Can one construct a sequence $(a_k)_{k\geqslant 0}$ of rational numbers such that, for every positive integer $n$ the polynomial $a_nX^n+a_{n-1}X^{n-1}+\cdots +a_0$ has exactly $n$ distinct rational roots ?

If we cannot construct it explicitly, can we show that such a sequence exists?

PS: One can show (not easily) that the polynomial $\displaystyle \sum_{k=0}^n\frac{1}{3^{k^2}}X^k$ has $n$ distinct real roots.

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  • Doesn't [Vieta's fomulas](http://en.wikipedia.org/wiki/Vieta's_formulas) do exactly this? Assuming all roots are rational, the sequence of the polynomial coefficients will also be rational, no? – Daniel R Sep 13 '13 at 07:26
  • @DanielR $x^2-2=0$ but no rational roots; $\sqrt{2}x-\sqrt{2}=0$ has a rational root. – Arash Sep 13 '13 at 09:12
  • @ArashBeh: Sure, but is that really an objection? The question is if there exists a rational sequence of polynomial coefficients such that the roots of the polynomial are rational and distinct. By going "backwards" from rational roots and constructing the coefficients using Vieta's formulas, a rational polynomial will be obtained. – Daniel R Sep 13 '13 at 09:20
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    @DanielR, I just wanted to say that having rational coefficients is neither sufficient nor necessary. By the way the suggested answer for $n=2$ does not seem to have rational roots. – Arash Sep 13 '13 at 09:46
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    @DanielR The problem is that having chosen coefficients which work in degree $n$ you have to find one additional coefficient (keeping the others the same) which gives all rational roots in degree $n+1$ - but you can't keep the $n$ roots you had before. – Mark Bennet Sep 13 '13 at 09:51
  • @MarkBennet: Right, it's the sequence part that is troublesome, I think I get it now. – Daniel R Sep 13 '13 at 09:57
  • @user84673, In the example you gave at the end, do you mean real roots or rational roots? Do you look for real or rational roots in general? – Arash Sep 16 '13 at 15:14
  • @Arash Beh, in the example I mean real roots. In the question I look for rational roots with rational coefficients. – user84673 Sep 17 '13 at 02:10
  • I think it is really unlikely that such a sequence exists. Assume that the denominator of $a_n$ is divided by a prime number $p$ that does not divide the denominator of $a_j$ for any $j – Jack D'Aurizio Dec 07 '13 at 17:01
  • Hence we can take a prime number $q$ that is greater than any prime number that divides some of the $a_i$. All the (rational) roots of $p_n(x)$ can be seen as elements of $\mathbb{F_q}$, and the sequence $\{a_n\}_{n\in\mathbb{N}}$ as a sequence of elements of $\mathbb{F}_q$. Now, I strongly believe that for any prime number $q$, we cannot find an infinite sequence of polynomials in $\mathbb{F}_q[x]$, having the stated form, such that every polynomial completely splits over $\mathbb{F}_q$. – Jack D'Aurizio Dec 07 '13 at 17:52
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    I don't understand this argument. A polynomial with a $p$-fractional leading coefficient and all other coefficients $p$-integral can still factor completely over $\bf Q$. For example, consider $p^{-1} \prod_{i=1}^n (x-x_i)$ with $p\mid x_i$ for each $i$. And even if all the $a_k$ are $p$-integral, they could eventually all be multiples of $p$, and again there would be no $p$-adic contradiction. – Noam D. Elkies Dec 08 '13 at 03:41
  • @NoamD.Elkies: the point is that there is a dichotomy, since you have that either $p$ divides all the numerators of $a_1,\ldots,a_k$, or $p$ divides all the roots of $p_k(x)=\sum_{j=0}^{k}a_j x^j$, so divides all the denominators of $a_1,\ldots,a_k$, and this happens for an infinite number of primes. If for any $a_i=\frac{n_i}{d_i}$ we consider $W(a_i)=\omega(n_i)+\omega(d_i)$, $W(a_i)$ grows unbounded unless $a_i=0$. – Jack D'Aurizio Dec 08 '13 at 11:21
  • Back to the original argument: from now on, fix $a_i=\frac{u_i}{l_i}$, $a_0=1$ and $p_k(x)=\sum_{j=0}^k a_j x^j$. If a prime $p$ divides $l_i$ but none of the $l_k$ with $k – Jack D'Aurizio Dec 08 '13 at 16:45
  • I suspect that http://en.wikipedia.org/wiki/Rational_root_theorem could be of use in deriving a contradiction. – Pol van Hoften Dec 10 '13 at 19:40
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    @JackD'Aurizio Did you solve it? – leo Dec 11 '13 at 00:43
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    What if we we had $(x-a_1)(x-a_2)...$ with $a_i$ to be distinct and rational for all $i=1,2,...$. then for any natural number $n$, pick $n$ distinct numbers $\lbrace a_1, \ldots, a_n \rbrace$ and just consider the polynomial $\Pi_{i=1}^{n} (x-a_i)$. ?? – AXH Jan 10 '14 at 19:33
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    @ Arbias : But the values of the coefficients changes as you raise the polynomial. Note the constant term of the polynomial is (-1)^n product of roots in every polynomial, so what is $a_0$ for n - deg polynomial in your construction is different for the (n+1) - deg polynomial. – DiffeoR Jan 11 '14 at 14:13
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    Do you even have an example of such sequence of length $5$, i.e. for the degree 4? I have problems finding one... – user68061 Jan 11 '14 at 14:36
  • I only pointed out that this construction cannot work. The difficulty in finite case also the same as in case of infinite case . – DiffeoR Jan 11 '14 at 14:53
  • Is the problem any easier if indexes are swapped? It seems to me letting $a_n$ being the free term is an equivalent formulation. Interesting challenge, I wish I could add some more points to the bounty... – carlosayam Jan 12 '14 at 01:53
  • @caya It makes no difference - if the roots of one polynomial are $r_i$, the roots of the other are $\frac1{r_i}$. I think the formulation in the question is a bit easier; the indices 'match up' cleanly and the transition from one polynomial to the next is just adding a single term, rather than multiplying by $X$ and adding a constant. – Steven Stadnicki Jan 12 '14 at 02:43
  • You can divide by $a_0$, then replace $X$ with $X/a_1$ so that wlog your first two coefficients are $1$. Finding a degree-2 requires you to make a choice of rational number (which produces a Pythagorean triple); if I didn't make a mistake, finding an appropriate degree-3 was equivalent to finding a rational solution of $r^2 + s^2 + 3t^2 = 3$. I don't have intuition for how hard solving the next would be. – Tyler Lawson Jan 12 '14 at 03:10
  • @StevenStadnicki, I preferred to multiply by X and add a free term. But still no luck... as said, interesting challenge. – carlosayam Jan 12 '14 at 11:12
  • Can you provide some more support to your last claim? Some source or reference? – leo Jan 16 '14 at 23:44
  • amazing [ http://www.wolframalpha.com/input/?i=solve+sum_%7Bk%3D0%7D%5E20%28x%5Ek%2F3%5E%28k%5E2%29%29%3D%3D0 ] – janmarqz Jan 28 '14 at 18:47
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    Well, there's the stupid answer, (0). My quick program to solve for small upper orders was "very helpful" in finding this case... – Eric Towers Feb 21 '14 at 03:45
  • ... so all the polynomials's discriminants are positive squares of rational numbers. That *sounds* rather strict. – Eric Towers Feb 21 '14 at 04:59
  • @caya: The reversed version is equivalent. Consider the coefficient sequence of the polynomial $x^n p(1/x)$. And this polynomial has roots that are reciprocals of those of $p$. (This requires none of the $a_i$ are zero, which is promised in the problem statement by the requirement that each polynomial has $n$ roots. I.e., every polynomial has nonzero leading coefficient.) – Eric Towers Feb 21 '14 at 05:13
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    One small thing that I don't think anyone has noted explicitly yet: the $a_n$ must have unbounded height (where I'm using $\max(|p_n|, |q_n|)$ as my specific definition of height for $a_n=\frac{p_n}{q_n}$). Otherwise, by the rational root theorem there would be a bounded number of possible roots of the polynomial, and so once the degree gets high enough some root would have to be multiple. – Steven Stadnicki Mar 09 '14 at 20:29
  • Related: http://math.stackexchange.com/questions/146288/polynomials-all-of-whose-roots-are-rational – Watson Sep 17 '16 at 10:21

1 Answers1


This does not answer the OP's question, rather gives a partial result. Anyhow, it is too long to fit into a comment. From now on, I work with the Zariski topology.

Let $p(x)=a_0+a_1x+\cdots+a_nx^n=(x-x_1)\cdots(x-x_n)\in\mathbb{Q}[x]$.

First, let us describe the set of coefficients $(a_0,\ldots,a_n)\in\mathbb{A}^{n+1}_{\mathbb{Q}}$ for which $p$ has distinct rational roots. Relations between the roots and coefficients (Vieta's formulae) tell us that


for $i=0,\ldots,n-1$, where $s_j$ denotes the $j$-th elementary symmetric polynomial. Now, let $X=Y=\mathbb{A}^{n+1}_{\mathbb{Q}}$, and consider the regular map $\varphi:X\longrightarrow Y$ defined by


I claim that the set of coefficients we are looking for is precisely $\varphi(X)\cap D(\Delta_p)\cap D(a_n)$, where $\Delta_p$ denotes the discriminant of $p$. Indeed, while $\varphi(X)$ guarantees rational roots, $D(\Delta_p)$ that they are distinct, $D(a_n)$ ensures that the leading coefficient is not zero.

Lemma. The map $\varphi:X\longrightarrow Y$ is dominant, that is, $\overline{\varphi(X)}=Y$.

Proof. We argue by contradiction. Suppose that $Z\subset Y$ is a proper closed subset containing $\varphi(X)$. Without loss of generality, we may assume that $Z=V(f)$ for some non-zero polynomial $f\in\mathbb{Q}[y_0,\ldots,y_n]$. Define $f_{a_n}=f(y_0,\ldots,y_{n-1},a_n)$. Now, by assumption, $f\circ\varphi$ vanishes identically on $X$, but then $f_{a_n}=0$ for all $0\neq a_n\in\mathbb{Q}$ since (and this is the key point) the elementary symmetric polynomials are algebraically independent over $\mathbb{Q}$. This implies that $f=0$, a contradiction.

As a dominant map, $\varphi$ contains a non-empty open set $U\subset Y$ in its image (the proof uses Noether normalisation). Therefore, $U\cap D(\Delta_p)\cap D(a_n)$ is a non-empty open set of coefficients for which $p$ has distinct rational roots. Let us denote this set by $D_n$, emphasising the degree of the polynomial. Now, if $(a_0,\ldots,a_n)$ is in the non-empty open set

$$V_n=\bigcap_{k=0}^n D_k\times\mathbb{A}^{n-k}_{\mathbb{Q}}\subset\mathbb{A}^{n+1}_{\mathbb{Q}},$$

then $p_k(x)=a_0+a_1x+\cdots+a_kx^k$ has distinct rational roots for all $k=0,\ldots,n$.

Therefore, in degree $n$, we have very many solutions.

Unfortunately, it is not clear how we can construct an infinite sequence in this way since $V_n$ might have empty intersection with the closed set $\{a_0\}\times\cdots\times\{a_{n-1}\}\times\mathbb{A}^1_{\mathbb{Q}}$ for a given $(a_0,\ldots,a_{n-1})\in V_{n-1}$. If we could bound the dimension of the complement of an open subset of $\varphi(X)\cap D(\Delta_p)\cap D(a_n)$ independently of $n$, then one could argue inductively. However, all we know for sure is that $\dim(U^c)\leq n-1$.

EDIT. I have just realised that, though it is dominant, $\varphi$ might not have any non-empty open set in its image, for $\mathbb{Q}$ is not algebraically closed. Unfortunately, this means that the conclusion I drew is wrong. However, the first part still makes sense.

  • I managed to prove that for $n=2$, there is no non-empty open subset of $Y$ contained in $\varphi(X)$. Therefore, the method above does not work. –  Mar 15 '14 at 20:09
  • And $D$? What does it mean / denotes? – leo Mar 22 '14 at 05:23
  • $D(f)$ is the complement of $V(f)$. It is the set of those points where $f$ does not vanish. –  Mar 22 '14 at 12:12