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 Square Crossed Sequence (Posted on 2013-12-09)
A sequence {Bp} of positive integers is such that:

B1 = 20, B2 = 30, and:

Bp+2 = 3*Bp+1 – Bp, whenever p ≥ 1.

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

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

 No Solution Yet Submitted by K Sengupta No Rating

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 Possible solution Comment 3 of 3 |

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

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