# Integer-valued polynomial

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Integer-valued polynomial

In mathematics, an integer-valued polynomial "P(t)" is a polynomial taking an integer value "P(n)" for every integer "n". Certainly every polynomial with integer coefficients is integer-valued. There are simple examples to show that the converse is not true: for example the polynomial

:"t(t + 1)/2"

giving the triangle numbers takes on integer values whenever "t = n" is an integer. That is because one out of "n" and "n" + 1 must be an even number.

In fact integer-valued polynomials can be described fully. Inside the polynomial ring "Q" ["t"] of polynomials with rational number coefficients, the subring of integer-valued polynomials is a free abelian group. It has as basis the polynomials

:"Pk"("t") = "t"("t" − 1)...("t" − "k" + 1)"/k!"

for "k" = 0,1,2, ... .

Fixed prime divisors

This concept may be used effectively to solve questions about fixed divisors of polynomials. For example, the polynomials "P" with integer coefficients that always take on even number values are just those such that "P"/2 is integer valued. Those in turn are those expressed as sums of the basic polynomials, with even coefficients.

In questions of prime number theory, such as Schinzel's hypothesis H and the Bateman-Horn conjecture, it is a matter of basic importance to understand the question when "P" has no fixed prime divisor (this has been called "Bunyakovsky's property", for Viktor Bunyakovsky). By writing "P" in terms of the basic polynomials, we see the highest fixed prime divisor is also the highest common factor of the coefficients in such a representation. So Bunyakovsky's property is equivalent to coprime coefficients.

As an example, the pair of polynomials "n" and "n"2 + 2 violates this condition at "p" = 3: for every "n" the product

:"n"("n"2 + 2)

is divisible by 3. Consequently there cannot be infinitely many prime pairs "n" and "n"2 + 2. The divisibility is attributable to the alternate representation

:"n"("n" + 1)("n" − 1) + 3"n".

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