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I am reposting the second question from here (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 $k \gg 0$?

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  • $\begingroup$ The role of $G(x,y)$ is unclear. Why don't you simply define $g(x)$ as $a_0(x) + a_1(x)(x-b_1) + a_2(x)(x-b_1)(x-b_2) + \dots$? $\endgroup$ – GH from MO Dec 12 '17 at 17:03
  • $\begingroup$ You are right, it is just an artifact of where the problem came from. The $\lambda$ in the previous question was part of the original question but of course if we fix $\lambda$, we might as well take $\lambda = 1$ and scale the $a_i$ accordingly. But then, as you note, we might as well do away with $G(x,y)$ altogether. $\endgroup$ – Asvin Dec 12 '17 at 17:08
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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.

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    $\begingroup$ You should be more ambitious! Your idea can show that, for any function $v$ defined on $\Bbb Q$ (including every polynomial function), there is a unique sequence of monic linear polynomials $a_{m;v}(x) = x-c_{m;v}$ such that $g(x)=v(x)$ for every $x\in\Bbb Q$. $\endgroup$ – Greg Martin Dec 12 '17 at 19:18

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