Finding $(a+\sqrt b)^n+(a-\sqrt b)^n$ where $n$ is natural For the expression $\left(a+\sqrt{b}\right)^n+\left(a-\sqrt{b}\right)^n$ where $n \in \mathbb{N}$, and $a,b, \in \mathbb{Q}$, the radical is always ends up cancelled, and the result is always in $\mathbb{Q}$.  Is there any way that this could be reexpressed with the assumption that n is always a positive integer without the use of a square root operation, such as an alternative closed form for even $n$, and another closed form for odd $n$, or am I always stuck having to calculate a square root?

Note: $(x + y)^n = x^n + {n \choose 1} x^{n-1}y + \dotsc + y^n$
While $(x - y)^n = x^n - {n \choose 1} x^{n-1}y + \dotsc \pm y^n$ 
The even terms are opposite parity while the odd terms are the same so
$(x + y)^n + (x - y)^n = 2x^n + 2{n \choose 2}x^{n-2}y^2 + \dotsc$
So $(a + \sqrt{b})^n + (a - \sqrt{b})^n = 2(a^{n} + {n \choose 2}a^{n -2}\sqrt{b}^{2} + \dotsc)$
So we are only taking $\sqrt{b} $ to even powers, and $\sqrt{b}^{2k} = b^k$.
So $\displaystyle(a + \sqrt{b})^n + (a - \sqrt{b})^n = 2 \sum_{i = 0; \,i \text{ even}}^n {n  \choose i}a^{n - i}b^{i/2}$.
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Or if you don't like taking a sum with only even summands:
$\displaystyle(a + \sqrt{b})^n + (a - \sqrt{b})^n = 2\sum_{k = 0}^{\lfloor n/2 \rfloor} {n  \choose 2k}a^{n - 2k}b^{k}$
