Find-the-value-of-a-2-b-2-for-a-b-real-number-such-that-a-b-1-a-1-b-1-a-and-b-a-1-b-1-a-1-b-

Question Number 136039 by liberty last updated on 18/Mar/21

F i n d t h e v a l u e o f a^{2} + b^{2}

f o r a, b r e a l n u m b e r s u c h t h a t

a = b + \frac{1}{a + \frac{1}{b + \frac{1}{a + \dots}}}

a n d b = a - \frac{1}{b + \frac{1}{a - \frac{1}{b + \dots}}}

Commented by MJS_new last updated on 18/Mar/21

a^{2} + b^{2} = \sqrt{5}

Commented by liberty last updated on 18/Mar/21

h o w s i r ?

Answered by EDWIN88 last updated on 18/Mar/21

(1) a = b + \frac{1}{a + \frac{1}{a}}; a = b + \frac{a}{a^{2} + 1}

a (a^{2} + 1) = b (a^{2} + 1) + a; a^{3} = a^{2} b + b

\Rightarrow b = \frac{a^{3}}{a^{2} + 1}

(2) b = a - \frac{1}{b + \frac{1}{b}}; b = a - \frac{b}{b^{2} + 1}

b^{3} + b = a (b^{2} + 1) - b; b^{3} + 2 b = a (b^{2} + 1)

\Rightarrow a = \frac{b^{3} + 2 b}{b^{2} + 1} = \frac{b (b^{2} + 2)}{b^{2} + 1}

\Rightarrow a = \frac{a^{3}}{a^{2} + 1} . (\frac{b^{2} + 2}{b^{2} + 1}) \Rightarrow (a^{2} + 1) (b^{2} + 1) = a^{2} (b^{2} + 2)

\Rightarrow a^{2} b^{2} + a^{2} + b^{2} + 1 = a^{2} b^{2} + 2 a^{2}

\Rightarrow a^{2} - b^{2} = 1

Answered by MJS_new last updated on 18/Mar/21

(1) a = b + \frac{1}{a + \frac{1}{a}} \Leftrightarrow b = \frac{a^{3}}{a^{2} + 1}

(2) b = a - \frac{1}{b + \frac{1}{b}} \Leftrightarrow a (b^{2} + 1) - b (b^{2} + 2) = 0

insert (1) into (2)

- \frac{a (a^{4} - a^{2} - 1)}{{(a^{2} + 1)}^{3}} = 0

a = 0 [rejected]

a^{4} - a^{2} - 1 = 0

{(a^{2})}^{2} - (a^{2}) - 1 = 0

a^{2} = \frac{1 - \sqrt{5}}{2} [rejected because a \in R] \lor a^{2} = \frac{1 + \sqrt{5}}{2}

b^{2} = \frac{a^{6}}{{(a^{2} + 1)}^{2}} = - \frac{1 - \sqrt{5}}{2}

a^{2} + b^{2} = \sqrt{5}

Answered by mr W last updated on 18/Mar/21

f r o m (1) : a = b + \frac{1}{a + \frac{1}{a}} = b + \frac{a}{a^{2} + 1}

a - b = \frac{a}{a^{2} + 1} = k s a y

{k a}^{2} - a + k = 0

a^{2} = \frac{a}{a - b} - 1

f r o m (2) : b = a - \frac{1}{b + \frac{1}{b}} = a - \frac{b}{b^{2} + 1}

a - b = \frac{b}{b^{2} + 1} = k

{k b}^{2} - b + k = 0

b^{2} = \frac{b}{a - b} - 1

a, b a r e r o o t s o f {k x}^{2} - x + k = 0

a + b = \frac{1}{k}

a b = 1

s i n c e a - b = k

{(a + b)}^{2} - 4 a b = k^{2}

\frac{1}{k^{2}} - 4 = k^{2}

k^{4} + 4 k^{2} - 1 = 0

k^{2} = - 2 + \sqrt{5} > 0 (- 2 - \sqrt{5} < 0 r e j e c t e d)

k^{2} + 2 = \sqrt{5}

a^{2} + b^{2} = \frac{a + b}{a - b} - 2 = \frac{1}{k^{2}} - 2 = k^{2} + 2 = \sqrt{5} ✓

a^{2} - b^{2} = (a + b) (a - b) = \frac{1}{k} \times k = 1 ✓

Answered by ajfour last updated on 19/Mar/21

a - b = \frac{1}{a + \frac{1}{a}} = \frac{1}{b + \frac{1}{b}}

c l e a r l y a \neq b

b u t a + \frac{1}{a} = b + \frac{1}{b}

\Rightarrow a b = 1

n o w (a - \frac{1}{a}) (a + \frac{1}{a}) = 1

\Rightarrow a^{2} - \frac{1}{a^{2}} = 1

a^{4} - a^{2} - 1 = 0

a^{2} = \frac{\sqrt{5} + 1}{2}

b^{2} = \frac{1}{a^{2}} = \frac{2}{\sqrt{5} + 1} = \frac{\sqrt{5} - 1}{2}

a^{2} + b^{2} = \sqrt{5}

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