# Monic polynomials whose roots are their remaining coefficients

I was bored in class one day and wondered to myself if there were any quadratics $x^2+ax+b$ such that $a$ and $b$ are the zeros. I found two: $x^2+x-2,$ and $x^2 -{1\over2}x -{1\over2}$. The comments suggested $x^2+0x+0$, though this seems trivial. I wonder if this applies to other degree polynomials. Clearly it never works for a linear, except for $x+0=0$, as if $x+a=0$, $x=-a$, not $a$. What about cubics, quadratics, or even higher powers? In general, $x^n$ works. What about non-trivial solutions?

• Actually $x^2+0x+0$ is also one.
– user304329
Mar 13, 2018 at 14:06
• Also $x+0$ for the linear case. Mar 13, 2018 at 14:08
• Indeed, this problem has been studied for a long while. There's a question on this on MathOverflow. Mar 13, 2018 at 15:06
• Notice that $x^2-\frac{1}{2}x-\frac{1}{2}$ only 'works', because it has roots $-\frac{1}{2}$ and $1$, while $1$ is not one of the polynomial's coefficients. Mar 14, 2018 at 22:17
• @Frozn The minor variant of this question, "monic polynomials whose remaining coefficients are all roots," is slightly broader than the current answers then. Mar 15, 2018 at 2:28

For the quadratic case $x^2+ax+b=0$, by Viète's formulas we have $$-a=a+b\qquad b=ab$$ The first formula implies $b=-2a$. Substituting this into the second equation gives $-2a=-2a^2$ or $a=a^2$, so $a=0,1$. These give corresponding $b$ values of 0 and $-2$ respectively, so the unique non-trivial quadratic "auto-solving" polynomial is $x^2+x-2=0$.

For the cubic case $x^3+ax^2+bx+c$, the same formulas give $$-a=a+b+c\qquad b=ab+bc+ca\qquad-c=abc$$ If $c=0$ the situation is just the quadratic case with another zero root. In general, if $p(x)$ is auto-solving, so is $p(x)x^k$ for any $k>0$, since only extra zeros are introduced to both the coefficient and root list. Thus we have for example $x^3+x^2-2x$, but these are trivial.

There is a nice non-trivial auto-solving cubic polynomial, $x^3+x^2-x-1$, but also a very ugly one given by $$a=0.56519771\dots$$ $$b=-1.76929235\dots$$ $$c=0.63889691\dots$$ (It turns out that these values are in the OEIS: A273065, $-$A273066, A273067.)

And this is just the all-real solution! There are two solutions with complex coefficients. One is given by $$a={-0.78259885\dots}-0.52171371\dots i$$ $$b={0.88464617\dots}-0.58974280\dots i$$ $$c={0.68055154\dots}+1.63317024\dots i$$ and the other is obtained by taking the complex conjugates of the values above.

For quartics, there is only one non-trivial real auto-solving polynomial, $x^4+x^3+ax^2+bx+c$ where $$a=-1.75487766\dots$$ $$b=-0.56984029\dots$$ $$c=0.32471795\dots$$ Even more startlingly, there are no auto-solving real polynomials of degree five or higher. This and the uniqueness of the quartic example were proved by Paul R. Stein in "On Polynomial Equations with Coefficients Equal to Their Roots" (March 1966), American Mathematical Monthly 73 (3), pp. 272-274.

• Is $2a=-(b+c)$ valid for all the cubic polynomials? Mar 13, 2018 at 15:27
• @NetravatPendsey Yes, by the same Viète's relations I posted earlier in the answer. Mar 13, 2018 at 15:28
• Ohh. Missed it. Derived the same relationship by a very long method. Mar 13, 2018 at 15:32
• Upvoted for the reference. Mar 13, 2018 at 15:54
• That ending. Plus one. Mar 14, 2018 at 3:00

Not a full solution to the general case, but at least two solutions to the general case and all solutions to the cubic.

Well, we know $(x-p)(x-q)=x^2-(p+q)x+pq$, so if $a,b$ are zeroes of $x^2+ax+b$, then $a=-a-b$ and $b=ab$. From the last, we see $b=0$ or $a=1$, and so we get solutions $(a,b)=(0,0)$ and $(a,b)=(1,-2)$. Those are all.

For higher degree polynomials, we know $(a_n,\cdots,a_0)=(0,\cdots,0)$ is a solution (to the $n+1$-degree polynomial) so let's assume they're not all $0$. We have (again, our polynomial is degree $n+1$)

$$\prod_{k=0}^{n+1}(x-a_k)=x^{n+1}+\sum_{k=0}^{n}a_kx^k$$

This seems quite hard, so let's just try it for cubics first.

Cubics.

Allright, so, let's jump right into it. We need:

$$x^3+ax^2+bx+c=(x-a)(x-b)(x-c)$$

and working out the parentheses on the right gives us the following equations:

\begin{align} a&=-a-b-c\\ b&=ab+bc+ca\\ c&=-abc \end{align}

So let's see what happens when $c=0$; then, $a=-a-b$ and $b=ab$. Hey! This is the solution to the second degree equation! Let's keep that in mind, maybe we can prove something about that for the general case. Now we proceed to assume $c\neq 0$. Then, $ab=-1$. Now let's see if we can express $a$ in terms of $b$ and $c$ to hopefully eliminate that variable:

\begin{align} a&=-\tfrac12(b+c)\\ a&=\frac{b-bc+1}{c}\\ a&=-\frac1b \end{align}

So now we express $c$ in terms of $b$ by pairing the last equation to the first two:

\begin{align} c&=\frac{2-b^2}{b}\\ c&=\frac{-b-b^2}{1-b^2} \end{align}

And so we get

$$b^4+b^3-2b^2+2=0$$

Ouch! That's ugly. But, it has a solution $b=-1$ (corresponding to $(a,b,c)=(1,-1,-1)$). The other three solutions are pretty ugly though; one real one and two complex ones (I won't work them out because they're too chaotic and seeing them won't accomplish anything. Feel free to use wolframalpha to get them though).

As for our little trick with $c=0$ obtaining the solutions from the quadratic equation, that's actually quite simple to see. If $x^{n+1}+a_nx^n+\cdots+a_0$ has roots $a_n,\cdots,a_0$, then surely multiplying by $x$ doesn't change that: $x^{n+2}+a_nx^{n+1}+\cdots+a_0x+0$ now also has those roots, and since we multiplied by $x$, it has an extra root $0$. As a result, we may assume the constant term is nonzero, since if it is, we may divide the polynomial by $x$ to get to a lower degree case.

• Shouldnt $c=-abc$ Mar 13, 2018 at 14:35
• Whoops, you're right. I'll edit it.
– user304329
Mar 13, 2018 at 14:37
• For the record: I did not use Wolfram Alpha to get the ugly cubic solutions. I used SymPy to completely and exactly solve the general cubic case, then mpmath to numerically evaluate the results. Mar 13, 2018 at 15:41