Fibonacci

Squares in the Fibonacci series (end)

We now know that every square Fibonacci number must have the form F_12k. It is not so difficult to prove that k must be even, for, by making use of the preceding method we see inmediately that F_24p+12 and -F₁₂ (= -144) have the same remainder when divided by L_2^r, because 24p+12 = 12k+q = 2^r+1(2t+1)+q with k=2p and q=12 and r>1.
This shows that if F_24p+12 = x², then L_2^r divides x²+144.
144 is always a factor of F_24p+12 ( to be shown with formula 15.) or look here ), hence if F_24p+12 is a square, then there is a y so that F_24p+12 = (12y)².
En as L_2^r is always odd (for it always of the form 4p+3), L_2^r divides y²+1, and that is, as we know, inpossible.
F_24p+12 cannot be written as 2x², because that implies that F_24p+12 does not only have a factor 144, but even a factor 288. That is never the case ( to be shown with formula 15.) or look here ).

In conclusion we show that there are no square Fibonacci numbers with index larger than 12.

Suppose S = { m>1 | F_12m is a square or 2 times a square}. We will prove that this set is empty, by showing that S has no smallest element. Suppose M is the smallest element of S. Then M is even, and M>3, because F₂₄ = 144*322. Now F_12M = F_6ML_6M. F_6M and L_6M can only have a factor 2 in common, which is clear from the general formula
L_n² = 5F_n² + 4(-1)ⁿ.
F_12M is a square or 2 times a square. Thus F_6M is a square or 2 times a square.
M is even and M>3, so M/2 is an element of S, which is impossible. (After a proof of J.H.E. Cohn)