How to show that the following function isn't a polynomial over Q? 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?.
 A: You can adapt the same approach as follows:  Say $p_n(x)$ is an $n$-th degree polynomial matching $f(x)$ at all rational points, then in particular,
$$
p_n(b_1) = 0\\
p_n(b_2) = (b_2-b_1)^2 \\ \vdots\\
p_n(b_k) = \sum_{i\in\Bbb Z+, i<k}\prod_{j\in\Bbb Z+, j\leq i} (b_k-b_i)^2
$$
For a given fixed sequencing of the rationals as $b_1, b_2, \cdots$, and for any given $k$, the latter expression is just some fixed rational number.  
So $p(n)$ is fixed by its values at $b_1 \ldots b_n$, and now consider $-f(b_{n+1}-p_n(b_{n+1}))$.  Since all the terms past the  $n+2$ term in $f(b_{n+1})$ are zero,  $$f(b_{n+1}) = p_n(b_{n+1}) + \prod_{i\leq n}(b_{n+1}-b_i)^2 > p_n(b_{n+1})$$ which contradicts the statement that $f$ matches $p_n$ at all rationals.
A: 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 $≥2n$.
A: 
"If $f(x)$ were defined as $(x−b_1) + (x−b_1)(x−b_2) + …,$ then this question is not so hard."

Does taking the derivative of the given function get you to this simpler case?

"At any particular $x$, only finitely many terms are non zero"

But at irrational values of $x$, none of the terms are zero. Now, you may respond "But I'm talking about it over $\mathbb Q.$" But unless you're going to claim that $f$ isn't continuous,  if you take a sequence of irrationals approaching a rational, the function evaluated at that rational must be the limit of the function evaluated at those irrational numbers. You could also construct a sequence of rationals approaching $b_1$ such that there exists some $\epsilon > 0$ such that for all $x$ in the sequence, $f(x) > \epsilon$. If $f(x)$ is continuous, $f(b_1)$ must be $> 0$, but clearly by the definition of $f$, $f(b_1) = 0$. I suppose you'll still have to argue that $f$ must be continuous, but that should follow from it being a polynomial, even with the restriction to $\mathbb Q$.
