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If $a$ and $b$ are two roots of $x^4 + x^3 - 1 = 0$ prove that $ab$ is a root of $x^6 + x^4 + x^3 - x^2 - 1$.

Students and I are unsure how to go about this problem. Also will this be a problem I can solve and prove in front of a class in 20 minutes?

Aeolian
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user61646
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  • Where did the problem come from? – Jonas Meyer Feb 13 '13 at 04:14
  • Are $a,b$ just randomly selected among those 4 roots? – JSCB Feb 13 '13 at 04:17
  • Not sure it helps or if I'm just pointing out the obvious, but since $a$ and $b$ are roots of the equation, we know $a^4+a^3=1$ and $b^4+b^3=1$. – Clayton Feb 13 '13 at 04:22
  • I'm unsure where the problem comes from. Students bring me problems they are unsure of and we attempt to work them out. – user61646 Feb 13 '13 at 04:22
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    Note $\ $ The statement is true only for distinct roots $, a\ne b.:$ The squares of the roots satisfy $, x^4 - x^3 - 2 x^2 + 1 = 0.\ \ $ – Math Gems Feb 13 '13 at 06:47
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    Thanks for posting this. I enjoyed tinkering around with it, and was pleased to learn that whenever $x^3+bx^2+cx+d$ has the roots ${p,q,r}$, then $x^3 -cx^2 +bdx-d^2$ has the roots ${pq,qr,pr}$. – MJD Feb 14 '13 at 18:44
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    @JonasMeyer USAMO https://artofproblemsolving.com/wiki/index.php/1977_USAMO_Problems/Problem_3 – crxyz Aug 23 '22 at 23:10

3 Answers3

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There is probably a shorter way, but I think this way sheds some light on why it's true:

Let $a,b,c,d$ be the four roots of $x^4+x^3-1$, so that $x^4+x^3-1=(x-a)(x-b)(x-c)(x-d)$. Now set $$ g(x) = (x-ab)(x-ac)(x-ad)(x-bc)(x-bd)(x-cd). $$ This polynomial is symmetric in the roots $a,b,c,d$, and so its coefficients will be rational numbers (since the coefficients of $x^4+x^3-1$ are rational). This explains why there is such a sextic; to work out its coefficients, we need to mess around with symmetric polynomials.

The coefficients of the original polynomial tell us that $a+b+c+d=-1$, $abcd=-1$, and $ab+ac+ad+bc+bd+cd=abc+abd+acd+bcd=0$. Therefore the coefficient of $x^5$ in $g(x)$ equals $0$; the coefficient of $x^4$ equals $$ abac+abad+abbc+abbd+abcd+acad+acbc+acbd+accd+adbc+adbd+adcd+bcbd+bccd+bdcd = (a + b + c + d) (a b c + a b d + a c d + b c d) - abcd = 1; $$ and so on - the coefficient of $x^0$ equals $(abcd)^3=-1$.

(One can play the same game with $f(x)=(x-a^2)(x-b^2)(x-c^2)(x-d^2)$ to obtain the polynomial $x^4−x^3−2x^2+1$.)

Greg Martin
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  • +1 IMHO this is the way to go. There may be some tricks helping us calculate the coefficients of the sextic, but manipulating the symmetric polynomials on the four roots is guaranteed to work, so :-) – Jyrki Lahtonen Feb 13 '13 at 08:35
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The resultant $$h(z)=\text{Res}( \ \text{Res} ( \ z-xy, \ x^4+x^3-1, \ x), \ y^4+y^3-1, \ y)$$ is a $16^{th}$degree polynomial in $z$ with roots $\{a_ia_j:i,j=1,2,3,4\}$, where $a_1,a_2,a_3,a_4$ are the roots of $x^4+x^3-1$.

By expanding the above resultant we get that the polynomial $h(z)$ is $$h(z)= \left( {z}^{4}-{z}^{3}-2\,{z}^{2}+1 \right) \left( {z}^{6}+{z}^{4}+{z}^{3}-{z}^{2}-1 \right)^2.$$

The roots of the polynomial $\left( {z}^{4}-{z}^{3}-2\,{z}^{2}+1 \right)$ are $a_i^2$ and the roots of $\left( {z}^{6}+{z}^{4}+{z}^{3}-{z}^{2}-1 \right)$ are $a_ia_j$ with $i<j$.

P..
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The statement holds true only for distinct roots $\rm\, a\ne b.\:$ As such, the proof will require use of $\rm\, a\ne b,\:$ e.g. by cancelling $\rm\,a-b.\:$ Let $\rm\,f(x)\,$ be the quartic and let $\rm\,g(x)\,$ be the sectic. Then $\rm\,g(ab)\,$ can be reduced to $\,0\,$ by using $\rm\,\color{#C00}{f(a)} = 0 = \color{#C00}{f(b)}\ $ and $\rm\ \color{#C00}{h(a,b)} = \dfrac{f(a)-f(b)}{a-b} =\, 0,\,$ namely

$$\rm\begin{eqnarray} g(ab)\,\ &=&\ \rm (a^5\!+ a^4b+ a^3b^2\!+\!a^2 b^2\!+a^3\!+1)\ \color{#C00}{f(a)} \\ && \ + \rm\ (a^6 b^2\!-\!a^6 b+a^6\!+a^4)\ \color{#C00}{f(b)}\\ &&\ -\ \rm (a^6\!+a^4\!-\!a^3)\ \color{#C00}{ h(a,b)}\\ &=&\ \ 0\quad\rm \end{eqnarray}$$

Remark $\ $ There may well be a simpler such expression (e.g. a symmetric one). I have not attempted to simplify it.

Math Gems
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