Very different aspect, but interesting to me at least. The odd index Fibonacci numbers are one branch in the Markov tree,

http://en.wikipedia.org/wiki/Markov_number 

http://oeis.org/A002559 

Taking notation from the book by Cusick and Flahive, we have the Markov (or Markoff) equation
$$m^2 + m_1^2 + m_2^2 = 3 m m_1 m_2,$$
where it follows from the process that generates the tree that 
$$ \gcd(m, m_1) = \gcd (m_1, m_2) = \gcd(m_2,m) = 1.$$
Then C+H define $u$ on pages 10 and 18 as the smallest positive integer such that 
$ \pm m_1 u \equiv m_2 \pmod m.$ Then we define an integer $v$ by $mv = u^2 + 1.$ 

It follows from $mv = u^2 + 1$ that every Markov number is represented, and primitively, as the sum of two squares. 

At some point I had a proof (well, I think I did) that the Markov discriminants $$ 9 m^2 - 4 $$ were also the sum of two squares, although sometimes not primitively because these are divisible by 4 when $m$ is even. I cannot recall the proof but it was short and elementary.

PROOF: That was fun. We have $ m^2 + m_1^2 + m_2^2 = 3 m m_1 m_2$ in positive integers. We know from
$ v m = u^2 + 1$ that $m$ is not divisible by any prime $q \equiv 3 \pmod 4,$ simply because
any $k^2 +1$ is never divisible by such a prime, since $(-1|q)=-1.$  Now, take $\delta = \pm 1,$ and assume
$$ 3 m \equiv 2 \delta \pmod q.$$
Then  $ m^2 + m_1^2 + m_2^2 \equiv 2 \delta m_1 m_2 \pmod q,$ so then 
 $ m^2 + m_1^2 - 2 \delta m_1 m_2 +  m_2^2 \equiv 0 \pmod q,$ finally
 $$ m^2 + (m_1 -  \delta  m_2)^2  \equiv 0 \pmod q.$$
But this is a contradiction, as $m$ is not divisible by $q.$ So, in fact
$ 3 m \neq \pm 2 \pmod q,$ then $9 m^2 - 4 \neq 0 \pmod q$ for any prime
$q \equiv 3 \pmod 4.$