Suppose I have this binary recurrent sequence $\{B_{n}\}$ called balancing sequence, satisfying $B_{n} = 6B_{n-1}-B_{n-2}$ with initial terms $(B_{0},B_{1}) = (0,1).$ This sequence has following properties :
$1. 2|B_{n} \iff 2|n,$
$2.B_{2n+1}-B_{2n-1} \equiv 2 ~(mod~4)$ and $B_{2n+1}B_{2n-1} \equiv 3 ~(mod~4)$
Similar results can be found with any other modulus.
Suppose we have the following system of Diophantine equations
$B_{2n-1}X+1 = B_{2n+1}(B_{2n+1}-B_{2n-1})A^2,$
$4B_{2n}X+1 = B_{2n+1}(B_{2n+1}-4B_{2n})C^2,$
$B_{2n+1}X+1 = (B_{2n+1}-4B_{2n})(B_{2n+1}-B_{2n-1})D^2,$
where $X,n,A,C,D$ are integer variables.
My aim is to show that above system does not yield any solution. I have shown that when $n \equiv 1 ~(mod~3)$, for (eq-3) $LHS \equiv 1 ~(mod~7)$ but $RHS \equiv 0,3,5,6 ~(mod~7).$ I am failing to show for the case when $n \equiv 0,2 ~(mod~3).$ In which way can I show so?