Integer solution to a polynomial

Question

For which real numbers $q$ is $n = 1/20 q^5 – 1/4 q^3 + 1/5 q$ an integer?

Modified, it reads $n = 1/20(q^5 - 5q^3 +4q)$ and it directly follows that $q=1$ is an answer. But it seems that $n$ is an integer for all $q$ that are integers too. How can I prove this?

Answer 1

This is one explanation for why it holds for all integer values of $q$:

$n = 1/20 (q^5 - 5q^3 + 4q)$ can be factored on the right

$n = 1/20 (q) (q^4 - 5 q^2 + 4) = 1/20 (q) (q^2 - 4)(q^2 - 1)$

$ = 1/20 (q) (q + 2)(q - 2)(q + 1)(q - 1)$ and then rearranging some factors we get

$ = 1/20 (q - 2) (q - 1)(q)(q + 1)( q + 2)$

So in $q-2$, $q - 1$,$q$, $q+1$, and $q + 2$ we have five consecutive integers.

For the cases of $q = -2$, $q = -1$, $q = 0$, $q = 1$, and $q = 2$, we get the $0$ for $n$, clearly an integer.

Otherwise, we get either five consecutive negative or five consecutive positive integers for those factors. One will be a multiple of $5$ and at least one will be a multiple of $4$. So their product is guaranteed to be a multiple of $20$, and $1/20$ of that product will therefore be an integer.

Answer 2

Since rational function $$f(x)={1\over 29(x^5-5x^3+4x)}$$ has range $\mathbb{R}\setminus\{0\}$ we see than $n$ can be any nonzero integer.

Answer 3

Hint: try factoring $q^5-5q^3+4q$ and try to figure out why that implies that $n$ is an integer whenever $q$ is. (Second hint: $20= 4\cdot 5$.)

Answer 4

An integer-valued polynomial (also known as a numerical polynomial) $P(t)$ is a polynomial whose value at integers are all integers.

It is known that a polynomial $P(t)$ with rational coefficients is an integer-valued polynomial if and only if it is an integer combination of polynomials of the form

$$P_k(t) = \frac{t(t-1)\cdots(t-k+1)}{k!}\quad\text{ for }\quad k = 0,1,2,\ldots$$

This is just the binomial coefficients $\binom{t}{k}$ when we view it as a polynomial in $t$.

For the polynomial at hand, we have

$$\frac{1}{20}(q^5-5q^3+4q) = 6\binom{q}{5} + 12\binom{q}{4} + 6\binom{q}{3}$$

Since it is an integer combination of binomial coefficients in $q$, it is an integer-valued polynomial.

For completeness, let me illustrate why $\binom{t}{n}$ are integer-valued polynomials.

1. $\binom{x}{0} = 1$ is clearly a integer-valued polynomial.

2. let's say we have shown $\binom{x}{n}$ is an integer-value polynomial for some $n \ge 0$, then for the case $n+1$. We can use following two properties on binomial coefficients $$\binom{0}{n+1} = 0\quad\text{ and }\quad \binom{x+1}{n+1}-\binom{x}{n+1} = \binom{x}{n}$$ and induction on $x$ ( and $-x$ ) to deduce $\binom{x}{n+1}$ is an integer for positive ( and negative ) integers $x$.

3. Finally, another induction over $n$ will establish $\binom{x}{n}$ is an integer-valued polynomial for all $n$.