Omid Hatami wrote: Let $ u$ be an odd number. Prove that $ \frac {3^{3u} - 1}{3^u - 1}$ can be written as sum of two squares. Call p is a prime divisor of $ \frac {3^{3u} - 1}{3^u - 1}$ We will prove that \[ p\equiv 1 (\mod 4) \] From condtion we have \[ p|3^{2u} + 3^u + 1 \] Therefore \[ p |(2.3^u + 1)^2 + 3 \] \[ p|(3^u + 1)^2 - 3^u \] And because $ u$ is odd so $ (\frac {3}{p}) = (\frac { - 3}{p}) = 1$ It gives $ (\frac { - 1}{p}) = 1$ , therefore p have form $ 4k + 1$ It is a well know result that if n have all primes divisor in the form $ 4k + 1$ can represent as sum of two square . So problem claim.

The problem is somehow more obvious: \[ \frac {3^{3u} - 1}{2^u - 1} = 3^{2u} + 3^u + 1 = (3^u - 1)^2 + (3^{\frac {u + 1}2})^2 \]

That settles the case that $ u \equiv 3 \bmod 4$, but what about the case that $ u \equiv 1 \bmod 4$?

The solution works for all $ u$ (after some correction )

Lucas cyclotomic formula ( http://www.mathlinks.ro/viewtopic.php?t=146020 ) strikes again.