Showing $\left(a + \frac{1}{2}\right)^N + \left(b + \frac{1}{2}\right)^N \in \mathbb{Z}$ for finite amount of natural numbers $N$ If $a$ and $b$ are positive integers, how would I go about showing that$$\left(a + \frac{1}{2}\right)^N + \left(b + \frac{1}{2}\right)^N \in \mathbb Z $$ for only a finite amount of natural numbers $N$? I've tried messing around with mods, but I haven't got anywhere with that...
 A: Let $v_p(x)$ be the greatest power in which a prime $p$ divides $x$. It is easy to prove $$v_2(x^k + y^k) = v_2(x + y)$$ when $k$ is odd and $x, y$ are integers. $($*$)$
We have$$\left(a + {1\over2}\right)^N + \left(b + {1\over2}\right)^N = {1\over{2^N}}\left((2a + 1)^N + (2b + 1)^N\right).$$It suffices to show for sufficiently large positive integers $N$, $$ 2^N\nmid \left((2a+1)^N+(2b+1)^N\right).$$ By $($*$)$, it suffices to check only powers of $2$. Thus consider $$ (2a+1)^{2^N}+(2b+1)^{2^N}\equiv 0\text{ }\left(\text{mod }2^{2^N}\right).$$But $$ (2a+1)^{2^N}+(2b+1)^{2^N}\equiv 2\text{ }(\text{mod }8),$$contradiction for $N \ge 1$. Hence we are done.
A seemingly more interesting question arises when we replace $a + 1/2$, $b + 1/2$ with arbitrary rational numbers, since the fact that every odd integer squares to $1\text{ }(\text{mod }8)$ kills this problem. The modified problem is killed by $``$Lifting the Exponent$"$ though $($described here: Lifting the Exponent Lemma (LTE)$)$, so perhaps it is not that much more interesting...
