Home Competitions SE-SKOLORNAS 1996 Problem 3
SE-SKOLORNAS 1996 Problem 3
For every positive integer n n n , we define the function p n p_n p n for x ≥ 1 x\ge 1 x ≥ 1 by p n ( x ) = 1 2 ( ( x + x 2 − 1 ) n + ( x − x 2 − 1 ) n ) . p_n(x) = \frac12 \left(\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n\right). p n ( x ) = 2 1 ( ( x + x 2 − 1 ) n + ( x − x 2 − 1 ) n ) . Prove that p n ( x ) ≥ 1 p_n(x) \ge 1 p n ( x ) ≥ 1 and that p m n ( x ) = p m ( p n ( x ) ) p_{mn}(x) = p_m(p_n(x)) p mn ( x ) = p m ( p n ( x )) .
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