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Does there exist a sequence $(a_n)_{n\in \Bbb{N}}$ of rationals such that for all $n\in \Bbb{N}$, $a_n\neq 0$ and the polynomial $a_0+a_1X+\cdots+a_nX^n$ is split over $\Bbb{Q}$?

I was asked this question by myself but I am unable to find a solution, does anyone have any ideas please?

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  • $\begingroup$ Well the sequence must be finite. Then how about $a_{0}=1$, $a_{1}=1$ and $a_{n}=0$ for all $n\geq 2$. $\endgroup$
    – TheNumber23
    Commented Jul 9, 2014 at 13:41
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    $\begingroup$ A related (terrible) question: math.stackexchange.com/questions/492326/… $\endgroup$
    – Jack D'Aurizio
    Commented Jul 11, 2014 at 21:31
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    $\begingroup$ You had me at "I was asked this question by myself". $\endgroup$
    – Martin Brandenburg
    Commented Jul 15, 2014 at 11:42

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You can recursively define the coefficients in order that $$ p_n(x) = a_0 + a_1 x + \ldots + a_n x^n $$ has a root in $x=n$, for instance: $$ a_n = -\frac{1}{n^n}\sum_{j=0}^{n-1}a_j\, n^j.$$ If you meant that every $p_n(x)$ must completely split over $\mathbb{Q}$ for a sequence with infinite non-zero terms, it is believed that there is no chance, but the linked question is still open.

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    $\begingroup$ Split = factors into linear factors = all the roots are rational $\endgroup$
    – Jyrki Lahtonen
    Commented Jul 11, 2014 at 21:39
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    $\begingroup$ I just realized the possibility of such interpretation, but in this case the question is a duplicate of math.stackexchange.com/questions/492326/…. $\endgroup$
    – Jack D'Aurizio
    Commented Jul 11, 2014 at 21:40
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    $\begingroup$ It is, indeed, close to being an exact duplicate. Apparently here we won't insist that the roots should be distinct. $\endgroup$
    – Jyrki Lahtonen
    Commented Jul 11, 2014 at 21:42

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