Pólya's theorem on integer-valued entire functions

Remarks

The idea of the proof uses the Newton series $f(z) = \sum_{k \ge 0} \Delta^k f(0) \binom{z}{k}$, where $\Delta g(z) = g(z+1) - g(z)$ and $\binom{z}{k}$ is the binomial coefficient polynomial. The differences $\Delta^k f(0) = \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} f(j)$ are integers. A contour integral estimate shows that the growth bound forces $\Delta^k f(0) \to 0$, so $\Delta^k f(0) = 0$ for all large $k$, and the series is a polynomial. The constant $\log 2$ is sharp: $f(z) = 2^z = e^{z \log 2}$ takes integer values at the nonnegative integers, is not a polynomial, and has $\log M(r) = r \log 2$.

Going deeper

For a student

The condition says that $f$ is of exponential type less than $\log 2$. The polynomials that the theorem produces are integer-valued polynomials, which are exactly the integer combinations of the polynomials $\binom{z}{k}$. The theorem started a series of arithmetic results on entire functions. In these results, integrality conditions on values combined with growth bounds force algebraic behaviour.