 & \implies S_4 = 0 \\ P_n & = S_1 P_{n-1} - S_2P_{n-2} + S_3P_{n-3} - S_4P_{n-4} = 2P_{n-1} - P_{n-3} \end{aligned}$

We can use the linear recurrence relation $P_n = 2P_{n-1} - P_{n-3}$ to solve progressively from $P_5$ to $P_{25}$ . Alternatively, we solve the linear recurrence relation and we get $P_n = \varphi^n + (1-\varphi)^2 + 1$ , where $\varphi = \dfrac {1+\sqrt 5}2$ is the golden ratio (see note). Then $P_{25} = \varphi^{25} + (1-\varphi)^{25} + 1 = \boxed {167762}$ .

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Note: The characteristic equation of $P_n = 2P_{n-1} - P_{n-3}$ is:

$\begin{aligned} r^3 & = 2r^2 - 1 \\ r^3-r^2 - 1 & = 0 \\ (r-1)(r^2 - r -1) & = 0 \\ \implies r & = 1, \varphi, 1 - \varphi \\ \implies P_n & = c_1 + c_2 \varphi^n+c_3(1 - \varphi)^n \end{aligned}$

$\begin{array} {rl} P_1: & c_1 + c_2 \varphi+c_3(1 - \varphi) = 2 \\ P_2: & c_1 + c_2 \varphi^2+c_3(1 - \varphi)^2 = c_1 + c_2(1+\varphi)+c_3(2 - \varphi) = 4 \\ P_3: & c_1 + c_2 \varphi^3+c_3(1 - \varphi)^3 = c_1 + c_2(1+2\varphi)+c_3(3-2\varphi) = 5 \\ P_2-P_1: & c_2 + c_3 = 2 \\ P_3-P_2: & c_2\varphi + c_3(1-\varphi) = 1 \end{array}$

$\begin{array} {rll} (1-\varphi)(P_2-P_1)-(P_3-P_2): & c_2(1-2\varphi) = 1-2\varphi & \blue{\implies c_2 = 1} \\ P_2-P_1: & 1 + c_3 = 2 & \blue{\implies c_3 = 1} \\ P_1: & c_1 + \varphi+ 1 - \varphi = 2 & \blue{\implies c_1 = 1} \end{array}$

From @Chew-Seong Cheong 's solution, we can also know that $S_1=2, S_2=0, S_3=-1, S_4=0$ , so $a,b,c,d$ are the roots of the equation $x^4-2x^3+x=0$ and the roots are $0, 1, \phi, 1-\phi$ . Substitute them in and we get our answer.