n=1 positive integer of the form $4k+1$. Prove that $x+y\in A$ and $x+y\not \in A$.

I think that in Romania $0$ is not a positive integer so $k\neq0$ :

Already posted by me last year... http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Moldavian&t=46004

A little problem with solution given there. maxal wrote: because $u^2 + n*v^2 \equiv 0\pmod p$ has no solutions case $u\equiv v\equiv0$ failes entire solution.

A very easy solution: $x=2^{n+1}+2$, $y=2^n+1$, $x+y\equiv3\mod 4$, hence $x+y\notin A$. $x^n+y^n=((2^n+1)^{2k+1})^2\in A$.