How do I show that the sum $(a+\frac12)^n+(b+\frac12)^n$ is an integer for only finitely many $n$?
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 $2^n\gt 2a+2b+2$.
Hint $\ $ Suppose that $\rm\:\left(\dfrac{2a+1}{2}\right)^2 + \left(\dfrac{2b+1}{2}\right)^n = m\in \mathbb Z.\:$ Then scaling by $\rm\:2^n\:$ we deduce
$$\rm 2^{n-2}(2a+1)^2 + (2b+1)^n =\, m\, 2^n$$
which yields a contradiction if $\rm\:n > 2\!:\:$ LHS = even + odd = odd, but RHS is even.
There are much more elegant ways to prove the result, but here's a quasi-experimental approach that shows how you might attack such a problem. (And it actually proves that there are no positive integers $n$ for which the sum is an integer.)
First note that $\left(a+\frac12\right)^2=a^2+a+\frac14$ is always $\frac14$ more than an integer. Thus, $$\left(a +\frac12\right)^2 + \left(b+\frac{1}{2}\right)^n$$ is an integer if and only if $\left(b+\frac12\right)^n$ is $\frac14$ less than an integer, i.e., if and only if there is an integer $m$ such that $\left(b+\frac12\right)^n=m-\frac14$. If this is the case, then $4\left(b+\frac12\right)^n=4m-1$.
Is this possible when $n=1$? No: $4\left(b+\frac12\right)^1=4b+2$ is an integer, but it's an even integer, and $4m-1$ is odd.
What if $n=2$? Still no: $4\left(b+\frac12\right)^2=4b^2+4b+1=4(b^2+b)+1$, which is odd, but it's one more than a multiple of $4$, and $4m-1$ is one less than a multiple of $4$.
For $n=3$ matters are still worse: $4\left(b+\frac12\right)^3=4b^3+6b^2+3b+\frac12$, which isn't even an integer.
For $n=4$ we have $4\left(b+\frac12\right)^4=4b^4+8b^3+6b^2+2b+\frac14$, which is worse yet.
Now look at the constant terms in (1)-(4): $2,1,\frac12$, and $\frac14$. They should suggest the conjecture that the constant term in $4\left(b+\frac12\right)^n$ is $\dfrac1{2^{n-2}}$. If you can prove this, you're practically done. Equivalently, try to prove the
Conjecture: The constant term in $\left(b+\frac12\right)^n$ is $\dfrac1{2^n}$.
If you already know the binomial theorem, you can get this immediately from it. Otherwise, you can prove the conjecture by induction on $n$.