x-1-x-3-x-R-x-2020-1-x-2020-mod-10-y-y-2-1-

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