Given p=1+i(√5) and q=1−i(√5) prove that p^6 +q^6 = 2^5 ×11


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Question Number 138014 by bobhans last updated on 09/Apr/21

$G i v e n p = 1 + i \sqrt{5} a n d q = 1 - i \sqrt{5}$ $p r o v e t h a t p^{6} + q^{6} = 2^{5} \times 11$
	Answered by bobhans last updated on 09/Apr/21
	$from the given we have ⇒x^2 −2x+6 = 0 has the roots { ((x_1 = p=1+i(√5))),((x_2 = q =1−i(√5))) :} ⇒by Vieta′s rule { ((p+q=2)),((pq = 6)) :} we want to find p^6 +q^6 . (p^3 )^2 +(q^3 )^2 = [ p^3 +q^3 ]^2 −2(pq)^3 = [ (p+q)^3 −3pq(p+q)]^2 −2(pq)^3 = [ 8−3(6)(2)]^2 −2(6)^3 = [ −28 ]^2 −2(216) =784−432 = 352 = 2^5 ×11$
	$f r o m t h e g i v e n w e h a v e$ $\Rightarrow x^{2} - 2 x + 6 = 0 h a s t h e r o o t s$ ${\begin{cases} x_{1} = p = 1 + i \sqrt{5} \\ x_{2} = q = 1 - i \sqrt{5} \end{cases}$ $\Rightarrow b y {V i e t a}^{'} s r u l e {\begin{cases} p + q = 2 \\ p q = 6 \end{cases}$ $w e w a n t t o f i n d p^{6} + q^{6} .$ ${(p^{3})}^{2} + {(q^{3})}^{2} = {[p^{3} + q^{3}]}^{2} - 2 {(p q)}^{3}$ $= {[{(p + q)}^{3} - 3 p q (p + q)]}^{2} - 2 {(p q)}^{3}$ $= {[8 - 3 (6) (2)]}^{2} - 2 {(6)}^{3}$ $= {[- 28]}^{2} - 2 (216)$ $= 784 - 432 = 352 = 2^{5} \times 11$
	Commented by bobhans last updated on 09/Apr/21

	$r e q u e s t m r L i k i$
	Commented by SLVR last updated on 09/Apr/21

	$s i r w h a t i s V i e t a R u l e . . p l e a s e$
	Answered by Rasheed.Sindhi last updated on 09/Apr/21

	$p + q = (1 + i \sqrt{5}) + (1 - i \sqrt{5}) = 2$ $p q = (1 + i \sqrt{5}) (1 - i \sqrt{5}) = 1 + 5 = 6$ ${(p + q)}^{2} = {(2)}^{2}$ $p^{2} + q^{2} + 2 p q = 4$ $p^{2} + q^{2} = 4 - 2 (6) = - 8$ ${(p^{2} + q^{2})}^{3} = {(- 8)}^{3} = - 512$ $p^{6} + q^{6} + 3 {(p q)}^{2} (p^{2} + q^{2}) = - 512$ $p^{6} + q^{6} = - 512 - 3 {(6)}^{2} (- 8)$ $= - 2^{9} - 3 . (3^{2} . 2^{2}) (- 2^{3})$ $= - 2^{9} + 3^{3} . 2^{5} = 2^{5} (- 2^{4} + 3^{3})$ $= 2^{5} (- 16 + 27) = 2^{5} . 11$
	Answered by mathmax by abdo last updated on 10/Apr/21

	$p^{6} + q^{6} = {(1 + i \sqrt{5})}^{6} + {(1 - i \sqrt{5})}^{6} = \sum_{k = 0}^{6} C_{6}^{k} {(i \sqrt{5})}^{k} + \sum_{k = 0}^{6} C_{6}^{k} {(- i \sqrt{5})}^{k}$ $= \sum_{k = 0}^{6} C_{6}^{k} (i^{k} + {(- i)}^{k}) {(\sqrt{5})}^{k}$ $= \sum_{p = 0}^{3} 2 C_{6}^{2 p} {(- 1)}^{p} . 5^{p} = 2 \sum_{p = 0}^{3} {(- 1)}^{p} C_{6}^{2 p} . 5^{p}$ $rest to verify that this Σ is equal to 2^{5} \times 11 . . . .$
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