# $a,b,c,d\ne 0$ are roots (of $x$) to the equation $x^4 + ax^3 + bx^2 + cx + d = 0$

Find all quadruplet(s) of non-zero real numbers $(a,b,c,d)$ such that $a,b,c$ and $d$ are roots (of $x$) to the equation $x^4 + ax^3 + bx^2 + cx + d = 0$.

My friend found a set of irrational roots that satisfy this condition through experimentation.

And no, I don't know how to solve this question properly.

This question was inspired by another math question in another math platform.

My useless attempt: By Vieta's formula, $2a+b+c+d = 0, ab+ac+ad+bc+bd+cd = b, abc + abd + acd + bcd = -c , abc =1$. That's all....

EDIT: Fixed typo/redundancy.

EDIT2: Changed the word "rational" to "irrational", because I'm a klutz.

• What happens if you eliminate $c,d$ by using $abc=1,2a+b+c+d=0$? – πr8 May 1 '16 at 13:44
• @πr8 If I'm not mistaken, we arrive at $2b^3a^4 + (b^4 + 2*b)a^3 + 3b^2a^2 + (b^3 - b + 1)a + b=0$ as condition for $a,b$. Playing around it seems that e.g. $b=1$ allows rational solutions for $a$ namely $a=1|-1|\frac12$, then $c=-1|1|2$. Only the last case leads to non-zero $d$. I have not guessed other solutions yet – Hagen von Eitzen May 1 '16 at 14:06
• Maple only finds the trivial solution $a = b = c = d = 0$, $a = 1, b = -2, c = 0, d = 0$, $a = 1, b = -1, c = -1, d = 0$ and the irrational solution $a = \alpha, b = -2 \alpha^2 - 2 \alpha, c = \alpha^2, d = 0$, where $2 \alpha^3 - 2 \alpha^2 - 1 = 0$. (It also finds two complex solutions whose components are algebraic of degree $14$ over $\Bbb Q$.) – Travis Willse May 1 '16 at 14:38
• Given that the reduced condition in @HagenvonEitzen's answer is quartic (without any obvious simplification available), it may be hard to prove that a list of all solutions is exhaustive. J.-P. Serre's famous problem of determining the rational points of $x^4 + y^4 = 17$ was only resolved in the last 15 or 20 years. – Travis Willse May 1 '16 at 14:46
• I have added an Edit to my answer. – Jean Marie May 1 '16 at 17:12

Hereafter, I use two powerful algebraic tools:

I have relied on a Computer Algebra System (Mathematica) for final computations.

Using Vieta's formulas, the issue is equivalent with the following system of 4 equations in the four unknowns $$a,b,c,d$$:

$$\begin{cases}a+b+c+d&=&-a&(V_1)\\ ab+ac+ad+bc+bd+cd&=& \ b&(V_2)\\ bcd+acd+abd+abc&=&-c&(V_3)\\ abcd&=& \ d&(V_4)\\ \end{cases}$$

Using $$(V_1)$$ and $$(V_4)$$, one obtains

$$c=\dfrac{1}{ab} \ \ \text{and} \ \ d=-2a-b-c=-2a-b-\dfrac{1}{ab}$$

Therefore, the two other Vieta's formulas yield two equations with variables $$a$$ and $$b$$ only, under the form:

$$\begin{cases}f(a,b)&=&-1 - 2\,a^2\,b - a\,b^2 - 2\,a^4\,b^2 - a^2\,b^3 - 2\,a^3\,b^3 - a^2\,b^4&=&0 & \ \ \ (1)\\ g(a,b)&=&-a - b + a\,b - 2\,a^3\,b - 3\,a^2\,b^2 - a\,b^3 - 2\,a^4\,b^3 - a^3\,b^4&=&0 & \ \ \ (2)\end{cases}$$

1. First approach: Groebner basis of ((1),(2)).

This is a set of equivalent equations to (1)+(2), simpler in a certain sense, as we will see, at the price of degree elevation.

I obtained (as I said, using Mathematica with GroebnerBasis[...] function) the two following equations (factorization of (1') has in fact been made in a second step):

$$\begin{cases}(1') \ \ \ &P_1(b)P_2(b)P_3(b)&=&0 \ \ \text{where}\\ &P_1(b)&=&1+b\\ &P_2(b)&=&1+b+2b^2+b^3\\ &P_3(b)&=&4 - 8\,b + 4\,b^2 + 4\,b^4 - 6\,b^5 - 4\,b^6 + 6\,b^7\\ &&&+ 4\,b^8 - b^{10} - 4\,b^{11} - b^{12} + 2\,b^{13} + b^{14} \ \ \ \text{and}\\ (2') \ \ \ &82012\,a &=& 27660 - 66204\,b - 24260\,b^2 + 20952\,b^3 \\ &&& + 121422\,b^4 + 52670\,b^5 - 94920\,b^6 - 9194\,b^7 \\ &&& + 79461\,b^8 + 45735\,b^9 + 19696\,b^{10} - 34429\,b^{11}\\ &&& - 53258\,b^{12} + 3176\,b^{13} + 35041\,b^{14} + 21135\,b^{15} \\ &&& + 5820\,b^{16} + 809\,b^{17}\end{cases}$$

Two immediate remarks:

• equation (1') is a polynomial equation in the single parameter $$b$$,

• equation (2') determines in a unique manner the value of $$a$$ once a value of $$b$$ is given.

Let us concentrate our study on (1'), considering the three cases:

• $$P_1(b)=0$$ giving $$b=-1$$. Using (2'), we deduce that $$a=1$$, giving
$$c=1/b=-1$$, and, as a consequence, $$d=-2a-b-c=0$$, thus non acceptable.

• $$P_2(b)=0$$ is a third degree equation with a unique real solution (this can be established rigorously by studying variations of function $$P_2$$). This solution is $$b_0=-\dfrac{1}{3}\left(2+\sqrt[3]{\dfrac{25+3 \sqrt{69}}{2}}+\sqrt[3]{\dfrac{25-3 \sqrt{69}}{2}}\right) \approx -1.7548776662.$$

• $$P_3(b)=0$$ has no real solution, as can be "seen" either on the numerical values of its roots, all complex non real, or by having a look at its graphical representation as a function. But, till now, I haven't been able to prove it rigorously.

Plugging the numerical value of $$b_0$$ in (2'), one obtains $$a_0=1.$$. As a consequence:

$$c_0=\dfrac{1}{a_0b_0}\approx -0.5698402910 \ \ \ \text{and} \ \ \ d_0 = -2a_0-b_0-c_0 \approx 0.3247179572$$ that can be checked as the roots of the function $$y=x^4+a_0x^3+b_0x^2+c_0x+d_0$$ (see graphical representation below)

This "quatuor" $$(a_0,b_0,c_0,d_0)$$ is the unique solution of the problem for which $$abcd \neq 0$$.

Note: we have nothing to check because conditions (1') and (2') (which are equivalent to (1)+(2)) are fulfilled.

1. 2nd approach, using resultants:

A resultant (equal to $$0$$) is a necessary and sufficient compatibility condition between to parametric polynomials (here $$f(a,b)$$ and $$g(a,b)$$) for them to have at least a common root. There are two ways to consider this issue (with notations of "abstract algebra": $$K[ a,b ] \approx K [a ] [ b ] \approx K [ b ] [ a ]$$).

• either $$a$$ is considered as the main variable and $$b$$ is a parameter. In this case the resultant will generate a condition on $$b$$, which is exactly the same as condition (1').

• or, in a reverse way ($$b$$ the main variable and $$a$$ the parameter):

$${\left( -1 + a \right) }^4\,a^6\, \left( 1 + 3\,a + 6\,a^2 + 6\,a^3 + 3\,a^4 - 12\,a^5 - 34\,a^6 - 44\,a^7 - 28\,a^8 + 4\,a^9 + 48\,a^{10} + 80\,a^{11} + 80\,a^{12} + 48\,a^{13} + 16\,a^{14} \right)=0 \ \ \ (3)$$

which is an interesting relationship giving solutions $$a=1$$, $$a=0$$ (not possible) and no other solution because the third factor (the 14th degree polynomial) has no real roots. The interesting thing in (3) is that $$a$$ can - theoretically - play the main rôle in the computation of the solution as $$b$$ has played upwards.

A remark: combining equations (1) and (2):

$$h(a,b):=g(a,b)-a h(a,b)=b\left( a - 1 \right) \,\left( 1 + 2\,a^2\,b + a\,b^2 + 2\,a^3\,b + a^2\,b^2 + 2\,a^4\,b \right)=0 \ \ \ (4)$$

The first two factors $$b(a-1)$$ account for special values $$b=0$$ (not considered) and $$a=1$$, which has been the final value for the solution.

The last factor equal to $$0$$ is a quadratic equation in variable $$b$$; it can be solved with $$a$$ as a parameter (with a certain interval restriction on the values of $$a$$ for the discriminant to be positive). We will not do any computation: it suffices to say that $$b$$ is determined as a function of $$a$$ ...

Here is a graphical representation of the fourth degree polynomial:

$$y= x^4 + x^3 -1.7548776662316001862 x^2 - 0.5698402909809427985 x + 0.32471795721254298472$$

with roots: $$x_1 = 1$$ , $$x_2 = -1.7548776662316001862...$$ , $$x_3 = -0.5698402909809427985...$$ $$x_4=0.32471795721254298472...$$.

• This is a pretty impressive job! ThankYou – GohP.iHan May 8 '16 at 11:48
• 1) I have corrected a typo in (*) without any consequence: $b^3+2b^2+b+1$ instead of $b^3+3b^2+b+1$ 2) I will improve the demonstration using a Groebner basis – Jean Marie May 8 '16 at 16:04
• Looking forward to it!!! =D – GohP.iHan May 9 '16 at 5:47
• I just posted the "final" version, shedding - I think - a completely different light on the issue. – Jean Marie May 9 '16 at 6:56
• Isn't $V_2$ missing two terms ? – Yves Daoust May 9 '16 at 8:25