Find a real numbers $a,b$ such $a^n+b^n$ is rational Question:

prove or disprove :there exsit real numbers $a,b$ such follow two condition:

(1):$a+b$ is irrational
(2): for any postive integer $n\ge 2$, then $a^n+b^n$ is rational.
I have know if 
$n=2k$ case is true,because I  let $a=\sqrt{2}+1,b=\sqrt{2}-1$,so
$$a^{2k}+b^{2k}=(\sqrt{2}+1)^{2k}+(\sqrt{2}-1)^{2k}\in Q$$
But for $n=2k+1$,I can't find a example.(if you can't find,can you prove when$n=2k+1$,there can't exsit?)  Thank you for help
 A: Suppose that $a,b$ are real numbers such that $a^n+b^n$ is rational for all integers $n \ge 2$. 
Since $a^2+b^2$ and $a^4+b^4$ are rational, $\dfrac{(a^2+b^2)^2-(a^4+b^4)}{2} = a^2b^2$ is also rational. 
Then, since $(a^5+b^5)-(a^2+b^2)(a^3+b^3)+a^2b^2(a+b) = 0$, we have that 
$\dfrac{(a^2+b^2)(a^3+b^3)-(a^5+b^5)}{a^2b^2} = a+b$ is rational, as desired. 
EDIT: The above is only valid if $a^2b^2 \neq 0$, but the case where $a = 0$ or $b = 0$ is easy. 
A: Theorem.  There are no real numbers $a,b$ such that
$$\hbox{$a+b$ is irrational}$$
and
$$\hbox{$a^2+b^2$, $a^3+b^3$, $a^4+b^4$ and $a^6+b^6$ are rational.}$$
Proof.  Suppose that there are such $a,b$; clearly they are non-zero.  Then
$$2a^2b^2=(a^2+b^2)^2-(a^4+b^4)$$
and
$$2a^3b^3=(a^3+b^3)^2-(a^6+b^6)$$
are rational, and so is
$$ab=\frac{2a^3b^3}{2a^2b^2}\ .$$
Therefore
$$(a+b)^2=(a^2+b^2)+2ab$$
is rational, and since
$$(a+b)^3=(a^3+b^3)+3ab(a+b)$$
we have
$$a+b=\frac{a^3+b^3}{(a+b)^2-3ab}$$
which is rational (note that with real $a,b$, the denominator of this last fraction cannot be zero).
