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Question Number 68960 by ajfour last updated on 17/Sep/19

Commented by TawaTawa last updated on 17/Sep/19

$Wow, weldone sir, God bless you sir$
Commented by ajfour last updated on 17/Sep/19
$a=b=c=d=e=−1 λ^2 −λ−1=0 ⇒ λ=((1±(√5))/2) A=−λ^2 −λ−1=−(2+λ) =−(((5±(√5))/2)) Q_1 =q^2 −q−1 , Q_2 =−q^2 −q−1 L_1 =2λq++b(λ+q)+2d =2(((1±(√5))/2))q−((1±(√5))/2)−q−2 =±(√5)q−(((5±(√5))/2)) L_2 =2aλq+c(λ+q)+2e =−2(((1±(√5))/2))q−((1±(√5))/2)−q−2 =−(2±(√5))q−(((5±(√5))/2)) Now AQ_1 ^2 −L_1 L_2 Q_1 +L_1 ^2 Q_2 =0 −(((5±(√5))/2))(q^2 −q−1)^2 −[±(√5)q−(((5±(√5))/2))][−(2±(√5))q−(((5±(√5))/2))] ×(q^2 −q−1) +[±(√5)q−(((5±(√5))/2))]^2 (−q^2 −q−1)=0 ⇒ −2(5±(√5))(q^2 −q−1)^2 +[±2(√5)q−(5±(√5))][2(2±(√5))q+(5±(√5)] ×(q^2 −q−1) −[±2(√5)q−(5±(√5))]^2 (q^2 +q+1)=0 ⇒ let 5±(√5)=n ⇒ −2n(q^2 −q−1)^2 +[−10q+2nq−n] ×[−6q+2nq+n](q^2 −q−1) −[−10q+2nq−n]^2 (q^2 +q+1)=0 ⇒ −2n(q^2 −q−1)^2 + [(2n−10)(2n−6)q^2 −4nq−n^2 ] ×(q^2 −q−1)+ −[(2n−10)^2 q^2 −2n(2n−10)q+n^2 ] ×(q^2 +q+1) = 0 ⇒ −2n(q^4 −2q^3 −q^2 +2q+1)+ +{4(n−5)(n−3)q^4 −4q^3 [(n−5)(n−3)+n] −q^2 [4(n−5)(n−3)+n^2 −4n] +q(n^2 +4n)+n^2 } −[(2n−10)^2 q^2 −2n(2n−10)q+n^2 ] ×(q^2 +q+1)=0 ⇒ q^4 {−2n+4n^2 −32n+60 −4n^2 +40n−100} +q^3 {4n−4n^2 +28n−60 −4n^2 +40n−100 +4n^2 −20n} +q^2 {2n−5n^2 +36n−60 −4n^2 +40n−100 +4n^2 −20n−n^2 } +q{−4n+n^2 +4n 4n^2 −20n−n^2 } (−2n+n^2 −n^2 )=0 Reducing (6n−40)q^4 −(4n^2 −52n+160)q^3 −(6n^2 −58n+160)q^2 +4(n^2 −5n)q−2n=0 Again dividing by 2 (3n−20)q^4 −(2n^2 −26n+80)q^3 −(3n^2 −29n+80)q^2 +(2n^2 −10n)q−n=0 to remind again n=5±(√5) taking n_1 =5+(√5) n^2 (−2q^3 −3q^2 +2q)+ n(3q^4 +26q^3 +29q^2 −10q−1)+ 1(−20q^4 −80q^3 −80q^2 )=0 since n=5±(√5) (n−5)^2 =5 n^2 −10n+20=0 n^2 =10n−20 So, i can write n(−20q^3 −30q^2 +20q) −20(−2q^3 −3q^2 +2q) n(3q^4 +26q^3 +29q^2 −10q−1)+ 1(−20q^4 −80q^3 −80q^2 )=0 ⇒ n(3q^4 +6q^3 −q^2 +10q−1) +(−20q^4 −40q^3 −20q^2 −40q)=0 ⇒ n(3q^4 +6q^3 −q^2 +10q−1) −20(q^4 +2q^3 +q^2 +2q)=0 ________________________ (3n−20)(q^4 +2q^3 +q^2 +2q) −n(4q^2 −4q+1) = 0 _______________________ 3n−20 = −5±3(√5) = N 6n−40=−10±6(√5) −n−20 = −25∓(√5) = −((N+80)/3) 10n−40 = 10±10(√5) = 10(((N+8)/3)) ⇒ q^4 +2q^3 −(((N+80)/(3N)))q^2 +(((10N+80)/(3N)))q−(((N+20)/(3N)))=0$
$a = b = c = d = e = - 1$ $λ^{2} - λ - 1 = 0 \Rightarrow λ = \frac{1 \pm \sqrt{5}}{2}$ $A = - λ^{2} - λ - 1 = - (2 + λ)$ $= - (\frac{5 \pm \sqrt{5}}{2})$ $Q_{1} = q^{2} - q - 1, Q_{2} = - q^{2} - q - 1$ $L_{1} = 2 λ q + + b (λ + q) + 2 d$ $= 2 (\frac{1 \pm \sqrt{5}}{2}) q - \frac{1 \pm \sqrt{5}}{2} - q - 2$ $= \pm \sqrt{5} q - (\frac{5 \pm \sqrt{5}}{2})$ $L_{2} = 2 a λ q + c (λ + q) + 2 e$ $= - 2 (\frac{1 \pm \sqrt{5}}{2}) q - \frac{1 \pm \sqrt{5}}{2} - q - 2$ $= - (2 \pm \sqrt{5}) q - (\frac{5 \pm \sqrt{5}}{2})$ $N o w {A Q}_{1}^{2} - L_{1} L_{2} Q_{1} + L_{1}^{2} Q_{2} = 0$ $- (\frac{5 \pm \sqrt{5}}{2}) {(q^{2} - q - 1)}^{2}$ $- [\pm \sqrt{5} q - (\frac{5 \pm \sqrt{5}}{2})] [- (2 \pm \sqrt{5}) q - (\frac{5 \pm \sqrt{5}}{2})]$ $\times (q^{2} - q - 1)$ $+ {[\pm \sqrt{5} q - (\frac{5 \pm \sqrt{5}}{2})]}^{2} (- q^{2} - q - 1) = 0$ $\Rightarrow$ $- 2 (5 \pm \sqrt{5}) {(q^{2} - q - 1)}^{2}$ $+ [\pm 2 \sqrt{5} q - (5 \pm \sqrt{5})] [2 (2 \pm \sqrt{5}) q + (5 \pm \sqrt{5}]$ $\times (q^{2} - q - 1)$ $- {[\pm 2 \sqrt{5} q - (5 \pm \sqrt{5})]}^{2} (q^{2} + q + 1) = 0$ $\Rightarrow$ $l e t 5 \pm \sqrt{5} = n \Rightarrow$ $- 2 n {(q^{2} - q - 1)}^{2} + [- 10 q + 2 n q - n]$ $\times [- 6 q + 2 n q + n] (q^{2} - q - 1)$ $- {[- 10 q + 2 n q - n]}^{2} (q^{2} + q + 1) = 0$ $\Rightarrow$ $- 2 n {(q^{2} - q - 1)}^{2} +$ $[(2 n - 10) (2 n - 6) q^{2} - 4 n q - n^{2}]$ $\times (q^{2} - q - 1) +$ $- [{(2 n - 10)}^{2} q^{2} - 2 n (2 n - 10) q + n^{2}]$ $\times (q^{2} + q + 1) = 0$ $\Rightarrow$ $- 2 n (q^{4} - 2 q^{3} - q^{2} + 2 q + 1) +$ $+ {4 (n - 5) (n - 3) q^{4}$ $- 4 q^{3} [(n - 5) (n - 3) + n]$ $- q^{2} [4 (n - 5) (n - 3) + n^{2} - 4 n]$ $+ q (n^{2} + 4 n) + n^{2}}$ $- [{(2 n - 10)}^{2} q^{2} - 2 n (2 n - 10) q + n^{2}]$ $\times (q^{2} + q + 1) = 0$ $\Rightarrow$ $q^{4} {- 2 n + 4 n^{2} - 32 n + 60$ $- 4 n^{2} + 40 n - 100}$ $+ q^{3} {4 n - 4 n^{2} + 28 n - 60$ $- 4 n^{2} + 40 n - 100$ $+ 4 n^{2} - 20 n}$ $+ q^{2} {2 n - 5 n^{2} + 36 n - 60$ $- 4 n^{2} + 40 n - 100$ $+ 4 n^{2} - 20 n - n^{2}}$ $+ q {- 4 n + n^{2} + 4 n$ $4 n^{2} - 20 n - n^{2}}$ $(- 2 n + n^{2} - n^{2}) = 0$ $R e d u c i n g$ $(6 n - 40) q^{4} - (4 n^{2} - 52 n + 160) q^{3}$ $- (6 n^{2} - 58 n + 160) q^{2}$ $+ 4 (n^{2} - 5 n) q - 2 n = 0$ $A g a i n d i v i d i n g b y 2$ $(3 n - 20) q^{4} - (2 n^{2} - 26 n + 80) q^{3}$ $- (3 n^{2} - 29 n + 80) q^{2}$ $+ (2 n^{2} - 10 n) q - n = 0$ $t o r e m i n d a g a i n n = 5 \pm \sqrt{5}$ $t a k i n g n_{1} = 5 + \sqrt{5}$ $n^{2} (- 2 q^{3} - 3 q^{2} + 2 q) +$ $n (3 q^{4} + 26 q^{3} + 29 q^{2} - 10 q - 1) +$ $1 (- 20 q^{4} - 80 q^{3} - 80 q^{2}) = 0$ $s i n c e n = 5 \pm \sqrt{5}$ ${(n - 5)}^{2} = 5$ $n^{2} - 10 n + 20 = 0$ $n^{2} = 10 n - 20 S o, i c a n w r i t e$ $n (- 20 q^{3} - 30 q^{2} + 20 q)$ $- 20 (- 2 q^{3} - 3 q^{2} + 2 q)$ $n (3 q^{4} + 26 q^{3} + 29 q^{2} - 10 q - 1) +$ $1 (- 20 q^{4} - 80 q^{3} - 80 q^{2}) = 0$ $\Rightarrow$ $n (3 q^{4} + 6 q^{3} - q^{2} + 10 q - 1)$ $+ (- 20 q^{4} - 40 q^{3} - 20 q^{2} - 40 q) = 0$ $\Rightarrow n (3 q^{4} + 6 q^{3} - q^{2} + 10 q - 1)$ $- 20 (q^{4} + 2 q^{3} + q^{2} + 2 q) = 0$ $________________________$ $(3 n - 20) (q^{4} + 2 q^{3} + q^{2} + 2 q)$ $- n (4 q^{2} - 4 q + 1) = 0$ $_______________________$ $3 n - 20 = - 5 \pm 3 \sqrt{5} = N$ $6 n - 40 = - 10 \pm 6 \sqrt{5}$ $- n - 20 = - 25 \mp \sqrt{5} = - \frac{N + 80}{3}$ $10 n - 40 = 10 \pm 10 \sqrt{5} = 10 (\frac{N + 8}{3})$ $\Rightarrow$ $q^{4} + 2 q^{3} - (\frac{N + 80}{3 N}) q^{2}$ $+ (\frac{10 N + 80}{3 N}) q - (\frac{N + 20}{3 N}) = 0$
Commented by TawaTawa last updated on 17/Sep/19

$Is this for only a = b = c = d = e = - 1 ?$
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