Integer-valued polynomial

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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

P_k(t) = t(t − 1)...(t − k + 1)/k!

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

[edit] 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 n² + 2 violates this condition at p = 3: for every n the product

n(n² + 2)

is divisible by 3. Consequently there cannot be infinitely many prime pairs n and n² + 2. The divisibility is attributable to the alternate representation

n(n + 1)(n − 1) + 3n.