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Question Number 115871 by bemath last updated on 29/Sep/20

Commented by bemath last updated on 29/Sep/20

$(\begin{matrix} m n \\ n m \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 1 \\ \frac{3 m n}{2 m^{2} + n^{2}} \end{matrix})$ $(\begin{matrix} x \\ y \end{matrix}) = \frac{1}{m^{2} - n^{2}} (\begin{matrix} m - n \\ - n m \end{matrix}) (\begin{matrix} 1 \\ \frac{3 m n}{2 m^{2} + n^{2}} \end{matrix})$ $(\begin{matrix} x \\ y \end{matrix}) = \frac{1}{m^{2} - n^{2}} (\begin{matrix} m - \frac{3 {m n}^{2}}{2 m^{2} + n^{2}} \\ - n + \frac{3 m^{2} n}{2 m^{2} + n^{2}} \end{matrix})$
	Answered by MJS_new last updated on 29/Sep/20

	$m x + n y = 1$ $n x + m y = \frac{3 m n}{2 m^{2} + n^{2}}$ $\Rightarrow x = \frac{2 m}{2 m^{2} + n^{2}} \land y = \frac{n}{2 m^{2} + n^{2}}$ $\Rightarrow$ $(x^{2} + y^{2}) (2 m^{2} + 2 n^{2}) = \frac{2 (m^{2} + n^{2}) (4 m^{2} + n^{2})}{{(2 m^{2} + n^{2})}^{2}} = α$ $transforming leads to$ $4 m^{4} + \frac{2 (2 α - 5)}{α - 2} m^{2} n^{2} + n^{4} = 0$ $let m = \frac{\sqrt{2 p}}{2} \land n = \sqrt{q}; p, q ⩾ 0$ $p^{2} + \frac{2 α - 5}{α - 2} p q + q^{2} = 0$ ${(p + q)}^{2} = \frac{p q}{α - 2} \land {(p - q)}^{2} = - \frac{4 α - 9}{α - 2} p q$ $\Rightarrow α \neq 2 \land \frac{1}{α - 2} ⩾ 0 \land - \frac{4 α - 9}{α - 2} ⩾ 0$ $\Rightarrow 2 < α ⩽ \frac{9}{4}$ $but maybe I^{'} m wrong . . .$
	Commented by MJS_new last updated on 29/Sep/20

	$yes, at least if we stay in R$
	Commented by bemath last updated on 29/Sep/20

	$t h a n k y o u s i r . i t d o e s m e a n t t h e a l l$ $c h o i c e a n s w e r n o t c o r r e c t$
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