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Relevant wiki: Newton's Identities

Let $P_n = a^n+b^n+c^n+d^n$ , where $n$ is a positive integer, $S_1 = a+b+c+d$ , $S_2=ab+ac+ad+bc+bd+cd$ , $S_3=abc + abd+acd+bcd$ , and $S_4 = abcd$ . By Newton's sums or Newton's identities,

$\begin{aligned} P_1 & = S_1 = 21 \\ P_2 & = S_1P_1 - 2S_2 = 21(21) - 2S_2 = 13 & \small \color{#3D99F6} \implies S_2 = 214 \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 = 21(13) - 214(21) + 3S_3 = 12 & \small \color{#3D99F6} \implies S_3 = 1411 \\ P_4 & = S_1P_3 - S_2P_2 + S_3P_1 - 4S_4 = 21(12) - 214(13) + 1411(21) - 4S_4 = 97 & \small \color{#3D99F6} \implies S_4 = 6751 \\ P_5 & = S_1P_4 - S_2P_3 + S_3P_2 - S_4P_1 = 21(97) - 214(12) + 1411(13) - 6751(21) = \boxed{-123959} \end{aligned}$

Since $\tfrac12(21^2 - 13) = 214$ , we see that $a,b,c,d$ satisfy a quartic equation $X^4 - 21X^3 + 214X^2 - \alpha X + \beta = 0$ for some $\alpha,\beta$ to be determined. This means that $S_{n+4} - 21S_{n+3} + 214S_{n+2} - \alpha S_{n+1} + \beta S_n \; = \; 0 \hspace{2cm} n \in \mathbb{Z}$ where $S_n \; = \; a^n + b^n + c^n + d^n \hspace{2cm} n \in \mathbb{Z}$ Since $S_{-1} = \tfrac{\alpha}{\beta}$ , we see that (using the case $n=-1$ ) $12 - 21 \times 13 + 214 \times21 - 4\alpha + \alpha \; = \; 0$ and hence $\alpha = 1411$ . But (using the case $n=0$ ) $97 - 21\times12 + 214\times13 - 1411\times21 + 4\beta \; = \; 0$ and hence $\beta = 6751$ . Thus (putting $n=1$ ) $S_5 \; = \; 21S_4 - 214S_3 + 1411S_2 - 6751S_1 \; = \; \boxed{-123959}$