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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.

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    $\begingroup$ $f(x) = 2020$? Some correction is required. Also, if $f(x) = x^2$, then $f(\sqrt 2) = 2$ is not irrational. $\endgroup$ – Teresa Lisbon Jul 17 at 6:11
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    $\begingroup$ Sorry 😞🙏 that would be f(0)=2020. I have edited it. You can help me out now. $\endgroup$ – Rhitankar_21 Jul 17 at 6:33
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    $\begingroup$ I suspect $f(x)=x+2020$ is the only such polynomial. $\endgroup$ – Gerry Myerson Jul 17 at 6:46
  • $\begingroup$ Yes f(x)= x+2020 can definitely be a solution but in case it is the only solution, please provide me a proof that no other solution exists. $\endgroup$ – Rhitankar_21 Jul 17 at 7:22
  • $\begingroup$ "putting an irrational $x$ would always produce an irrational value of $f(x)$ quite obviously": try $f(x)=x^2$ with $x=\sqrt 2$. $\endgroup$ – TonyK Jul 17 at 12:23
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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.

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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 $p(x) = x+2020$.

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  • $\begingroup$ Another way to end the argument is that if $a$ is large enough then $p(x)+a$ has at most one real zero, and that zero is simple, so $p(x)+a$ can't split over the rationals, if its degree exceeds one. $\endgroup$ – Gerry Myerson Jul 18 at 12:56
  • $\begingroup$ @Gerry This is not quite true. For instance, if $p(x)$ is an even function, the zeroes of $p(x)-a$ must come in pairs. So I would have to throw in some treatment of “special cases” (eg: $p(x)$ is even of degree $2$) to my argument in order to rigorously implement your idea, although your statement is morally correct that the number of solutions is “typically” much smaller than the degree of $p$ for a generic $a \in \Bbb{Z}$. $\endgroup$ – Rivers McForge Jul 18 at 13:29
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    $\begingroup$ $p$ is, by hypothesis, monic. If $p$ is even, or just of even degree, then if $a$ is large enough, $p(x)+a$ (note – not $p(x)-a$, but $p(x)+a$) will have no real roots. $\endgroup$ – Gerry Myerson Jul 19 at 0:48
  • $\begingroup$ Excellent point! That “$+a$” was doing more work than I realized in your comment. $\endgroup$ – Rivers McForge Jul 19 at 5:13

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