For even values of n, this doesn't divide. For them, there's no denominator, but the logic below still applies.
For odd values of n: Suppose you have a composite n = k*m, with k>=3. Then,
(a^{n}+b^{n})/(a+b) = (a^{km}+b^{km})/(a+b) =
= (a^{k}+b^{k})/(a+b)*(a^{(k1)m}  b^{m}*a^{(k2)m} + b^{2m}*a^{(k3)m} + ... + b^{(k1)m})
where (a^{k}+b^{k})/(a+b) is > 1. QED, composite.
For proper Mersenne numbers a+b = 2+(1) = 1, which makes for an interesting case  there's no denominator.
When b=1, a^{n}+b^{n} is a proper Fermat number, and for b>1 and odd (prime) n, (a^{n}+b^{n})/(a+b) has the property that you sought for.
Last fiddled with by Batalov on 20121027 at 18:24
