<s>[INST] How do I show that the sum $(a+\frac12)^n+(b+\frac12)^n$ is an integer for only finitely many $n$? 
Show that if $a$ and $b$ are positive integers, then $$\left(a +\frac12\right)^n + \left(b+\frac{1}{2}\right)^n$$is an integer for only finitely many positive integers $n$.

I tried hard but nothing seems to work. :( 
 [/INST] First of all, rewrite the equation as others have, to yield $(2a+1)^n+(2b+1)^n = m\cdot 2^n$, or in other words $(2a+1)^n+(2b+1)^n \equiv 0\pmod {2^n}$.  Now, if $n=2k$ is even then for the equation to hold $\bmod 2^{2k}$ it must certainly hold $\bmod 4$; i.e., $\bigl((2a+1)^k\bigr)^2 + \bigl((2b+1)^k\bigr)^2\equiv 0\pmod 4$.  But this can't work, because both of the squares on the left must be congruent to $1 \bmod 4$ and so their sum is congruent to $2$.  Therefore, $n$ must be odd, say $n=2k+1$.
Now, the left side can be factored using the classic formula for $\frac{x^n-y^n}{x-y}$ (substitute $x=2a+1, y=-(2b+1)$), yielding 
$$(2a+1)^{2k+1}+(2b+1)^{2k+1} = (2a+2b+2)\cdot(x^{2k}+x^{2k-1}y+x^{2k-2}y^2+\cdots+y^{2k})$$
But the factor on the right is odd (it's the sum of $2k+1$ terms each of which is odd), so for the LHS to be $0\bmod 2^n$, we must have $2a+2b+2\equiv 0 \pmod {2^n}$, and this can only be true for finitely many $n$; it becomes impossible as soon as $2^n\gt 2a+2b+2$.
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