$Given :\ u+v \ is \ rational, \ u^2 + v^2 =1 \ , prove \ v^n + u^n \ is \ rational$.
What I have done so far is proving that $uv$ is rational by expanding $(u+v)^2$. I expanded $(u+v)^n$ using binomial expansion but I do not think this is the proper way.
Any hint about how to prove it please?
On
Related to another answer and comment, but perhaps less "magic" and more motivated.
It is given that $u+v$ is rational, and you have proved that $uv$ is rational. By the binomial theorem, $$\eqalign{u^n+v^n &=(u+v)^n-\binom n1u^{n-1}v-\binom n2u^{n-2}v^2-\cdots -\binom n{n-2}u^2v^{n-2}-\binom n{n-1}uv^{n-1}\cr &=(u+v)^n-\binom n1(uv)(u^{n-2}+v^{n-2}) -\binom n2(uv)^2(u^{n-4}+v^{n-4})-\cdots\ .\cr}$$ Now every term on the RHS is either rational by what you know already, or can be assumed rational by induction.
From the given condition, $(u+v)^2=u^2+v^2+2uv=1+2uv$ is rational implies $uv$ is rational. $u^{n+2}+v^{n+2} = (u+v)(u^{n+1}+v^{n+1})-uv(u^n+v^n) $
since $u+v$ and $uv$ are rational, and the statement is true for $n=1,2$, $u^n+v^n$ is rational for all positive integers $n$ by induction.