- Integer-valued polynomial
In

mathematics , an**integer-valued polynomial**"P(t)" is apolynomial taking aninteger value "P(n)" for every integer "n". Certainly every polynomial with integercoefficient s 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 number s takes on integer values whenever "t = n" is an integer. That is because one out of "n" and "n" + 1 must be aneven number .In fact integer-valued polynomials can be described fully. Inside the

polynomial ring "Q" ["t"] of polynomials withrational number coefficients, thesubring of integer-valued polynomials is afree abelian group . It has as basis the polynomials:"P

_{k}"("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 theBateman-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", forViktor Bunyakovsky ). By writing "P" in terms of the basic polynomials, we see the highest fixed prime divisor is also thehighest 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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