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  "body": "\n## Statement\n\nLet $f$ be an [[entire function|0419794R0T]] such that $f(n) \\in \\Z$ for every integer $n \\ge 0$.\nLet $M(r)$ be the maximum of $\\abs{f(z)}$ on the circle $\\abs{z} = r$.\nIf\n$$\\limsup_{r \\to \\infty} \\frac{\\log M(r)}{r} \u003c \\log 2,$$\nthen $f$ is a polynomial.\n\n## Remarks\n\nThe 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|092GEF38YD]] polynomial.\nThe differences $\\Delta^k f(0) = \\sum_{j=0}^{k} (-1)^{k-j} \\binom{k}{j} f(j)$ are integers.\nA 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.\nThe 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 $\\log M(r) = r \\log 2$.\n\n## Explanations\n\n### L1 human\n\nPolynomials such as $z(z-1)/2$ take integer values at all nonnegative integers.\nSo does $2^z$, which grows exponentially.\nPólya's theorem says that any entire function with integer values at $0, 1, 2, \\dots$ that grows more slowly than $2^z$ must be a polynomial.\nSo the growth of $2^z$ is the smallest possible for such a function that is not a polynomial.\n\n### L2 human\n\nThe condition says that $f$ is of [[exponential type|9P8D8YAHQS]] less than $\\log 2$.\nThe polynomials that the theorem produces are integer-valued polynomials, which are exactly the integer combinations of the polynomials $\\binom{z}{k}$.\nThe theorem started a series of arithmetic results on entire functions.\nIn these results, integrality conditions on values combined with growth bounds force algebraic behaviour.\n",
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