Question Number 101247 by 175 last updated on 01/Jul/20

Answered by 1549442205 last updated on 06/Jul/20

Note :the integral part of a real number  x be the grearest integer which don′t exceed x  and is denoted by symbol [x]  Put A_n = (2+(√2))^n +(2−(√2))^n .Then we   prove that A_n  is even ∀n∈N .Indeed,  i)For n=0 ⇒ A_0 =2.For n=1⇒A_1 =4  A_n =[(2+(√2))^n +(2−(√2))^n ]=  [(2+(√2))^(n−1) +(2−(√2))^(n−1) ][(2+(√2))+(2−(√2))]  −(2−(√2))(2+(√2))^(n−1) −(2+(√2))(2−(√2))^(n−1)   =4.A_(n−1) −2(2+(√2))^(n−2) −2(2−(√2))^(n−2)   =4.A_(n−1) −2A_(n−2) which shows that  A_n is an even number ∀n∈N  Now we see that 0<m=(2−(√2))^n <1   due to 2−(√2)<1.It follows that   (2+(√2))^n =A_n −(2−(√2))^n =2M−m  ⇒[(2+(√2))^n ]=2M−1  which means that [2+(√2))^n ]is odd(q.e.d)