Let $P(x)$ be an $n^{th}$ degree real polynomial taking integer values at $n+1$ integer points. Is it true that $P(x)$ is an integer for all integer values ?
$n^{th}$ degree polynomial taking integral values at $n+1$ integer points
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real-analysis
elementary-number-theory
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0If I am right, it is more often the converse. – 2017-01-07
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0Just to generalize $(0.5x+0.5)^n$ is integer when x is odd and not an integer when x is even – 2017-01-07
1 Answers
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The polynomial $P(x)=\dfrac{x^n}{n}$ takes integer values at all integer points of the form $x=kn$ but not at integer points $x=kn+1$.
If $P$ takes integer values at $n+1$ consecutive integer points $a, a+1, \dots$, then it is true that $P$ takes integer values at all integer points.
Indeed, Newton's interpolation formula gives $$ f(n+a) = d_0 \binom{n}{0} + d_1 \binom{n}{1} + d_2 \binom{n}{2} + d_3 \binom{n}{3} +\cdots $$ where $d_i$ are the numbers in the first column of the repeated differences array. These $d_i$ (and the entire array) are integers and so $f(n)$ is an integer for all $n$.
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0Thanks ! But is consecutive integers a necessary condition, or maybe co-prime will do ? – 2017-01-08