x + (1/x) = 3 , x ∈ R (x^(2020) + (1/x^(2020) )) mod (10) = y y^2 − 1 = ?


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Question Number 78880 by naka3546 last updated on 21/Jan/20

$x + \frac{1}{x} = 3, x \in R$ $(x^{2020} + \frac{1}{x^{2020}}) m o d (10) = y$ $y^{2} - 1 = ?$
	Answered by mind is power last updated on 21/Jan/20

	$l e t U_{n} = x^{n} + \frac{1}{x^{n}}$ $U_{0} = 2, U_{1} = 3$ $U_{n + 1} = (x + \frac{1}{x}) U_{n} - U_{n - 1}$ $\Rightarrow U_{n + 1} = 3 U_{n} - U_{n - 1}$ $U_{2} = 7, U_{3} = 18 \Rightarrow U_{3} = 8 (10)$ $U_{4} = 7 (10), U_{5} = 3 (10), U_{6} = 2 (10), U_{7} = 3 (10)$ $U_{8} = 7 (10), U_{9} = 8 (10), U_{10} = 7 (10)$ $W e g o t a C y c l e$ $U_{9} = U_{3} = 8 (10), U_{10} = U_{4} = 7 (10)$ $U_{3 + 6 k} = U_{3} = 8 (10), U_{4 + 6 k} = U_{4} = 7 (10)$ $U_{5 + 6 k} = U_{5} = 3 (10), U_{6 (k + 1)} = U_{6} = 2 (10), U_{6 (k + 1) + 1} = U_{7} = 3 (10)$ $U_{6 k + 8} = 7 (10)$ $2020 = 6 . (336) + 4 = 6 k + 4$ $U_{2020} = U_{6 k + 4} = U_{4} = 7 (10), y = 7$ $y^{2} - 1 = 48$
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