Enumerate the rationals as $b\_1,b\_2,\dots$ and define the (set) function:
$$f(x) = (x-b\_1)^2 + (x-b\_1)^2(x-b\_2)^2 + \dots.$$
At any particular $x$, only finitely many terms are non zero so this is perfectly well defined as a (set) function but surely, it is not equal to any polynomial! (or is it?) How do I show that there is no polynomial $p(t) \in \mathbb Q[t]$ such that $p(x) = f(x)$ for all $x \in \mathbb Q$?


If $f(x)$ were defined as $(x-b\_1) + (x-b\_1)(x-b\_2) + \dots$, then this question is not so hard. If $p(x)$ has degree $n$, then testing on $b\_1,\dots,b\_n$ would show that $p(x)$ is necessarily $(x-b\_1) + (x-b\_1)(x-b\_2) + \dots + (x-b\_1)\dots(x-b\_n)$ but then $x=b\_{n+1}$ derives a contradiction.


I don't know how to adapt this approach. Trying to guess the polynomial seems hard even if we think $p(x)$ is degree $1$.


I posted a follow up to this question here: [(Variation of an old question) Are these functions polynomials?](https://mathoverflow.net/questions/288290/variation-of-an-old-question-are-these-functions-polynomials).



How to show that the following function isn't a polynomial over Q?
For each positive integer $n$ and any rational $x$, we have  

$$f(x)\geq (x-b\_1)^2(x-b\_2)^2\dots(x-b\_n)^2.$$
For large $x$, we then have $f(x)\gg x^{2n}$, which implies that if $f$ is a polynomial, it must have degree $≥2n$.


