If $f_{n-1}^2=(f_n/2)^2+h^2$ then $n=6$ How can I prove that if $f_n$ is a term of the Fibonacci sequence divisible by $4$ and if $$f_{n-1}^2=(f_n/2)^2+h^2,$$ $h\in\Bbb Z^+$ then $n=6$? 
I know that since $\gcd(f_k,f_{k+1})=1$ for every $k\in\Bbb N$ then $h$ must be odd and that if $h=2m+1$ with $m\in\Bbb Z$ then $h=(m+1)^2-m^2$. But I cannot assume that $f_{n-1}=(f_n/2)+1$ so I don't know how to proceed. 
Any help is really appreciated. 
 A: We need $4h^2 = 4f_{n-1}^2-f_n^2 = (2f_{n-1}+f_n)(2f_{n-1}-f_n) = (4f_{n-2}+3f_{n-3})f_{n-3}$. 
Since $\gcd(f_{n-2},f_{n-3}) = 1$, we have $\gcd(4f_{n-2}+3f_{n-3},f_{n-3}) = \gcd(4f_{n-2},f_{n-3}) \in \{1,2,4\}$. 
Let $v_p(n)$ denote the largest integer $k$ such that $p^k \mid n$. 
Then, if $p \ge 3$ is a prime such that $p \mid f_{n-3}$, then $p \not\mid (4f_{n-2}+3f_{n-3})$, and so we have $v_p(f_{n-3}) = v_p((4f_{n-2}+3f_{n-3})f_{n-3}) = v_p(4h^2) = v_p(h^2) = 2v_p(h)$, which is even. 
If $p \ge 3$ is a prime such that $p \not\mid f_{n-3}$, then trivially, $v_p(f_{n-3}) = 0$, which is even. 
Therefore, $v_p(f_{n-3})$ is even for all primes $p \ge 3$. 
Hence $f_{n-3}$ is either a perfect square or twice a perfect square. 
By this link the only Fibonacci numbers which are perfect squares are $0,1,144$, and the only Fibonacci numbers which are twice a perfect square are $0,2,8$. So it remains to check all of these $5$ cases and see which work. 
A: Recall the pythagorean triples are of the form $(u^2-v^2,2uv,u^2+v^2)$. As you pointed out $h = 2m+1$ is the odd member of the triplet, that makes $\dfrac{f_n}{2} = 2uv$ and $f_{n-1} = u^2+v^2$. Thus, $f_{n-3} = 2(u-v)^2$.
The only Fibonacci Number $f_k$ which is twice a perfect square are when $k = 0,3,6$.
${n-3}$ cannot be $0,6$. Thus, $n = 6$.
A: Note that
$$h^2=f_{n-1}^2-(f_n/2)^2=(f_{n-1}-f_n/2)(f_{n-1}+f_n/2)$$
Now
\begin{align*}\gcd(f_{n-1}-f_n/2,f_{n-1}+f_n/2)&=\gcd(f_{n-1}-f_n/2,f_n)=\gcd(f_{n-1}-f_n/2,f_n/2)\\&=\gcd(f_{n-1},f_n/2)=1\end{align*}
(the second step is allowed since $f_{n-1}-f_n/2$ must be odd and you already know the last step)
This means $f_{n-1}-f_n/2=f_{n-1}/2-f_{n-2}/2=f_{n-3}/2$ must be a perfect square, but it is well known this is only possible for $n=3$, $n=6$, or $n=9$. But only for $n=6$ is $f_{n-1}+f_n/2$ also a square.
