Find all monic polynomials $f(x)$ having integer coefficients such that $f(0)=2020$ and for any irrational $x$, $f(x)$ is also irrational.

Question

I have tried to approach the sum with some help from the rational root theorem but apparently it seems to me that for a polynomial with integer coefficients, putting an irrational $x$ would always produce an irrational value of $f(x)$ quite obviously but I cannot prove it rigorously. Can anybody please help me with the solution.

Answer 1

We will show there are no such polynomials of degree greater than $1$.

For the sake of contradiction, let's assume there is such $f(x)$ with $\deg f \geq 2$. Since $f(x)$ is monic (particularly because the leading coefficient is positive), $f(x)-m$ will have at least one real root for all sufficiently large integers $m$. If such root $\alpha$ is irrational, then it means $f(\alpha)=m$ is rational, impossible (by assumed property of $f$). Hence $\alpha$ is rational, but that means $f(x)-m$ is reducible over $\mathbb{Z}$. However, we can choose arbitrary large $m$ such that $p=2020-m$ is a prime (in absolute value) and this will guarantee that $f(x)-m$ is irreducible. Indeed, let $$f(x)=x^n+a_{n-1}x^{n-1}+\dots+a_1x+2020,$$ then for any prime $p=|2020-m|$ such that $p>1+|a_1|+\dots+|a_{n-1}|$, the $f(x)-m$ is irreducible. This is because all of its (complex) roots lie outside of the unit circle and its constant coefficient is prime (this is common argument and has been used on the site quite a few times, see for example Show that $x^4 + 8x - 12$ is irreducible in $\mathbb{Q}[x]$. ). So we have reached a contradiction, and so $\deg f \leq 1$.

Only monic polynomial with given constant coefficient and $\deg f \leq 1$, is $f(x)=x+2020$, which indeed works.

Answer 2

Let $p(x)$ be a monic polynomial with integer coefficients such that $p(0) = 2020$. Let $a \in \Bbb{Z}$ be an arbitrary integer. By hypothesis, $y$ irrational implies $p(y)$ irrational, so $p(x) - a$ splits over $\Bbb{Q}$ (and therefore over $\Bbb{Z}$, since it has integer coefficients) regardless of the value of $a$. This means $p(x)$ is a surjective function from $\Bbb{Z}$ onto $\Bbb{Z}$, since $p(x) - a$ splits over $\Bbb{Z}$ only if $a \in p(\Bbb{Z})$.

Now suppose $p(x)$ has degree $n$. The set $$S_k := \{ y \in \Bbb{Z}: |y| \leq k \text{ and } y \in p(\Bbb{Z}) \}$$ grows at the rate $|S_k| = \mathcal{O}(\sqrt[n]{k})$ for $k \geq 1$, and if $p$ is surjective, $|S_k| = 2k+1$. These orders of growth are only compatible if $n=1$, so $p$ is linear, and the only monic linear polynomial with $p(0) = 2020$ is $p(x) = x+2020$.