# Polynomials with rational zeros

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.
• you could have tried at least for small $n$ like $n=1$ and $n=2$
– user87543
Dec 7, 2013 at 7:40
• @PraphullaKoushik The generalization to $n$ seems to be a difficult task Dec 7, 2013 at 7:41
• Why not experiment by going in the opposite direction, i.e., expand the expression : $(x-(a_1/b_1))(x-(a_2/b_2))...(x-(a_n/b_n))$ and see if you can see a pattern ? Of course, use a computer, or maybe Wolfram to do the expansion. Dec 7, 2013 at 7:53
• In addition to saying what you've tried - homework is given to help you learn by doing - it would help to say where you encountered this. What theorems did your course just cover? Dec 7, 2013 at 9:07
• I suspect that the polynomials in question has distinct roots (that is, the discriminant does not vanish). If that is the case, using the rational root test, one can check whether there are enough rational roots at all. As an indication, it seems there are no solutions for $n\geq 4$, and only a few for $n=1,2,3$.
– user37391
Mar 10, 2014 at 2:22

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.

• The hypotheses can evidently be weakened quite a bit without affecting the answer too much. For instance, it seems we could just assume that the roots are all rational and the coefficients are integers between $0$ and $n$ and still get a small list of polynomials. Sep 9, 2014 at 0:32