Fibonacci

Squares in the Fibonacci series (part 2)

We have to look for another way to continue. The next method will lead up to the conclusion that the Fibonacci numbers with index 12k-11, 12k-10 or 12k-1 are not squares. After that the only possible square Fibonacci numbers have an index of the form 12k.
The three beforementioned indices may be summarized by writing 12k+q with q = -1, 1 or 2 and k≥0. F_q = 1.
It is apprehensible to look for a useful formula.
Let's consider formula 1:
F_m+n + (-1)ⁿF_m-n = F_mL_n (*)
For m=6k+q and n=6k this yields F_12k+q + 1 = F_6k+qL_6k

If F_12k+q is a square (say x²), then 2(x² + 1) can be divided by L_6k. All divisors of 2(x² + 1) have the form 4m+1 or 2(4m+1) (proof), thus L_6k should also have that form.
When we are able to proof that L_6k does not have that form then we have a contradiction and our assumpion that F_12k+q is a square would be false, which is exactly what we want to proof.
But unfortunately, L_6k is always of the form 4m+1 or 2*(4m+1). We don't give up because, when we look closely at the Lucas numbers, we notice that no Lucas number L_2^r (with r>0) has that form.
(Proof: (by induction) L₂ = 4*0+3. Suppose we know that L_2^r = 4p+3, then (see 18.)) L_2^(r+1) = L_2^r²-2 = (4p+3)²-2 = 4p'+3 for certain p'. This proves the induction).
We will see how L_6k can be replaced by L_2^r.

Notation: There are positive integers r and t such that 12k = 2^r+1(2t+1).
Now, for the sake of readability, we define a(s) = 2^r(4t+s)+q for all integers s.
Notice that 12k+q = a(2).
Substitution of n = 2^r and m = a(s) in (*) yields:
F_a(s+1) + F_a(s-1) = F_a(s)L_2^r.

Now, F_12k+q + 1 =
F_12k+q + F_q =
F_a(2) + F_a(-4t) = {{telescopic sum}}
(F_a(2) + F_a(0)) - (F_a(0) + F_a(-2)) + (F_a(-2) + F_a(-4)) - (F_a(-4) + F_a(-6)) + ... + (F_a(-4k+2) + F_a(-4k)) =
F_a(1)L_2^r - F_a(-1)L_2^r + F_a(-3)L_2^r - F_a(-5)L_2^r + ... + F_a(-4k+1)L_2^r

This last expression can be divided by L_2^r.
Thus F_12k+q + 1 can be divided by L_2^r (without remainder). When F_12k+q is a square (say x²), then x² + 1 can be divided by L_2^r, and that is, as we know, incorrect.
Ergo F_12k+q can never be a square.