The Determinant

We have $s_j = j$ for $1 \le j \le 4$ and $s_0=4$ , where $s_k = \sum_{\mathrm{cyc}}a^k$ . We also have the elementary symmetric polynomials $e_1 =s_1 =1$ , $e_2 = \tfrac12(s_1^2 - s_2) = -\tfrac12$ , $e_3 = \sum_{\mathrm{cyc}}abc$ and $e_4 = abcd$ . Now $a,b,c,d$ are the roots of the quartic $X^4 - X^3 - \tfrac12X^2 - e_3X + e_4 = 0$ , and hence $0 \; = \;s_4 - s_3 - \tfrac12s_2 - e_3s_1 + e_4s_0 = 4e_4 - e_3$ so that $e_3 = 4e_4$ . Also $-1 \; = \; s_1s_2 - s_3 \; =\; \sum_{\mathrm{cyc}}a^2b \; = \; e_1e_2 - 3e_3 \; =\; -\tfrac12 - 3e_3$ and hence $e_3 = \tfrac16$ and therefore $e_4 = \tfrac{1}{24}$ .

As I said (if a bit concisely) $s_j \;=\; a^j+b^j+c^j+d^j$ for any $j$ . Thus $s_0=4$ . Since $a,b,c,d$ are the roots of $(x-a)(x-b)(x-c)(x-d)=x^4 -e_1x^3+e_2x^2-e_3x+e_4=0$ we see that, for any $n \ge 0$ , $a^{n+4} - e_1 a^{n+3} + e_2 a^{n+2} - e_3 a^{n+1} + e_4 a^n = 0$ and similarly for $b,c,d$ , so $s_{n+4} -e_1 s_{n+3} +e_2 s_{n+2} - e_3 s_{n+1} + e_4 s_n =0$ For any $n \ge 0$ . Try this with $n=0$ .