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+...))))))


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

$F i n d t h e v a l u e o f a^{2} + b^{2}$ $f o r a, b r e a l n u m b e r s u c h t h a t$ $a = b + \frac{1}{a + \frac{1}{b + \frac{1}{a + . . .}}}$ $a n d b = a - \frac{1}{b + \frac{1}{a - \frac{1}{b + . . .}}}$
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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