Prove that $f_{2k} \cdot f_{(2k+4)} + 1$ is a perfect square, where $f_n$ is the $n$th Fibonacci number, $n\geq0$. I found a pattern that makes me conjecture that this is also true for $f_{2k+4}$, as well as $f_{2k+2}$. I actually accidentally found this pattern while thinking about the proof for $f_{2k+2}$! But I haven't been able to prove it yet. Any help would be appreciated, thanks!
 A: Nice to see another Fibonacci question from you, Danny! It's really cool you found that pattern. I think pattern finding is a really important skill in math in general. Keep it going!
Before you look at my full solution below, try and see if you can use my hint and figure it out.
Hint: Use a direct proof (no induction, which we used for the previous problem), applying basic properties of Fibonacci numbers, i.e. $ f_{n} = f_{n-1} + f_{n-2} $.
Full proof:
We aim to prove that $f_{2k} \cdot f_{2k+4} + 1$ is a perfect square, where $f_n$ is the $n$th Fibonacci number, $n\geq0$.
$f_{2k} \cdot f_{2k+4} + 1 = f_{2k} \cdot (f_{2k+2} + f_{2k+3}) + 1 = f_{2k} \cdot f_{2k+2} + f_{2k} \cdot f_{2k+3} + 1$.
Based on what we previously proved (for your previous question), this is equivalent to:
$ (f_{2k+1}^2 - 1) + f_{2k} \cdot f_{2k+3} + 1 = (f_{2k+1}^2 - 1) + f_{2k} \cdot (f_{2k+1} + f_{2k+2}) + 1 = f_{2k+1} \cdot (f_{2k+1} + f_{2k}) + f_{2k} \cdot (f_{2k+2}) - 1 + 1$.
By the definition of a Fibonacci number, $f_{2k+2} = f_{2k+1} + f_{2k}$, so this is equivalent to:
$f_{2k+1} \cdot f_{2k+2} + f_{2k} \cdot (f_{2k+2}) = (f_{2k+2}) \cdot (f_{2k+1} + f_{2k}) = (f_{2k+2})^2$.
Because every Fibonacci is by definition, an integer (by its recursive definition, it is the sum of two previous integers), $ (f_{2k+2})^2 $ is a perfect square, which proves the claim.
$ Q.E.D. $
Hope this helps!
