Polynomials with rational zeros

Question

Find all polynomials $F(x)={a_n}{x^n}+\cdots+{a_1}x+a_0$ satisfying

• $a_n \neq0$;
• $(a_0, a_1, a_2, \ldots ,a_n)$ is a permutation of $(0, 1, 2 ... n)$;
• all zeros of $F(x)$ are rational.

Answer 1

The complete list of such polynomials is

$$x, \quad 2x^2+x, \quad x^2+2x, \quad 2x^3+3x^2+x, \quad x^3+3x^2+2x.$$

We now prove this. The first observation to make is that for a polynomial satisfying your hypotheses, all rational roots are non-positive, and that $0$ occurs as a root at most once and only if the constant term is $0$. The second is that by Descartes' rule of signs, the constant term of the polynomial must be zero (otherwise there are only $n-2$ sign changes in the sequence of coefficients of $F(-x)$; not enough to account for $n-1$ negative rational roots).

So now we let $g(x)=F(x)/x$. The coefficients of $g(x)$ are a permutation of the numbers $1,2,\dots,n$, and all the roots of $g$ are negative rational numbers with denominators dividing $a_n$. If $n=1$ obviously $g(x)=1$ and there is nothing to prove. From now on we assume $n>1$. We factor $$g(x)/a_n=(x+r_1)(x+r_2) \cdots (x+r_{n-1})$$ where $r_1,r_2,\dots,r_{n-1}$ are positive rational numbers, each of the form $r_i=m_i/a_n$ for positive integers $m_i$. In particular we have $r_i \geq 1/a_n$ and hence $$r_1+r_2+\cdots+r_{n-1} \geq \frac{n-1}{a_n}.$$ Since $$g(x)/a_n=x^{n-1}+\frac{a_{n-1}}{a_n} x^{n-2}+\cdots $$ we obtain $$\frac{n-1}{a_n} \leq r_1+r_2+\cdots+r_{n-1}=\frac{a_{n-1}}{a_n}$$ and hence there are only two possibilities: $a_{n-1}=n-1$ or $a_{n-1}=n$.

In case $a_{n-1}=n-1$, we find that $r_i=1/a_n$ for all $i$, and hence $$g(x)=a_n(x+1/a_n)^{n-1}.$$ If $n>2$ the constant term of this polynomial can be an integer only if $a_n=1$, so $g(x)=(x+1)^{n-1}$. But for $n>1$ this polynomial has constant term equal to its leading coefficient, contradiction. Thus $n=2$ and $g(x)=2(x+1/2)$, so that $xg(x)$ is one of the degree two polynomials in our list above.

It remains to consider the case $a_{n-1}=n$. In this case $n-2$ of the roots $r_i$ are of the form $1/a_n$ and the other one is $2/a_n$. So $$g(x)=a_n(x+1/a_n)^{n-2}(x+2/a_n).$$ The constant term of $g(x)$ is $$a_1=\frac{2}{a_n^{n-2}}.$$ This is an integer only if $a_n=1$ or $a_n=2$ and $n$ is $2$ or $3$. These last two cases correspond to $g(x)=2(x+1)$, contradicting our hypothesis, and $g(x)=2(x+1/2)(x+1)$, for which $x g(x)=2x^3+3x^2+x$, a polynomial in our list.

We are finally reduced to the case $a_n=1$ and $$g(x)=(x+1)^{n-2}(x+2).$$ Now the coefficient of $x$ in $g(x)$ is $1+2(n-2)$. For $n \geq 4$ we have $$1+2(n-2)=1+2n-4 \geq 1+n > n,$$ contradiction. Thus $n \leq 3$. The polynomials $x (x+2)$ and $x(x+1)(x+2)$ appear in our list so we are done.