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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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 many solutions | Comment 2 of 3 |
B(p+2) = 3*30 - 20 = 70

B(p+2) = 70, B(p+3) = x, a square y^2

1 + 5*70*x = y^2

Integer solutions:
x = 2(175*n^2 - 349 n + 174),   y = 349 - 350*n
x = 2(175*n^2 - 251 n + 90),   y = 251 - 350*n
x = 2(175*n^2 - 99 n + 14),   y = 99 - 350*n
x = 2(175*n^2 - n),   y = 1 - 350*n
n is an element of Z

n = 0
in y = 1 - 350*n ---> y = 1 => x = 0
in y = 99 - 350*n --> y = 99 => x = 28
in y = 251 - 350*n ---> y = 251 => x = 180
in y = 349 - 350*n ---> y = 349 => x = 348

n = 1
we get negative numbers

n < 0
n = -1
y = 351, x = 352
y = 449, x = 201601
y = 601, x = 1032
y = 699, x = 1396

etc.

 Posted by Benny on 2013-12-09 19:21:26

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