A sequence {B_p} of positive integers is such that:

B₁ = 20, B₂ = 30, and:

B_p+2 = 3*B_p+1 – B_p, whenever p ≥ 1.

Determine all possible positive integer values of p such that:
1 + 5*B_p+1*B_p is a perfect square.

**** For an extra challenge, derive a non computer program assisted solution.

      Possibly the simplest method:

1.  Compute small values of Bp: 20,30,70,180,470,1230...
      
2.  Derive the corresponding recurrence relation:      
x(n) = 10(((1 + sqrt(5))/2)^(2n) + ((1 - sqrt(5))/2)^(2n))    
x(n+1) = 10(((1 + sqrt(5))/2)^(2n+2) + ((1 - sqrt(5))/2)^(2n+2)).
     
3.  Next:      
(a) compute small values again using the canonical formula this time.
(b) apply the function 1 + 5*Bp+1*Bp to them (p=n+1): ( C) 
(c) take the square root of C: (D)      
(d) reduce D to a integer (E) and deduct E^2 from C   
(e) note that outside the range -4<n<-3, C-E^2 = F is constant, at 501, since, deriving the corresponding recurrence relation again:

500 floor[(2/(1+sqrt(5)))^(-4n-2)]+2001 = C 
(500*floor[(2/(1+sqrt(5)))^(-4n-2)]+1500)^(1/2) = E      
So C-E^2 = 501  [1]* 
   
(f) Hence, by inspection, iff p=3 , then 1 + 5*Bp+1*Bp = 63001 = 251^2 is a perfect square (strictly, iff p={-2,3}, but a positive p was required).

*Note: 1701^2-1700^2 =3401

Edited on December 10, 2013, 7:08 am

Posted by broll on 2013-12-09 23:42:38