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
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 $q \equiv 3 \pmod 4.$