$n^a$ integral for all integer $n$ implies $a$ integral

Solution 1:

The idea is decreasing the exponent $a$ below $0$ by taking differences.

For every function $f$, let $(\Delta f)(x)=f(x+1)-f(x)$ be the usual forward difference.

$Lemma$. If $f(x)$ is a $k$ times differentiable function on $[n,n+k]$ then there is some $\xi\in(n,n+k)$ such that $$ f^{(k)}(\xi) = (\Delta^kf)(n) = \sum_{\ell=0}^k (-1)^{k-\ell}\binom{k}{\ell}f(n+\ell). $$

$Proof$. Induction on $k$. For $k=1$ this is exactly Lagrange's mean value theorem.

If the Lemma holds true for $k-1$ then apply it to the function $g(x)=\Delta f$. With some $\zeta\in(n,n+k-1)$, and then $\xi\in(\zeta,\zeta+1)\subset(n,n+k)$, we have $$ (\Delta^k f) (n) = (\Delta^{k-1} g)(n) = g^{(k-1)}(\zeta) = (\Delta f^{(k-1)})(\zeta) = f^{(k)}(\xi), $$ the lemma has been proved.

Now suppose that $a$ is not an integer; let $k$ be a positive integer with $k-1<a<k$. Applying the lemma with an arbitrary positive integer $n$, $$ a(a-1)\dots(a-k+1) \, \xi^{a-k} = \sum_{\ell=0}^k (-1)^{k-\ell}\binom{k}{\ell} (n+\ell)^a. $$ If $n$ is sufficiently large then the LHS is a number between $0$ and $1$. But we have some integer on the RHS, contradiction.