I am reposting the second question from [here](https://mathoverflow.net/questions/288290/variation-of-an-old-question-are-these-functions-polynomials) (after clarifying it) on the recommendation of user "GH from MO". 


Let $b\_1,b\_2,\dots$ be an enumeration of $\mathbb Q$. 


**Question 2:** Suppose I define $$G(x,y) = a\_0(y) + a\_1(y)(x-b\_1) + a\_2(y)(x-b\_1)(x-b\_2) + \dots$$ where the $a\_k(y)$ are polynomials in $y$ and $g(x) = G(x,x)$.


Suppose the $a\_k(y)$ are not identically zero for $k$ large enough. Otherwise, we clearly get polynomials. Is this the only way to get a polynomial? That is, if $g(x)$ is equal to a polynomial function, then is $a\_k = 0$ for $k \gg 0$?



Is the following function a polynomial?
The answer is no. There are uniquely determined linear polynomials $a\_m(x):=x-c\_m$ over $\mathbb{Q}$ such that $g(x)$ is identically zero on $\mathbb{Q}$. Indeed, for such polynomials the condition means that
$$ \sum\_{0\leq m\leq n-1}(b\_n-c\_m)\cdot(b\_n-b\_1)\dots(b\_n-b\_m) = 0,\qquad n\geq 1. $$
Here, the second product is meant to be $1$ when $m=0$. Now suppose that, for a given $n\geq 1$, the coefficients $c\_m\in\mathbb{Q}$ have been determined for $0\leq m<n-1$. Then, the above equation has a unique solution for $c\_{n-1}\in\mathbb{Q}$. So we obtained a recursion for the rational numbers $c\_0,c\_1,c\_2,\dots$, and we are done.


