Question Number 102474 by pticantor last updated on 09/Jul/20

Un=(1+(√2))^n   show that we have p_n ∈N /  U_n =(√p_n )+(√(p_n +1))

Answered by ~blr237~ last updated on 09/Jul/20

 observe that (1/U_n )=(−1)^n (1−(√2))^n   and that there exist   a_n  ,b_n ∈N  such as   U_n =a_n +b_n (√2)    and  (1/U_n )=(−1)^n (a_n −b_n (√2))  and  U_n ^2 =U_(2n)  .  So  (U_n ^ +(1/U_n ))^2 = 2+U_(2n) +(1/U_(2n) )=2 +2a_(2n)    (U_n −(1/U_n ))^2 =−2+U_(2n) +(1/U_(2n) ) =−2+2a_(2n)   we have  a_(n+1) +b_(n+1) (√2) =(1+(√2))(a_n +b_n (√2) ) lead to  a_(n+1) =a_n +2b_n  . So  such as a_1 =1 (odd ) , a_n  is always odd   So there exist  c_n  ∈N  / a_n =2c_n +1  Then    (U_n +(1/U_n ))^2 =2+2(2c_(2n) +1)=4(c_(2n) +1)   (U_n −(1/U_n ))^2 =−2+2(2c_(2n) +1)=4c_(2n)    By adding the two square root   U_n =(√c_(2n) ) +(√(c_(2n) +1))    take  p_n =c_(2n)