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Obviously, polynomials with integer coefficients will satisfy P(x)$\in$ Z or every x $\in$ Z. But how do we prove that those with rational coefficients can produce integer roots? For instance, I have a polynomial that is in the form of a newton's generalized binomial theorem.

$\binom{X}{m} = \frac{X(X-1)(X-2)...(X-m+1)}{m!}$

If a polynomial is in the form of P(x) = $a_{0} + a_{1}\binom{X}{1} + a_{2}\binom{X}{2} + a_{d}\binom{X}{d}$, how do we prove that this can have P(x) $\in$ Z for all x $\in$ Z.

user3467433
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The question in your title is different than the question you ask at the end of your post. So I will answer them both.

Firstly, it's not true that polynomials with rational coefficients have integer roots. For instance, $x^2 - 2$ has roots $\pm \sqrt 2$.

However, it is true that polynomials of the form $p(x) = a_0 + a_1 {X \choose 1} + \ldots + a_d {X \choose d}$, where $a_i \in \mathbb{Z}$, only take integer values. More generally, the monomial $a_k {X \choose k}$ only takes integer values when $X$ is an integer. This is equivalent to the denominator, which in this case is $k!$, dividing $X(X-1)(X-2)\dots(X-k+1)$ for any $X \in \mathbb{Z}$.

This last claim admits number theoretic and combinatorial proofs. Number theoretically, you might look at each prime in $k!$ and show that its multiplicity in the numerator is at least the multiplicity in the denominator. This is doable, but not particularly easy. Bill showed a completely arithmetic approach in his answer here.

Combinatorially, you simply notice that for integral $X$, the coefficient ${X \choose k}$ is the classical combination, and counts the number of ways of choosing $k$ things from $X$ choices, and is therefore an integer.

Regardless, as each $a_k {X \choose k}$ is an integer for each $X \in \mathbb{Z}$, you must have that polynomials formed from these monomomials also take integral values on the integers.

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davidlowryduda
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