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, CE^2 = F is constant, at 501, since, deriving the corresponding recurrence relation again:
500 floor[(2/(1+sqrt(5)))^(4n2)]+2001 = C
(500*floor[(2/(1+sqrt(5)))^(4n2)]+1500)^(1/2) = E
So CE^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: 1701^21700^2 =3401
Edited on December 10, 2013, 7:08 am

Posted by broll
on 20131209 23:42:38 