Given-p-1-i-5-and-q-1-i-5-prove-that-p-6-q-6-2-5-11-

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

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 \dots .